Download MATH-O-MANIA - Bookman India

5
1.
Revision
1. Complete the following :
Sol.
(a)
(b)
(Text Book Page No. 5)
8853195 = 8000000 + 800000 + 50000 + 3000 + 100 + 90 + 5
47258738 = 40000000 + 7000000 + 200000 + 50000 + 8000 + 700 + 30 + 8
(c) 231962857 = 200000000 + 30000000 + 1000000 + 900000 + 60000 + 2000 + 800
+ 50 + 7
(d) 374027961 = 300000000 + 70000000 + 4000000 + 0 + 20000 + 7000 + 900 + 60 + 1
2. Express the following numbers in words in the Indian system :
(Text Book Page No. 5)
Sol.
(a) 50112586
Five crores one lakh twelve thousands five hundreds and eighty six.
(b) 20225367
Two crores two laks twenty five thousands three hundreds and sixty seven.
(c) 876500148
Eighty seven crores sixty five lakhs one hundred and forty eight.
3. Express the following numbers in words in the International system :
(Text Book Page No. 5)
Sol.
(a) 80001212
Eighty millions one thousand two hundreds and twelve
(b) 959000089
Nine hundreds fifty nine millions and eighty nine
(c) 345678901
Three hundreds forty five millions six hundreds seventy eight thousands nine hundreds and
one
4. Compare the following numbers and put > or < in the boxes :
(Text Book Page No. 5)
Sol.
(a) 781000187
>
759857123
(b) 213913400
<
218965345
(c) 510000008
<
510001008
(d) 310101010
>
310100101
5. Fill up the blanks as per the given pattern :
Sol.
(Text Book Page No. 5)
(a) 7185768, 7185791, 7185814, 7185837, 7185860
(b) 97876757, 97876868, 97876979, 97877090, 97877201
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Math-O-Mania– 5
6. A factory produced 796765781 metres of yarn in one month, 810000015 metres
in the second month, 821501578 metres in the third month and 910015178
metres in the fourth month. Find the total length of the yarn produced in the four
months.
(Text Book Page No. 5)
Sol.
Total yarn produced in the = yarn produced in the 1st month + yarn produced in the 2nd month
four month
+ yarn produced in the 3rd month + yarn produced in the 4th month
= 796765781 + 810000015 + 821501578 + 910015178
= 3338282552 m
Ans.
7. Dhanpat Rai builts a showroom spending ` 899068 while Gurmeet Singh builts a
factory spending ` 7698567. Who spent more and by how much?
(Text Book Page No. 5)
Sol.
Money spent on building the showroom by Dhanpat Rai = ` 899068
Money spent on building the factory by Gurmeet Singh
= ` 7698567
As can be seen, spent on building the factory is more than money spent on building the showroom
and by
= ` 7698567 – ` 899068
= ` 6799499
\ Gurmeet Singh spent more by ` 6799499.
Ans.
8. Simplify :
Sol.
(Text Book Page No. 5)
(a) 6 × 768 × 23
(b) 5 × 108 × 21
= (23 × 6) × 768
= (21 × 5) × 108
= 138 × 768
= 105 × 108
= 105984
= 11340
(c) 8 × 663 × 29
(d) 10 × 100 × 801
= 8 × 663 × 29 = (29 × 8) × 663
= (10 × 100) × 801
= 232 × 663
= 1000 × 801
= 153816
= 801000
9. A group of 613 persons went on pilgrimage to Amarnath. After the completion of
the trip the share of each person came out to be of ` 9897. Find the total amount
spent by the group.
(Text Book Page No. 5)
Sol.
No. of persons that went on pilgrimage = 613
Share of each person
= ` 9897
\ Total amount spent by the group
= ` 9897 × 613
= ` 6066861
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3
Ans.
Math-O-Mania– 5
10. Divide and find the quotient and remainder in the following :
(Text Book Page No. 5)
Sol.
(a) 768 ÷ 37
(b) 3577 ÷ 27
20
37 768
– 74
28
– 0
28
132
27 3577
– 27
87
– 81
67
– 54
13
Quotient = 20
Quotient = 132
Remainder = 28
Remainder = 13
(c) 16233 ÷ 53
(d) 12246 ÷ 13
306
53 16233
– 159
33
–0
333
– 318
15
942
13 12246
– 117
54
– 52
26
– 26
0
Quotient = 306
Quotient = 942
Remainder = 15
Remainder = 0
11. Answer the following :
Sol.
(Text Book Page No. 5)
(a) Dividend = 65365, Quotient = 1867,
Divisor = 35,
Remainder = 20.
(b) Dividend = 87866,
Divisor = 441,
Remainder = 107.
Quotient = 199,
12. The teacher distributed 125 notebooks equally among 25 students. How many
notebooks were given to 13 students ?
(Text Book Page No. 5)
Sol.
No. of notebooks equally distributed by the teacher = 125
Total no. of students = 25
\ No. of notebooks each student gets = 125 ÷ 25
= 5
Thus no. of notebooks given to 13 students = 13 × 5
= 65
Ans.
13. Find all the factors of 39 and 169. List their common factors.
(Text Book Page No. 5)
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4
Math-O-Mania– 5
3 39
Sol.
13
13
13 169
13
1
39 = 3 × 13 × 1
13
1
169 =
13 × 13 ×1
Thus the factors of 39 are 3, 13 and 1.
The factors of 169 are 13, 13 and 1.
As we can see the common factors of 39 and 169 is 13.
Ans.
14. Which of the following numbers are divisible by 3 ?
Sol.
(a) 438192
(b) 214012
146064
3 438192
–3
13
– 12
18
– 18
01
–0
19
– 18
12
– 12
0
71337
3 214012
– 21
04
–3
10
–9
11
–9
22
– 21
1
As the remainder after division is 0,
so 438192 is divisible by 3.
As the remainder is not 0 after division,
so 214012 is not divisible by 3.
(c) 57018
19006
3 57018
–3
27
– 27
00
–0
01
–0
18
– 18
0
(d) 18252
6084
3 18252
– 18
02
–0
25
– 24
12
– 12
0
As the remainder after division is 0,
so 57018 is divisible by 3.
Runway
(Text Book Page No. 6)
As the remainder after division is 0,
so 18252 is divisible by 3.
5
Math-O-Mania– 5
15. Prime factorization of numbers are given here. Determine the numbers :
(Text Book Page No. 6)
Sol.
(a) 3 × 13 × 17
(b) 2 × 3 × 7 × 19
= 13 × 51
= 42 × 19
= 663
= 798
\ The number is 663.
\ The number is 798.
Ans.
16. Convert the following improper fractions into mixed numbers :
Sol.
(a)
11
2
= 3
3
3
(b)
24
4
= 4
5
5
(c)
(Text Book Page No. 6)
31
3
= 4
7
7
(d)
65
1
= 8
8
8
(e)
71
5
= 6
11
11
(f)
97
6
= 7
13
13
17. Which one is greater in the following fractions ?
Sol.
(a)
(Text Book Page No. 6)
13 , 15
To find out which fraction is greater, we have to first find the L.C.M. of 14 and 16.
14 16
2
14, 16
2
7,
8
2
7,
4
2
7,
2
7
7,
1
1,
1
L.C.M. =
2×2×2×2×7 =
Þ
13
13 × 8
104
=
=
14
14 × 8
112
Þ
15 × 7
105
15
=
=
16 × 7
112
16
As
104
105
>
112
112
\
112
13
15
>
14
16
(b) 7 , 24 To find out which fraction is greater, we have to first find the L.C.M. of 9 and 25.
29 25
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3
9, 25
3
3, 25
5
1, 25
5
1,
5
1,
1
6
Math-O-Mania– 5
L.C.M. =
(c)
3×3×5×5 =
Þ
7
9
Þ
15 × 7
216
24
=
=
16 × 7
225
25
As
216
175
>
225
225
\
24
>
25
=
225
7 × 25
175
=
9 × 25
225
7
9
11 , 31 To find out which fraction is greater, we have to first find the L.C.M. of 15 and 35.
15 35
3
15, 35
5
5, 35
7
1,
1,
7
1
L.C.M. =
3×5×7 =
Þ
11
11 × 7
77
=
=
15
15 × 7
105
Þ
31 × 3
93
31
=
=
35 × 3
105
35
As
93
77
>
105
105
\
105
11
31
>
15
35
(d) 41 57 To find out which fraction is greater, we have to first find the L.C.M. of 47 and 61.
,
47 61
47
47, 61
61
1, 61
1,
Þ
Clearly,
Runway
47 × 61 =
2867
41
41 × 61
2501
=
=
47
47 × 61 2867
Þ
\
1
L.C.M. =
57 × 47 2679
57
=
=
61 × 47 2867
61
2679
2501
>
2867
2867
is greater.
41
57
>
47
61
7
Math-O-Mania– 5
18. Simplify :
Sol.
(Text Book Page No. 6)
7
1
3
1
(a) 7
8
57
8
=
7
8
4
10
3
7
5
12
39
67
4
12
The L.C.M. of 8, 3, 4 and 12 is 24.
10
39
67
57
3
4
12
8
57 × 3 – 10 × 8 + 39 × 6 – 67 × 2
24
171 – 80 + 234 – 134
24
191
24
23
7
24
=
=
=
=
=
4
21
15
81
67
23
=
7
10
21
5
The L.C.M. of 5, 7, 10 and 21 is 210.
(b) 3
=
=
=
=
=
8
5
2
1
7
7
11
10
3
23
15
81
67
5
7
10
21
23 × 42 – 15 × 30 + 81 × 21 – 67 × 10
210
966 – 450 + 1701 – 670
210
1547
210
11
7
30
(c) 24.19 – 0.0029 + 5.191
= 24.1900 + 5.1910 – 0.0029
= 29.3810 – 0.0029
= 29.3781
(d) 31.53 + 4.0019 – 1.798
= 31.5300 + 4.0019 – 1.7980
= 35.5319 – 1.7980
= 33.7339
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Math-O-Mania– 5
19. Convert the following decimal fractions into common fractions :
(Text Book Page No. 6)
Sol.
(a) 0.891
=
891
1000
(b) 0.9590
=
9590
10000
(c) 0.0101
=
101
10000
(d) 0.0023
=
23
10000
(e) 0.05901
=
5901
100000
(f)
=
357
100000
0.00357
20. Express each of the following in terms of paise :
Sol.
(a) ` 447.03
(b) ` 501.97
1 ` = 100 paise
1 ` = 100 paise
\ ` 447.03
\ ` 501.97
= 447.03 × 100
= 501.97 × 100
= 44703 paise
= 50197 paise
(c) ` 597.78
(d) ` 965.34
1 ` = 100 paise
1 ` = 100 paise
\ ` 597.78
\ ` 965.34
= 597.78 × 100
= 965.34 × 100
= 59778 paise
= 96534 paise
21. Convert the following into g and cg :
Sol.
(Text Book Page No. 6)
(Text Book Page No. 6)
(a) 1411 dag
1 dag
=
10 g
\ 1411 dag
=
1411 × 10 g
=
14110 g
1 dag
=
1000 cg
\ 1411 dag
=
1411 × 1000
=
1411000 cg
1 dg
=
\ 2525 dg
=
=
1 g
10
2525 g
10
252.5 g
1 dg
=
10 cg
\ 2525 dg
=
2525 × 10
=
25250 cg
(b) 2525 dg
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Math-O-Mania– 5
(c) 6.75 mg
1 mg
=
1
g
1000
\ 6.75 mg
=
6.75
g
1000
=
0.00675 g
=
=
1 cg
10
0.675 cg
=
100 g
1 mg
= 6.75
10
(d) 14.235 hg
1 hg
\ 14.235 hg =
1 hg
14.235 × 100
=
1423.5 g
=
10000 cg
\ 14.235 hg =
=
14.235 × 10000
142350 cg
22. Add and express the sum in terms of centimeters :
Sol.
(Text Book Page No. 6)
(a) 251 m 300 mm, 2311 m 100 mm, 932 m 600 mm
251 m 300 mm
=
25130 cm
2311 m 100 mm
=
231110 cm
932 m 600 mm
=
93260 cm
\ 251m 300 mm + 2311 m 100 mm + 93260 mm
=
25130 cm + 231110 cm + 93260 cm
=
349500 cm
(b) 62 m 405 mm, 242 m 2597 mm, 111 m 9298 mm
62 m 405 mm
=
6200 + 40.5
=
6240.5 cm
242 m 2597 mm
=
24200 + 259.7
=
24459.7 cm
111 m 9298 mm
=
11100 + 929.8
=
12029.8 cm
=
6240.5 cm + 24459.7 cm + 12029.8 cm
=
42730 cm
23. Subtract :
Sol.
(Text Book Page No. 6)
(a) 25 l 615 ml from 66 l 832 ml.
l
66
– 25
41
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(b) 61 l 572 ml from 72 l 344 ml.
ml
832
615
217
l
72
– 61
10
10
ml
344
572
772
Math-O-Mania– 5
24. Classify the following triangles into scalene, isosceles or equilateral :
(Text Book Page No. 6)
Sol.
(a) See picture on page no 6.
As the length of all the three sides of the triangle are different therefore the D is scalene.
(b) See picture on page no 6.
As the two sides ( LM and LN) of the D are equal, so the D is isosceles.
(c) See picture on page no 6.
As all the three sides of the D are equal, so the D is equilateral.
25. Draw a circle of radius 4 cm. Now, draw 2 diameters in it. Measure them and say
if they are equal in length.
(Text Book Page No. 6)
Ans. Do Yourself
26. The perimeter of a rectangle is 160 m. Its length is 60 m. Find its breadth.
(Text Book Page No. 6)
Sol.
Perimeter of a rectangle
=
160 m
2 (l + b)
=
160 m
2 ( 60 + b)
=
160
Þ 60 + b
=
80
b
=
80 – 60
=
20 m
[ \ l = 60 (given) ]
Therefore the breadth of the rectangle is 20 m.
Ans.
27. How many vertices does a parallelogram have ?
(Text Book Page No. 6)
Ans. A parallelogram has 4 vertices.
28. Draw a rectangle whose length is 7 cm and breadth is 4 cm using a protractor and
a scale.
(Text Book Page No. 6)
Ans. Do Yourself
2.
Roman Numerals
Exercise 2
1. Write the following numbers in Roman System :
Sol.
(a) 12
= 10 + 2
(b) 37
= 10 + 1 + 1 = XII
Runway
(Text Book Page No. 8)
= 30 + 5 + 1 + 1
= XXXVII
11
Math-O-Mania– 5
(c) 48
= 40 + 8
(d) 59
= XLVIII
= 50 + 9
= LIX
(e) 650 = 500 + 100 + 50
(f)
1512 = 1000 + 500 + 12
= DCL
= MDXII
(g) 2312 = 1000 + 1000 + 100 + 100 + 100 + 12
(h) 1009 = 1000 + 9
= MMCCCXII
= MIX
2. Change the following Roman numerals into Hindu-Arabic numerals :
(Text Book Page No. 8)
Sol.
(a) IV
(b) CD
= 5 – 1
= 500 – 100
= 4
= 400
(c) XX
(d) CCI
= 10 × 1000 + 10 × 1000
= 100 + 100 + 1
= 20000
= 201
(e) MDXVII
(f)
CCCXXXIII
= 1000 + 500 + 10 + 5 + 1 + 1
= 100 + 100 + 100 + 10 + 10 + 10 + 1 + 1 + 1
= 1517
= 333
(h) VICCCLXVII
(g) CML
= 1000 – 100 + 50
= 5× 1000+ 1×1000+ 100+ 100+ 50+ 10 +7
= 950
= 6367
3. Write these Hindu-Arabic numerals in Roman System in more than one manner :
(Text Book Page No. 8)
Sol.
(a) 199
(b) 549
= 100 + 99
= 500 + 50 – 1
= CIC
= DIL
Also, 199
Also, 549
= 100 – 10 + 100 + 9
= 500 + 40 + 9
= CXCIX
= DXLIX
(c) 2499
Runway
(d) 149
= 1000 + 1000 + 500 – 1
= 100 + 50 – 1
= MMID
= CIL
Also, 2499
Also, 149
= 1000 + 1000 + 500 – 1 + 100 – 1
= 100 + 40 + 9
= MMCDIC
= CXLIX
12
Math-O-Mania– 5
4. Do these expressions have some meaning ?
Sol.
(a) VICL
(Text Book Page No. 8)
(b) LXIV
= No
= Yes (64)
(c) VVVV
(d) IIII
= No
= No
(e) LXIIVX
(f)
= No
IIXVV
= No
(g) LXXX
(h) XXXX
= Yes (80)
= No
5. Match with same meanings :
(Text Book Page No. 8)
(a) 56
(a) CI
(b) 101
(b) MD
(c) 1500
(c) XXXVII
(d) 37
(d) LVI
(e) 654
(e) DCLIV
Mental Maths (Text Book : Page No 9)
Ans : (Do yourself)
3.
Number System
Exercise 3.1
1. Write the following in figures :
Sol.
(Text Book Page No. 12)
(a) Sixty six lakhs fifty two thousands eight hundreds and one.
66,52,801
(b) Thirty two lakhs sixty eight thousands five hundreds seventy and two.
32,68,572
(c) Forty two lakhs twenty eight thousands four hundreds and six.
42,28,406
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Math-O-Mania– 5
(d) Three crores thirty one lakhs forty nine thousands seven hundreds and twenty eight.
3,31,49,728
(e) Fifty five crores eighty two thousands two hundreds and eleven.
55,00,82,211
2. Write in words using Indian System of writing :
Sol.
(Text Book Page No. 12)
(a) 5,72,820
Five lakhs seventy two thousands eight hundreds and twenty
(b) 74, 12, 358
Seventy four lakhs twelve thousands three hundreds and fifty eight
(c) 1,65,36,806
One crore sixty five lakhs thirty six thousands eight hundreds and six
(d) 36,00,64,521
Thirty six crores sixty four thousand five hundreds and twenty one
(e) 21,46,56,076
Twenty one crores forty six lakhs fifty six thousand and seventy six
(f)
4,76,35,56,309
Four arabs seventy six scores thirty five lakhs fifty six thousands three hundreds and nine
3. Write in words using International System of writing :
Sol.
(Text Book Page No. 12)
(a) 9,576,894
Nine millions five hundreds seventy six thousands eight hundreds and ninety four
(b) 13,068,214
Thirteen millions sixty eight thousands two hundreds and fourteen
(c) 336,792,403
Three hundreds thirty six millions seven hundreds ninety two thousands four hundreds and
three
(d) 5,108,372,224
Five billions one hundred eight millions three hundreds seventy two thousands two
hundreds and twenty four
(e) 76,537,432,140
Seventy six billions five hundreds thirty seven millions four hundreds thirty two thousands
one hundred and forty
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Math-O-Mania– 5
(f)
24,657,365,469
Twenty four billions six hundreds fifty seven millions three hundreds sixty five thousands
four hundreds and sixty nine
4. Write in figures :
Sol.
(Text Book Page No. 12)
(a) Six millions two hundreds fourteen thousands eight hundreds and forty four.
6,214,844
(b) Eighty millions one thousand and thirteen.
80,001,013
(c) Five millions nine hundreds seventy five thousands one hundred and seven.
5,957,107
(d) Two billions four hundreds thirty six millions five hundreds eighty two thousands nine hundreds
and sixty eight.
2,436,582,968
(e) Twenty nine billions one hundred ninety nine millions seven hundreds twenty one thousand and
eleven.
29,199,721,011
5. Compare the following numbers and mark < or > in the boxes provided :
(Text Book Page No. 12)
Sol.
(a) 54625
>
49768
(b) 7234470
>
6975224
(c) 5466394
>
4466394
(d) 313674495
>
41826574
(e) 98132452
>
94656102
(f)
<
24657425
24576350
6. Arrange the following in ascending order :
Sol.
(Text Book Page No. 12)
(a) 3,87,69,145
7,87,69,145
5,87,69,145
4,87,69,145
3,87,69,145
4,87,69,145
5,87,69,145
7,87,69,145
(b) 7,85,93,489
7,84,93,489
7,87,93,489
7,88,93,489
7,84,93,489
7,85,93,489
7,87,93,489
7,88,93,489
40,49,300
40,48,300
40,47,300
40,46,300
40,47,300
40,48,300
40,49,300
(d) 3,61,78,920
7,61,78,920
8,61,78,920
5,61,78,920
3,61,78,920
5,61,78,920
7,61,78,920
8,61,78,920
(c) 40,46,300
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Math-O-Mania– 5
Exercise 3.2
1. Write successor of following numbers :
Sol.
(Text Book Page No. 13)
(a) 8956312
Successor of 8956312 is
=
8956312 + 1
=
8956313
=
1467265 + 1
=
1467266
=
83205795 + 1
=
83205796
=
11511721 + 1
=
11511722
=
6810123216 + 1
=
6810123217
=
4561012 + 1
=
4561013
(b) 1467265
Successor of 1467265 is
(c) 83205795
Successor of 83205795 is
(d) 11511721
Successor of 11511721 is
(e) 6810123216
Successor of 6810123216 is
(f)
4561012
Successor of 4561012 is
2. Write the predecessor of following numbers :
Sol.
(Text Book Page No. 14)
(a) 567480
Predecessor of 567480 is
=
567480 – 1
=
567479
=
67342873 – 1
=
67342872
=
167280840 – 1
=
167280839
=
164382 – 1
=
164381
=
87643252 – 1
=
87643251
=
532467 – 1
=
532466
(b) 67342873
Predecessor of 67342873 is
(c) 167280840
Predecessor of 167280840 is
(d) 164382
Predecessor of 164382 is
(e) 87643252
Predecessor of 567480 is
(f)
532467
Predecessor of 532467 is
3. Find the place value and the face value of encircled digit in the following
(Text Book Page No. 14)
numbers :
Sol.
(a) 768 4 325
Place value of 4 in 7684325 =
=
4 × 1 thousand
4 × 1000
=
4000
Face value of 4 is 4.
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Math-O-Mania– 5
(b) 176432 8 8
Place value of 8 in 17643288 =
8 × 1 ten
=
8 × 10
=
80
=
1 × ten thousand
=
1 × 10000
=
10000
Face value of 8 is 8.
(c) 68 1 3452
Place value of 1 in 6813452
Face value of 1 is 1.
(d) 5 6443472
Place value of 5 in 56443472 =
5 × 1 crore
=
5 × 10000000
=
50000000
Face value of 5 is 5.
(e) 1896453 2
Place value of 2 in 18964532 =
2 × one
=
2 × 1
=
2
Face value of 2 is 2.
(f)
2 6432753
Place value of 2 in 26432753 =
2 × 1 crore
=
2 × 10000000
=
20000000
Face value of 2 is 2.
4. Give the expanded forms of the following numbers :
Sol.
(Text Book Page No. 14)
(a) 12397
1
2 3 9 7
Place value of 7 is 7.
Place value of 9 is 90.
Place value of 3 is 300.
Place value of 2 is 2000.
Place value of 1 is 10000.
Hence, it can be written in expanded form as :
10000 + 2000 + 300 + 90 + 7
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17
Ans.
Math-O-Mania– 5
(b) 3867589
3 8 6 7 5 8 9
Place value of 9 is 9.
Place value of 8 is 80.
Place value of 5 is 500.
Place value of 7 is 7000.
Place value of 6 is 60000.
Place value of 8 is 800000.
Place value of 3 is 3000000.
Hence, it can be written in expanded form as :
3000000 + 800000 + 60000 + 7000 + 500 + 80 + 9
Ans.
(c) 264327538
2 6 4 3 2 7 5 3 8
Place value of 8 is 8.
Place value of 3 is 30.
Place value of 5 is 500.
Place value of 7 is 7000.
Place value of 2 is 20000.
Place value of 3 is 300000.
Place value of 4 is 4000000.
Place value of 6 is 60000000.
Place value of 2 is 200000000.
Hence, it can be written in expanded form as :
200000000 + 60000000 + 4000000 + 300000 + 20000 + 7000 + 500 + 30 + 8
Ans.
(d) 567480
5 6 7 4 8 0
Place value of 0 is 0.
Place value of 8 is 80.
Place value of 4 is 400.
Place value of 7 is 7000.
Place value of 6 is 60000.
Place value of 5 is 500000.
Hence, it can be written in expanded form as :
500000 + 60000 + 7000 + 400 + 80 + 0
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18
Ans.
Math-O-Mania– 5
(e) 87643297
8 7 6 4 3 2 9 7
Place value of 7 is 7.
Place value of 9 is 90.
Place value of 2 is 200.
Place value of 3 is 3000.
Place value of 4 is 40000.
Place value of 6 is 600000.
Place value of 7 is 7000000.
Place value of 8 is 80000000.
Hence, it can be written in expanded form as :
80000000 + 7000000 + 600000 + 40000 + 3000 + 200 + 90 + 7
(f)
Ans.
534790
5 3 4 7 9 0
Place value of 0 is 0.
Place value of 9 is 90.
Place value of 7 is 700.
Place value of 4 is 4000.
Place value of 3 is 30000.
Place value of 5 is 500000.
Hence, it can be written in expanded form as :
500000 + 30000 + 4000 + 700 + 90 + 0
Ans.
5. Observe the pattern of skipping. Fill in the next four numbers :
(Text Book Page No. 14)
Sol.
(a) 123956, 123957, 123958, 123959, 123960, 123961, 123962
(b) 567902, 567912, 567922, 567932, 567942, 567952, 567962
(c) 7643890, 7643990, 7644090, 7644190, 7644290, 7644390, 7644490
(d) 6810129, 7810129, 8810129, 9810129, 10810129, 11810129, 12810129
Mental Maths (Text Book : Page No 15)
Ans :
(Do yourself)
Fun Activity (Text Book : Page No 15)
Ans :
Runway
(Do yourself)
19
Math-O-Mania– 5
4.
Operations on Large Numbers
Exercise 4.1
1. Add the following numbers :
Sol.
(a)
4653
2651
+7084
1 4 3 9 0
(d)
391
488
+549
1 4 30
(g)
(Text Book Page No. 17)
(b)
2
6
+3
1 3
3
8
6
7
(e)
48
64
+25
1 39
7284903
2452648
359234 1
+ 4269582
1 7 5 9 9 4 7 4
(h)
3958670
4257802
1962988
+ 4656769
1 4 8 3 6 2 2 9
8
4
9
2
7
3
6
6
356
907
986
25 0
537
874
758
1 7 0
9
6
7
3
7
2
6
6
4
5
9
9
4
6
8
8
8
2
8
9
2
8
5
6
(c)
0
9
7
6
1
7
+3
1 2
(f)
3
0
9
3
4
3
5
3
0
6
8
5
8
3
5
7
9
5
2
7
5
5
6
6
5703
4703
+1231
1 1 638
6
7
5
9
8
3
2
3
2. Fill in the circles with appropriate digits to make the sum statements true :
(Text Book Page No. 17)
Sol.
(a)
1
1
+ 7
9
2
2
3
8
5
4
2
2
9
3
4
7
(c)
5
6
1
+ 0
1 3
5
7
4
2
9
0
4
6
5
6
3
3
3
5
5
3
4
4
1
7
6
2
4
9
1
5
1
2
9
(b)
3
2
+ 4
9
3
1
2
7
6
5
5
7
7
1
9
9
5
9
8
2
(d)
3
2
+ 2
8
1
6
4
2
5
7
3
5
0
2
6
9
9
3
4
6
3. A man purchased a land for ` 3540290. He spent ` 19657600 in establishing a
factory and ` 516389 on the decoration of the office. How much did he spend in
all ?
(Text Book Page No. 17)
Sol.
Land was purchased by man for
=
` 3540290
Money spent is establishing a factory =
` 19657600
Money spent on the decoration
` 516389
Runway
=
20
Math-O-Mania– 5
3540290
19657600
+ 516389
23714279
\ In all the man spent = ` 23714279.
Ans.
4. A cloth mill made 2533288 m, 2635160 m and 2395290 m of cloth in three years
respectively. How much cloth did it make in all the three years ?
(Text Book Page No. 17)
Sol.
Quantity of cloth made by the mill in the 1st year
=
2533288 m
Quantity of cloth made by the mill in the 2nd year
=
2635160 m
Quantity of cloth made by the mill in the 3rd year
=
2395290 m
2533288
2635160
+ 2395290
7563738
\ Quantity of cloths made by the cloth mill in all the three years = 7563738 m
Ans.
5. A bank lent `12664570 in 1998, `13375288 in 1999, ` 23505746 in 2000 and
` 21656540 in 2001. How much money did the bank lend in four years ?
(Text Book Page No. 17)
Sol.
Money lent by the bank in 1998
=
` 12664570
Money lent by the bank in 1999
=
` 13375288
Money lent by the bank in 2000
=
` 23505746
Money lent by the bank in 2001
=
` 21656540
12664570
13375288
23505746
+ 21656540
71202144
\ Money lend by the bank in four years = ` 71202144
Ans.
6. In a state, there are 6354898 men, 6436369 women and 4570637 children. Find
the total population of that state.
(Text Book Page No. 18)
Sol.
No. of man
=
6354898
No. of women
=
6436369
No. of children =
4570637
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21
Math-O-Mania– 5
6354898
6436369
+ 4570637
17361904
\ Total population of the state = 17361904
Ans.
7. 346218 men, 339464 women and 112894 children live in a town. What is the
total population of the town ?
(Text Book Page No. 18)
Sol.
No. of men in the town
=
346218
No. of women in the town
=
339464
No. of children in the town
=
112894
346218
339464
+ 1 12894
798576
\ Total population of the town = 798576
Ans.
8. A dairy sold 1234567 l, 905289 l and 5324351 l of milk in three months. How
much milk did the dairy sell in all three months ?
(Text Book Page No. 18)
Sol.
Quantity of milk sold by the dairy in 1st month
=
1234567 l
Quantity of milk sold by the dairy in 2nd month =
905289 l
Quantity of milk sold by the dairy in 3rd month =
5324351 l
1234567
905289
+ 5324351
7464207
\ Quantity of milk sold by the dairy in the 3 months = 7464207 l
Ans.
9. A godown has 5035668 bags of rice, 10000816 bags of wheat and 19944775 bags
of sugar. Find the total number of bags in the godown.
(Text Book Page No. 18)
Sol.
No. of bags or rice
=
5035668
No. of bags of wheat
=
10000816
No. of bags of sugar
=
19944775
5035668
10000816
+ 19944775
34981259
\ Total no. of bags in the godown = 34981259
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22
Ans.
Math-O-Mania– 5
10. What will be the sum of 4335678 m, 6543244 m and 34000999 m ?
(Text Book Page No. 18)
Sol.
4335678
6543244
+ 34000999
4487992 1
\ The sum is 44879921 m.
Ans.
Exercise 4.2
1. Substract the following numbers :
Sol.
(Text Book Page No. 19)
(a)
3670535
– 1345897
2 3 2 4 6 3 8
(b)
2639586
– 1765349
0 8 7 4 2 3 7
(c)
5475998
– 3879078
1 5 9 6 9 2 0
(d)
2758510
– 1387658
1 3 7 0 8 5 2
(e)
60000606
– 19999175
4 0 0 0 1 4 3 1
(f)
75281605
– 59438418
1 5 8 4 3 1 8 7
2. Fill in the boxes with correct digits :
Sol.
(a)
3 0 0 0 0 0
–1 5 6 1 4 7
1 4 3 8 5 3
(b)
(d)
2 6 5 8 3 5 0 0
– 2 9 5 6 7 8 9
2 3 6 2 6 7 1 1
(e)
8 7 3 2 3
– 4 9 5 8 5
3 7 7 3 8
9 5 9 6 4 5 2
– 4 0 2 5 8 1 7
5 5 7 0 6 3 5
(Text Book Page No. 19)
(c)
7 4 3 5 1
– 2 8 5 6 3
4 5 7 8 8
(f)
9 6 5 5 6 7 0
– 7 2 6 5 4 2 8
2 3 9 0 2 4 2
3. The population of a city was 7447675 in 1981. It became 9865386 in 1991. Find
the increase in population ?
(Text Book Page No. 19)
Sol.
Population of the city in 1991
=
9865386
Population of the city in 1981
=
7447675
9865386
– 7447675
241 771 1
Increase in population in 10 years = 2417711
Runway
Ans.
23
Math-O-Mania– 5
4. The annual sale of a company is ` 54657361. The annual expenditure of it is
` 39506995. What will be the net profit of the company ?
(Text Book Page No. 19)
Sol.
Annual sale of a company
=
` 54657361
Annaul of expenditure of the company
=
` 39506995
54657361
– 39506995
15150366
\ Net profit of the company = 15150366
Ans.
5. A boy was asked to write 15206975. But he wrote 1526975 on his notebook. How
much less did he write from the original number ?
(Text Book Page No. 19)
Sol.
Original number
=
15206975
No. written
=
1526975
15206975
– 1526975
13680000
\ The boy wrote 1368000 less in his notebook.
6. What must be added to 85746352 to get 90000000 ?
Sol.
Ans.
(Text Book Page No. 19)
No. to be added to85746352 to get 90000000
90000000
– 85746352
4253648
\ No. of added is = 4253648
Ans.
7. In a city, there are 54670288 men and 49842700 women. How much more men
than women are there in that city ?
(Text Book Page No. 19)
Sol.
No. of men
=
54670288
No. of women
=
49842700
54670288
– 49842700
4827588
\ There are 4827588 more men than women in that city.
Ans.
8. What must be added to the greatest 7 digit number to get the greatest 8 digit
number ?
(Text Book Page No. 19)
Sol.
Greatest 7 digit number
=
9999999
Greatest 8 digit number
=
99999999
99999999
– 9999999
90000000
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24
Math-O-Mania– 5
\ The number to be added to the greatest 7 digit no. to get the greatest 8-digit no. is 90000000.
Ans.
9. The population of a town was 531784. There were 186276 men and 175901
women in the town. How many children lived in that town ? (Text Book Page No. 19)
Sol.
Total population of the town =
531784
No. of men in the town
=
186276
No. of women in the town
=
175901
No. of children
=
Total population – (No. of men + no. of women)
No. of men
=
186276
No. of women
=
175901
186276
+ 175901
362177
\ No. of children in the town
531784
– 362177
169607
\ The no. of children 169607.
Ans.
10. 141732 student appeared for All India Examination out of which 80497 were
boys. How many girls appeared in the examination ?
(Text Book Page No. 19)
Sol.
Total no. of students
=
141732
No. of boys
=
80497
141732
– 80497
61235
\ No. of girls 61235.
Ans.
Exercise 4.3
1. Find the products :
Sol.
(a)
Runway
16
×1
82
330
1654
2067
5
2
7
8
0
5
4
5
0
0
0
0
(Text Book Page No. 20)
(b)
23
×2
94
707
4714
5 51 5
25
5
3
2
1
0
3
7
4
8
0
0
8
(c)
576
×36
4035
34590
172950
21 1 5 7 5
5
7
5
0
0
5
Math-O-Mania– 5
(d)
×
2
17
266
1333
1620
4
3
2
7
7
5
2
4
6
2
8
0
0
0
4
4
2
0
0
0
2
5
5
5
0
0
0
5
(e)
695
×294
4171
27812
625770
1390600
2048353
3
6
8
0
0
0
8
(f)
3993
×4217
27951
39930
798600
15972000
16838481
2. Multiply the following :
Sol.
(Text Book Page No. 21)
(a) 234579 × 700
(b) 35284 × 20000
= 234579 × 7 × 100
= 35284 × 2 × 10000
Þ
Þ
Þ
234579
×7
1 6 4 2 0 5 3
Þ
1 6 4 2 0 5 3
×100
1 6 4 2 0 5 3 0 0
(c) 984978 × 10
35284
×2
7 0 5 6 8
7 0 5 6 8
×10000
7 0 5 6 8 0 0 0 0
(d) 76523 × 400
= 984978 × 10
= 76523 × 4 × 100
Þ
Þ
984978
×10
9 8 4 9 7 8 0
76523
×4
3 0 6 0 9 2
3 0 6 0 9 2
×100
3 0 6 0 9 2 0 0
(e) 33000 × 900
60531 × 10000
= 33000 × 9 × 100
= 60531 × 10000
Þ
Þ
Þ
Runway
(f)
33000
×9
2 9 7 0 0 0
60531
×10000
6 0 5 3 1 0 0 0 0
2 9 7 0 0 0
×100
2 9 7 0 0 0 0 0
26
Math-O-Mania– 5
(g)
10000 × 100000
Þ
(h) 4217 × 100
Þ
10000
×100000
1 0 0 0 0 0 0 0 0 0
4217
×100
4 2 1 7 0 0
3. The cost of a saree is ` 5385. Find the cost of 24 such sarees.
(Text Book Page No. 21)
Sol.
Cost of one saree
=
` 5385
Cost of 24 sarees
=
` 5385 × 24
5385
× 24
21540
107700
129240
\ Cost of 24 sarees ` 129240.
Ans.
4. The cost of a scooter is `36786. What will be the cost of 35 such scooters ?
(Text Book Page No. 21)
Sol.
Cost of one scooter
=
Cost of 35 such scooters =
` 36786
` 36786 × 35
36786
× 35
183930
1103580
1287510
\ Cost of 35 scooters ` 1287510.
Ans.
5. A factory makes 35750 m of cloth in one day. How many metre of cloth will be
made in a year ?
(Text Book Page No. 21)
Sol.
In one day quantity of cloth produced by the factory
=
35750 m
Quantity of cloths produced in 1 year
=
35750 × 365
(
\
1 year = 365 days)
35750
× 365
178750
2145000
10725000
13048750
\ Quantity of cloths produced in 1 year 13048750 m.
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27
Ans.
Math-O-Mania– 5
6. The cost of washing machine is ` 14999. Find the cost of 125 such machines.
(Text Book Page No. 21)
Sol.
Cost of one washing machine
=
` 14999
Cost of 125 washing machine
=
` 14999 × 125
14999
× 125
74995
299980
1499900
1874875
\ Cost of 125 washing machine is ` 1874875.
Ans.
7. In a school, there are 2687 students. The monthly fee of a student is ` 225. What
be the total monthly fee collection of 2687 students in that school ?
(Text Book Page No. 21)
Sol.
Monthly fee collection of one student =
` 225
No. of students in the school
2687
=
2687
× 225
13435
53740
537400
604575
\ Monthly fee collection of 268 students is ` 604575.
Ans.
8. A train covers a distance of 1385 km in one day. How many km will it cover in 180
days ?
(Text Book Page No. 21)
Sol.
In 1 day the train covers a distance
=
In 180 days the train covers a distance =
1385 km
180 × 1385 km
1385
× 180
0000
110800
138500
249300
\ In 180 days the train covers a distance of 249300 km.
Ans.
9. The cost of a dictionary is ` 450. What will be the cost of 5500 such dictionaries ?
(Text Book Page No. 21)
Runway
28
Math-O-Mania– 5
Sol.
Cost of 1 dictionary
=
` 450
Cost of 5500 such dictionaries
=
` 450 × 5500
5500
× 450
0000
275000
2200000
2475000
\ Cost of 5500 such dictionaries are ` 2475000.
Ans.
10. The cost of one bicycle is ` 1468. What will be the cost of 275 such bicycles ?
(Text Book Page No. 21)
Sol.
Cost of 1 bicycle
=
Cost of 275 such bicycle =
` 1468
` 1468 × 275
1468
× 275
7340
102760
293600
403700
\ Cost of 275 such bicycles are ` 403700.
Ans.
Exercise 4.4
1. Divide the following numbers and write the quotient and the remainder in each
(Text Book Page No. 21)
case :
Sol.
(a) 548079 ÷ 350
(b) 2459764 ÷ 225
1565
350 548079
– 350
1980
– 1750
2307
– 2100
2079
– 1750
329
Runway
10932
225 2459764
– 225
2097
– 2025
726
– 675
514
– 450
64
Quotient = 1565
Quotient = 10932
Remainder = 329
Remainder = 64
29
Math-O-Mania– 5
(c) 7058264 ÷ 445
15861
445 7058264
– 445
2608
– 2225
3832
– 3560
2726
– 2670
564
– 445
119
(d) 988470 ÷ 650
1520
650 988470
– 650
3384
– 3250
1347
– 1300
470
–0
470
Quotient = 15861
Quotient = 1520
Remainder = 119
Remainder = 470
(e) 9786255 ÷ 1625
(f)
6022
1625 9786255
– 9750
362
–0
3625
– 3250
3755
– 3250
505
7329842 ÷ 3145
2330
3145 7329842
– 6290
10398
– 9435
9634
– 9435
1992
–0
1992
Quotient = 6022
Quotient = 2330
Remainder = 505
Remainder = 1992
2. The price of 125 g of gold is ` 60625. Find the price of one gram of gold.
(Text Book Page No. 22)
Sol.
Price of 125 g of gold
=
` 60625
Price of 1g of gold
=
` 60625 ÷ 125
485
125 60625
– 500
1062
– 1000
625
– 625
0
Hence, the price of one gram of gold is ` 485.
Runway
Ans.
30
Math-O-Mania– 5
3. One carton can contain 244 eggs. How many cartons are required to pack
134200 eggs ?
(Text Book Page No. 22)
Sol.
Total number of eggs
=
134200
No. of eggs in one cartoon
=
244
No. of cartoons needed
=
134200 ÷ 244
550
244 134200
– 1220
1220
– 1220
00
–0
0
Hence, 550 cartoons are required.
Ans.
4. A train covers a distance of 115 km in one hour. How many hours will it take to
cover the distance of 269675 km at the same speed ?
(Text Book Page No. 22)
Sol.
No. of hours taken by the train to cover 115 km
=
No. of hours taken by the train to cover 269675 km =
1
269675 ÷ 115
2345
115 269675
– 230
396
– 345
517
– 460
575
– 575
0
Hence, 2345 hours are required.
Ans.
5. A cloth mill made 129348700 m of cloth in 365 days. How many metre of cloth
did it make in one day ?
(Text Book Page No. 22)
Sol.
In 365 days the cloth mill made cloth in quantity
=
129348700 m
In 1 day the mill made cloth in quantity
=
129348700 ÷ 365
Runway
31
Math-O-Mania– 5
354380
365 129348700
– 1095
1984
– 1825
1598
– 1460
1387
– 1095
2920
– 2920
00
–0
0
Hence, the mill made 354380 m cloth in one day.
Ans.
6. The price of 625 watches is ` 781250. Find the price of one watch.
(Text Book Page No. 22)
Sol.
Price of 625 watches
=
` 781250
Price of 1 watch
=
` 781250 ÷ 625
1250
625 781250
– 625
1562
– 1250
31250
– 31250
0
Hence, the price of 1 watch is ` 1250.
Ans.
7. What must be multiplied to 713 to get 4217554 ?
Sol.
(Text Book Page No. 22)
Let the required no. be x
x × 713
\
x
=
4217554
=
4217554 ÷ 713
5915223
713 4217554
– 3565
6525
– 64 1 7
1085
– 713
3724
Runway
32
contd ... page 33
Math-O-Mania– 5
3724
– 3565
1590
– 1426
1640
– 1426
2140
– 2139
1
Hence, 5915223 (approx.) must be multiplied to 713 to get 4217554.
Ans.
8. The product of two numbers is 26827710. If one of them is 5642. Find the other
number.
(Text Book Page No. 22)
Sol.
Let the required no. be x.
x × 5642
=
26827710
x
=
26827710 ÷ 5642
4755
5642 26827710
– 22568
42597
– 39494
31031
– 28210
28210
– 28210
0
Thus, the other number is 4755.
Ans.
9. Find the least number which when subtracted from 281365 becomes exactly
divisible by 445.
(Text Book Page No. 22)
Sol.
To find the required no. we have to divide 281365 by 445 and find the remainder. The remainder is the
required no.
632
445 281365
– 2670
1436
– 1335
1015
– 890
125
Thus, the least no is 125.
Runway
Ans.
33
Math-O-Mania– 5
10. A car travelled a total distance of 128750 km by moving round a track 103 times.
What was the length of the track ?
(Text Book Page No. 22)
Sol.
Total distance covered by the car =
128750 km
No. of times the car travelled
=
103
\ The length of the track
=
=
Total distance covered
No. of times the car travelled
128750 ÷ 103
1250
103 128750
– 103
257
– 206
5150
– 5150
00
–0
0
Hence, the length of the track was 1250 km.
Ans.
Mental Maths (Text Book : Page No 23)
(Do Yourself)
Ans :
Fun Activity (Text Book : Page No 23)
(Do Yourself)
Ans :
5.
Factors and Multiples
Exercise 5.1
1. Write the factors of following numbers :
Sol.
(Text Book Page No. 25)
(a) 128
= 1, 2, 4, 8, 16, 32, 64, 128
(b) 75
= 1, 3, 5, 15, 25, 75
(c) 10
= 1, 2, 5, 10
(d) 97
= 1, 97
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Math-O-Mania– 5
(e) 50
= 1, 2, 5, 10, 25, 50
(f)
52
= 1, 2, 4, 13, 26, 52
(g) 15
= 1, 3, 5, 15
(h) 64
= 1, 2, 4, 8, 16, 32, 64
2. Write down first 6 multiples of the following :
Sol.
(a) 6
(Text Book Page No. 25)
(b) 16
= 6 × 1
=
6
= 16 × 1 = 16
6 × 2 =
12
16 × 2 = 32
6 × 3 = 18
16 × 3 = 48
6 × 4 = 24
16 × 4 = 64
6 × 5 = 30
16 × 5 = 80
6 × 6 = 36
16 × 6 = 96
Thus, the first 6 multiples of 6 are 6, 12, 18,
Thus, the first 6 multiples of 16 are 16, 32, 48,
24, 30 and 36.
64, 80 and 96.
(c) 12
(d) 15
= 12 × 1
=
12
= 15 × 1 = 15
12 × 2 = 24
15 × 2 = 30
12 × 3 = 36
15 × 3 = 45
12 × 4 = 48
15 × 4 = 60
12 × 5 = 60
15 × 5 = 75
12 × 6 = 72
15 × 6 = 90
Thus, the first 6 multiples of 12 are 12, 24,
Thus, the first 6 multiples of 15 are 15, 30, 45,
36, 48, 60 and 72.
60, 75 and 90.
(e) 11
= 11 × 1
=
11
11 × 2 = 22
11 × 3 = 33
11 × 4 = 44
11 × 5 = 55
11 × 6 = 66
Runway
Thus, the first 6 multiples of 11 are 11, 22, 33, 44, 55 and 66.
35
Math-O-Mania– 5
Exercise 5.2
1. Ring all even numbers from the following numbers and also answer the questions
(Text Book Page No. 26)
that follow :
Sol.
1
3
4
16
18
21
30
37
47
52
53
74
89
90
92
94
68
12
61
10
43
78
20
70
40
98
24
42
100
35
45
61
(a) Are all even numbers prime ?
NO
(b) Are all odd numbers prime ?
NO
(c) Are composite numbers always odd ?
NO
(d) Which is an even prime number ?
2
(e) Which is the smallest odd natural number ?
1
2. Write all composite numbers between :
Sol.
(Text Book Page No. 27)
(a) 5 and 15
= 6, 8, 10, 12, 14
(b) 88 and 100
= 90, 92, 94, 96, 98
(c) 63 and 72
= 64, 66, 68, 70
(d) 121 and 135
= 122, 124, 126, 128, 130, 132, 134
3. Write all odd numbers between 30 and 60.
Sol.
(Text Book Page No. 27)
31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59
4. Write the pairs of prime numbers between 1 and 50 that differ by 2.
(Text Book Page No. 27)
Sol.
3 & 5, 5 & 7, 11 & 13, 17 & 19, 29 & 31, 41 & 43
5. 13 and 31 are both prime and contain same digits 1 and 3. Can you find much
such pairs ?
(Text Book Page No. 27)
Sol.
17 & 71, 37 & 73, 79 & 97
6. Write the prime number that comes just after :
Sol.
(a) 26 = 29
(b) 18 = 19
(c) 63 = 67
(d) 92 = 97
(e) 52 = 53
(f)
(g) 99 = 101
(h) 74 = 79
44 = 47
7. Write all 2 digit prime numbers.
Sol.
(Text Book Page No. 27)
(Text Book Page No. 27)
11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
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Math-O-Mania– 5
Exercise 5.3
1. Write the prime factors of :
Sol.
(Text Book Page No. 29)
(a) 108
2
2
3
3
3
(b) 64
108
54
27
9
3
1
2
2
2
2
2
2
64
32
16
8
4
2
1
\ 108 = 2 × 2 × 3 × 3 × 3
\ 64 = 2 × 2 × 2 × 2 × 2
(c) 735
(d) 225
3
5
7
7
735
245
49
7
1
3
3
5
5
225
75
25
5
1
\ 735 = 3 × 5 × 7 × 7
\ 225 = 3 × 3 × 5 × 5
(e) 168
(f)
2
2
2
3
7
168
84
42
21
7
1
10
2
5
10
5
1
\ 168 = 2 × 2 × 2 × 3 × 7
\ 10 = 2 × 5
(g) 99
(h) 69
3
3
11
99
33
11
1
3
23
\ 99 = 3 × 3 × 11
69
23
1
\ 69 = 3 × 23
2. Without performing actual division, test whether the following numbers are
divisible :
(Text Book Page No. 29)
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37
Math-O-Mania– 5
Sol.
(i)
By 2 :
(a) 17415
As this number does not have 2, 4, 6, 8 or 0 in units place have, it is not divisible by 2.
(b) 852572
As the number has 2 at its unit's place, therefore it is divisible by 2.
(c) 323225
As the number does not have 2, 4, 6, 8, 0 in its units place have, it is not divisible by 2.
(d) 360
As the number has 0 at its units place, hence it is divisible by 2.
(e) 501281
As the number does not have 2, 4, 6, 8 or 0 in its units place, hence it is not divisible by 2.
(f)
625320
As the number has 0 at its units place, hence it is divisible by 2.
(g) 42169
AS the number does not have 2, 4, 6, 8 or 0 at its unit's place therefore the no. is not divisible
by 2.
(h) 764348
As the no. has 8 at its unit's place, therefore the no. is divisible by 2.
(i)
94256
As the no. has 6 at its unit's place, therefore the no. is divisible by 2.
(ii) By 3 :
Sol.
(a) 624567
=
6 + 2 + 4 + 5 + 6 + 7
=
30
30, sum of digits is divisible by 3. Hence, the number is divisible by 3.
(b) 261111
=
2 + 6 + 1 + 1 + 1 + 1
=
12
12, sum of digits is divisible by 3. Hence, the number is divisible by 3.
(c) 3233325
=
3 + 2 + 3 + 3 + 3 + 2 + 5
=
21
21, sum of digits is divisible by 3. Hence, the number is divisible by 3.
(d) 804265
=
8 + 0 + 4 + 2 + 6 + 5
=
25
25, sum of digits is not divisible by 3, so the no. is not divisible by 3.
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Math-O-Mania– 5
(e) 776112
=
7 + 7 + 6 + 1 + 1 + 2
=
24
24, sum of digits is divisible by 3. Hence, the number is divisible by 3.
(f)
84257
=
8 + 4 + 2 + 5 + 7
=
26
26, sum of digits is not divisible by 3, so the no. is not divisible by 3.
(g) 101121
=
1 + 0 + 1 + 1 + 2 + 1
=
6
6, sum of digits is divisible by 3. Hence, the number is divisible by 3.
(h) 501282
=
5 + 0 + 1 + 2 + 8 + 2
=
18
18, sum of digits is divisible by 3. Hence, the number is divisible by 3.
(i)
179456
=
1 + 7 + 9 + 4 + 5 + 6
=
32
32, sum of digits is not divisible by 3, so the no. is not divisible by 3.
(iii) By 4 :
Sol.
(a) 123456
56 is divisible by 4. So the 123456 is divisible by 4.
(b) 235628
28 is divisible by 4. So the 235628 is divisible by 4.
(c) 776112
12 is divisible by 4. So the 235628 is divisible by 4.
(d) 234676
76 is divisible by 4. So the 234676 is divisible by 4.
(e) 247829
29 is not divisible by 4. So the 247829 is not divisible by 4.
(f)
123785
85 is not divisible by 4. So the 123785 is not divisible by 4.
(g) 723488
88 is divisible by 4. So the 723488 is divisible by 4.
(h) 534289
89 is not divisible by 4. So the 534289 is not divisible by 4.
(i)
672111
11 is not divisible by 4. So the 672111 is not divisible by 4.
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Math-O-Mania– 5
(iv) By 5 :
(a) 23490
As the digit at the unit's place is 0, so the no. is divisible by 5.
(b) 12345
As the digit at the unit's place is 5, so the no. is divisible by 5.
(c) 98760
As the digit at the unit's place is 0, so the no. is divisible by 5.
(d) 38597
As the digit at the unit's place is 7, so the no. is not divisible by 5.
(e) 34555
As the digit at the unit's place is 5, so the no. is divisible by 5.
(f)
534289
As the digit at the unit's place is 9, so the no. is not divisible by 5.
(g) 98765
As the digit at the unit's place is 5, so the no. is divisible by 5.
(h) 99878
As the digit at the unit's place is 8, so the no. is not divisible by 5.
(i)
5367813
As the digit at the unit's place is 3, so the no. is not divisible by 5.
(v) By 6 :
(a) 3810
As the digit at unit's place is 0, so the number is divisible by 2.
3 + 8 + 1 = 12, which is divisible by 3, so the no. 3810 is divisible by 3.
As the number 3810 is divisible by both 2 and 3 so this no. is also divisible by 6.
(b) 23456
As the digit at unit's place is 6, so the number is divisible by 2.
2 + 3 + 4 + 5 + 6 = 20, which is not divisible by 3.
As the number 23456 is not divisible by both 2 and 3 so this no. is not divisible by 6.
(c) 213060
As the digit at unit's place is 0, so the number is divisible by 2.
2 + 1 + 3 + 0 + 6 + 0 = 12, which is divisible by 3, so the no. 213060 is divisible by 3.
As the number 213060 is divisible by both 2 and 3 so this no. is also divisible by 6.
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Math-O-Mania– 5
(d) 2295
As the digit at unit's place is 5, so the number is not divisible by 2.
2 + 2 + 9 + 5 = 18, which is divisible by 3, so the no. 2295 is divisible by 3.
As the number 2295 is not divisible by both 2 and 3 so this no. is not divisible by 6.
(e) 66636
As the digit at unit's place is 6, so the number is divisible by 2.
6 + 6 + 6 + 3 + 6 = 27, which is divisible by 3, so the no. 66636 is divisible by 3.
As the number 66636 is divisible by both 2 and 3 so this no. is also divisible by 6.
(f)
48252
As the digit at unit's place is 2, so the number is divisible by 2.
4 + 8 + 2 + 5 + 2 = 21, which is divisible by 3, so the no. 48252 is divisible by 3.
As the number 48252 is divisible by both 2 and 3 so this no. is also divisible by 6.
(g) 9738
As the digit at unit's place is 8, so the number is divisible by 2.
9 + 7 + 3 +8 = 27, which is divisible by 3, so the no. 9738 is divisible by 3.
As the number 9738 is divisible by both 2 and 3 so this no. is also divisible by 6.
(h) 136383
As the digit at unit's place is 3, so the number is not divisible by 2.
1 + 3 + 6 + 3 + 8 + 3 = 24, which is divisible by 3, so the no. 136383 is divisible by 3.
As the number 136383 is not divisible by both 2 and 3 so this no. is not divisible by 6.
(i)
56789
As the digit at unit's place is 9, so the number is not divisible by 2.
5 + 6 + 7 + 8 + 9 = 35, which is not divisible by 3, so the no. 56789 is not divisible by 3.
As the number 56789 is not divisible by both 2 and 3 so this no. is not divisible by 6.
(vi) By 8 :
(a) 711524
The last 3 digits of the no. are 524 which forms the no. that is not divisible by 8. So the 711524
is not divisible by 8.
(b) 10488
The last 3 digits of the no. are 488 which forms the no. that is divisible by 8. So the 10488 is
divisible by 8.
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Math-O-Mania– 5
(c) 78988
The last 3 digits of the no. are 988 which forms the no. that is not divisible by 8. So the 78988
is not divisible by 8.
(d) 53176
The last three digits of the no. are 176 which forms the no. that is divisible by 8. So the 53176 is
divisible by 8.
(e) 69808
The last three digits of the no. are 808 which forms the no. that is divisible by 8. So the 69808
is divisible by 8.
(f)
20816
The last three digits of the no. are 816 which forms the no. that is divisible by 8. So the 20816
is divisible by 8.
(g) 721032
The last three digits of the no. are 032 which forms the no. that is divisible by 8. So the 721032
is divisible by 8.
(h) 34567
The last three digits of the no. are 567 which forms the no. that is not divisible by 8. So the
34567 is not divisible by 8.
(i)
30924
The last three digits of the no. are 924 which forms the no. that is not divisible by 8. So the
30924 is not divisible by 8.
(vii) By 9 :
(a) 8227638
Sum of the digits
=
8+2+2+7+6+3+8
=
36
Which is divisible by 9. So the no 8227638 is divisible by 9.
(b) 345672
Sum of the digits
=
3+4+5+6+7+2
=
27
Which is divisible by 9. So the no 345672 is divisible by 9.
(c) 780903
Sum of the digits
=
7+8+0+9+0+3
=
27
Which is divisible by 9. So the no 780903 is divisible by 9.
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Math-O-Mania– 5
(d) 5349
Sum of the digits
=
5+3+4+9
=
21
Which is not divisible by 9. So the no 5349 is not divisible by 9.
(e) 765432
Sum of the digits
=
7+6+5+4+3+2
=
27
Which is divisible by 9. So the no 765432 is divisible by 9.
(f)
123785
Sum of the digits
=
1+2+3+7+8+5
=
26
Which is not divisible by 9. So the no 123785 is not divisible by 9.
(g) 36081
Sum of the digits
=
3+6+0+8+1
=
18
Which is divisible by 9. So the no 36081 is divisible by 9.
(h) 543021
Sum of the digits
=
5+4+3+0+2+1
=
15
Which is not divisible by 9. So the no 543021 is not divisible by 9.
(i)
30924
Sum of the digits
=
3+0+9+2+4
=
18
Which is divisible by 9. So the no 8227638 is divisible by 9.
(viii)By 11 :
(a) 296153
Sum of the digits at odd places
=
3+1+9
=
13
Sum of the digits at even places
=
5+6+2
=
13
Difference
=
13 – 13
=
0
So the no 296153 is divisible by 11.
(b) 284713
Sum of the digits at odd places
=
3+7+8
=
18
Sum of the digits at even places
=
1+4+2
=
7
Difference
=
18 – 7
=
11
So the no 284713 is divisible by 11.
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43
Math-O-Mania– 5
(c) 360141
Sum of the digits at odd places
=
1+1+6
=
8
Sum of the digits at even places
=
4+0+3
=
7
Difference
=
8–1
=
7
So the no 360141 is not divisible by 11.
(d) 6336
Sum of the digits at odd places
=
6+3
=
9
Sum of the digits at even places
=
3+6
=
9
Difference
=
9–9
=
0
Sum of the digits at odd places
=
8+9+2
=
19
Sum of the digits at even places
=
5+8+6
=
19
Difference
=
19 – 19
=
0
So the no 6336 is divisible by 11.
(e) 628958
So the no 628958 is divisible by 11.
(f)
5367813
Sum of the digits at odd places
=
3+8+6+5 =
22
Sum of the digits at even places
=
1+7+3
=
11
Difference
=
22 – 11
=
11
So the no 5367813 is divisible by 11.
(g) 6215310
Sum of the digits at odd places
=
0+3+1+6 =
10
Sum of the digits at even places
=
1+5+2
=
8
Difference
=
10 – 8
=
2
So the no 6215310 is not divisible by 11.
(h) 729246
Sum of the digits at odd places
=
6+2+2
=
10
Sum of the digits at even places
=
4+9+7
=
20
Difference
=
20 – 10
=
10
So the no 729246 is not divisible by 11.
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Math-O-Mania– 5
(i)
672111
Sum of the digits at odd places
=
1+1+7
=
9
Sum of the digits at even places
=
1+2+6
=
9
Difference
=
9–9
=
0
So the no 672111 is divisible by 11.
3. Fill in the blank with smallest digit so that 831
1 is divisibly by 9.
(Text Book Page No. 29)
Sol.
Sum of the digits at present
=
8+3+1+1 =
13
For a no to be divisible by 9 we have to add all the digit of the no. and check whether the sum of the
digits is divisible by 9.
The sum 13 is not divisible by 9. The least sum that is divisible by 9 after 13 is 13 + 5 = 18.
Therefore, 831 5
1.
Ans.
4. Fill in the blank with smallest digit so that 20
2978 is divisible by 6.
(Text Book Page No. 29)
Sol.
Sum of the digits at present
=
2+0+2+9+7+8
=
28
For a no to be divisible by 6 it has to be divisible by both 2 and 3. The last digit of the given no is 8 so the
no is even. So this no is divisible by 2. The sum of digits to be added should be divisible by 3. But the
sum is 28. So 2 is needed in order for it to be divisible by 3.
Hence the smallest digit in the no is 20 2
978.
Ans.
5. What smallest number should be subtracted from 5678 to make it divisible by 4 ?
(Text Book Page No. 29)
Sol.
To find the no. we have to divide 5678 by 4 and the remainder will be required no.
1419
4 5678
–4
16
– 16
07
–4
38
– 36
2
So 2 must be subtracted from 5678 to make it divisible by 4.
Ans.
6. What smallest number should be added to 76319540 to make it divisible by 3 ?
(Text Book Page No. 29)
Sol.
For a no. to be divisible by 3, the sum of the digits of the number should be divisible by 3.
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45
Math-O-Mania– 5
7 + 6 + 3 + 1 + 9 + 5 + 4 + 0 = 35
35 is less than 36, which is divisible by 3. So the smallest number that should be added to 76319540 to
make it divisible by 3 is 1.
Ans.
Mental Maths (Text Book : Page No 30)
(Do yourself)
Ans :
Fun Activity (Text Book : Page No 30)
(Do yourself)
Ans :
6.
HCF and LCM
Exercise 6.1
1. Find out GCD of the following numbers by prime factorization method :
(Text Book Page No. 32)
Sol.
(a) 20 and 30
Finding Prime Factor
2
2
5
20
10
5
1
2
3
5
20 = 2 × 2 × 5
Hence, HCF or GCD
30
15
5
1
30 =
=
2 × 5
=
2
× 3 ×
5
10
Ans.
(b) 144 and 192
Finding Prime Factor
2
2
2
2
3
3
144
72
36
18
9
3
1
2
2
2
2
2
2
3
144 = 2 × 2 × 2 × 2 × 3 × 3
Hence, GCD
Runway
=
2 ×2 ×2 ×2 × 3 =
192 =
48
46
192
96
48
24
12
6
3
1
2 × 2 × 2 × 2 ×2×2× 3
Ans.
Math-O-Mania– 5
(c) 52, 78 and 130
Finding Prime Factor
2
2
13
52
26
13
1
2
3
13
52 = 2 × 2 × 13
Hence, GCD
=
2
5
13
78
39
13
1
78 = 2 × 3 × 13
2 × 13 =
130
65
13
1
130 = 2 × 5 × 13
26
Ans.
(d) 672 and 864
Finding Prime Factor
2
2
2
2
2
3
7
672
336
168
84
42
21
7
1
2
2
2
2
2
3
3
3
672 = 2 × 2 × 2 × 2 × 2 × 3 × 7
Hence, GCD
=
2×2×2×2×2×3
=
864
432
216
108
54
27
9
3
1
864 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
96
Ans.
(e) 51 and 85
Finding Prime Factor
3
17
51
17
1
5
17
51 = 3 × 17
Hence, GCD
(f)
85 =
=
5 × 17
17
Ans.
118 and 177
Finding Prime Factor
2
59
118
59
1
118 = 2 × 59
Hence, GCD
Runway
85
17
1
3
59
177 =
=
59
177
59
1
3 × 59
Ans.
47
Math-O-Mania– 5
(g) 8, 10 and 12
Finding Prime Factor
2
8
2
4
2
2
1
8
2
5
= 2 ×2×2
Hence, GCD
=
10
5
1
10 = 2 × 5
2
2
3
12
6
3
1
12 = 2 × 2 × 3
2
Ans.
(h) 60, 84 and 108
Finding Prime Factor
2
2
3
5
60
30
15
5
1
60 = 2 × 2 × 3 × 5
Hence, GCD
84
42
21
7
1
2
2
3
7
=
84 = 2 × 2 × 3 × 7
2×2 ×3 =
2
2
3
3
3
108
54
27
9
3
1
108 = 2 × 2 × 3 × 3 × 3
12
Ans.
2. Find the GCD of the following numbers by long division method :
(Text Book Page No. 32)
Sol.
(a) 120 and 140
120 140 1
– 120
20 120
– 120
×
6
Hence, the GCD of 120 and 140 is 20.
Ans.
(b) 48, 64 and 96
64
96 1
– 64
32 64
– 64
×
2
32
48 1
– 32
16 32
– 32
×
2
Hence, the GCD of 48, 64 and 96 is 16.
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Ans.
48
Math-O-Mania– 5
(c) 630 and 120
120 630 5
– 600
30 120
– 120
×
4
Hence, the GCD of 630 and 120 is 30.
Ans.
(d) 175, 225 and 555
225 555 1
– 450
105 225 2
– 210
15 105
– 105
×
7
15
175 11
– 15
25
– 15
10 15
– 10
5
1
10
– 10
×
Hence, the GCD of 175, 225 and 555 is 5.
2
Ans.
(e) 116, 290 and 580
290 580 2
– 580
×
116 290 2
– 232
58 116
– 116
×
Hence, the GCD of 116, 290 and 580 is 58.
(f)
2
Ans.
1191 and 2779
1191 2779 2
– 2382
397 1191 3
– 1191
×
Hence, the GCD of 1191 and 2779 is 397.
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Ans.
49
Math-O-Mania– 5
(g) 180, 252 and 774
252 774 3
– 756
18 252 14
– 18
72
– 72
×
18 180 10
– 180
×
Hence, the GCD of 180, 252 and 774 is 18.
Ans.
(h) 780 and 936
780 936 1
– 780
156 780 5
– 780
×
Hence, the GCD of 780 and 936 is 156.
Ans.
Exercise 6.2
1. Find the LCM of following numbers using prime factorization method :
(Text Book Page No. 34)
Sol.
(a) 15, 21 and 24
3
5
15
5
1
15 =
LCM
3 ×5
=
3×2×2×2×5×7
=
840
3
7
21 =
21
7
1
3 × 7
2
2
2
3
24 =
24
12
6
3
1
2 × 2 × 2 × 3
Ans.
(b) 56, 84 and 63
56
28
14
7
1
2
2
2
7
56 =
LCM
Runway
2 ×2 ×2 ×7
=
7×3×3×2×2×2
=
504
2
2
3
7
84 =
84
42
21
7
1
2 × 2 × 3 × 7
3
3
7
63 =
63
21
7
1
3 × 3 × 7
Ans.
50
Math-O-Mania– 5
(c) 78, 156 and 650
2
3
13
78
39
13
1
78 =
LCM
2 × 3 × 13
=
2 × 2 × 3 × 5 × 5 × 13
=
3900
2
2
3
13
156 =
156
78
39
13
1
2
5
5
13
2 × 2 × 3 × 13
650 =
650
325
65
13
1
2 × 5 × 5 × 13
Ans.
(d) 360, 540 and 900
360
180
90
45
15
5
1
2
2
2
3
3
5
360 =
LCM
2×2×2×3×3×5
=
2×2×2×3×3×3×5×5
=
5400
2
2
3
3
3
5
540
270
135
45
15
5
1
540 =
2×2×3×3×3×5
2
2
3
3
5
5
900 =
900
450
225
75
25
5
1
2×2×3×3×5×5
Ans.
(e) 66, 88 and 110
2
3
11
66
33
11
1
66 =
LCM
(f)
=
2 × 2 × 2 × 3 × 5 × 11
=
1320
2
5
11
110 =
110
55
11
1
2 × 5 × 11
Ans.
80, 96 and 105
2
2
2
2
5
Runway
2 × 3 × 11
88
44
22
11
1
88 = 2 × 2 × 2 × 11
2
2
2
11
80
40
20
10
5
1
2
2
2
2
2
3
96
48
24
12
6
3
1
51
3
5
7
105
35
7
1
Math-O-Mania– 5
80 =
LCM
2×2×2×2×5
96 =
=
2×2×2×2×2×5×3×7
=
3360
2×2×2×2×2×3
105 =
3×5×7
Ans.
(g) 650, 350 and 750
650
325
65
13
1
2
5
5
13
650 =
LCM
2
5
5
7
2 × 5 × 5 × 13
=
2 × 3 × 5 × 5 × 5 × 7 × 13
=
68250
350
175
35
7
1
350 =
750
375
125
25
5
1
2
3
5
5
5
2×5×5×7
750 =
2×3×5×5×5
Ans.
(h) 110, 132 and 154
2
5
11
110
55
11
1
110 =
LCM
132
66
33
11
1
132 = 2 × 2 × 3 × 11
2 × 5 × 11
=
2 × 2 × 3 × 5 × 7 × 11
=
4620
154
77
11
1
2
7
11
2
2
3
11
154 =
2 × 7 × 11
2. Find the LCM of the following numbers by long division method :
(Text Book Page No. 34)
Sol.
(a) 203, 329, 371
7
29
47
53
LCM
Runway
(b) 135, 105, 150
203, 329, 371
29, 47, 53
1, 47, 53
1,
1, 53
1,
1,
1
=
7 × 29 × 47 × 53
=
505673
2
3
3
3
5
5
7
135, 105, 150
135, 105, 75
45, 35, 25
15, 35, 25
1, 35, 25
1,
7,
5
1,
7,
1
1,
1,
1
LCM
52
=
2×3×3×3×5×5×7
=
9450
Math-O-Mania– 5
(c) 650, 350, 750
2
3
5
5
5
7
13
LCM
(d) 55, 308, 231
650, 350, 750
325, 175, 375
325, 175, 125
65, 35, 25
13,
7,
5
13,
7,
1
13,
1,
1
1,
1,
1
=
2 × 3 × 5 × 5 × 5 × 7 × 13
=
68250
2
2
3
5
7
11
LCM
(e) 90, 25, 180
(f)
2
2
3
3
5
5
90,
45,
45,
15,
5,
1,
1,
LCM
=
2×2×3×3×5×5
=
900
25, 180
25, 90
25, 45
25, 15
25,
5
5,
1
1,
1
2
2
2
2
3
3
3
5
7
LCM
LCM
Runway
2×2×2×2×3×3×3×5×7
=
15120
2 × 2 × 3 × 5 × 7 × 11
=
4620
96, 108, 120
48, 54, 60
24, 27, 30
12, 27, 15
6, 27, 15
3, 27, 15
1,
9,
5
1,
3,
5
1,
1,
5
1,
1,
1
=
2×2×2×2×2×3×3×3×5
=
4320
(h) 120, 216, 192
360, 168, 432
180, 84, 216
90, 42, 108
45, 21, 54
45, 21, 27
15,
7,
9
5,
7,
3
5,
7,
1
1,
7,
1
1,
1,
1
=
=
96, 108, 120
2
2
2
2
2
3
3
3
5
(g) 360, 168, 432
55, 308, 231
55, 154, 231
55, 77, 231
55, 77, 77
11, 77, 77
11, 11,
11
1,
1,
1
2
2
2
2
2
2
3
3
3
5
LCM
53
120, 216, 192
60, 108, 96
30, 54, 48
15, 27, 24
15, 27, 12
15, 27,
6
15, 27,
3
15,
9,
1
5,
3,
1
5,
1,
1
1,
1,
1
=
2×2×2×2×2×2×3×3×3×5
=
8640
Math-O-Mania– 5
Exercise 6.3
1. Complete the following table :
S.No.
(a)
(b)
(c)
(d)
(e)
1st Number
22
36
225
44
75
(Text Book Page No. 36)
2nd Number
121
48
125
188
155
GCD (HCF)
11
12
25
4
5
LCM
242
144
1125
2068
2325
2. The product of two numbers is 6552. If their HCF is 6, find the LCM.
(Text Book Page No. 36)
Sol.
Product of two numbers =
6552
=
LCM
=
=
HCF × LCM
6 × LCM
6552
6
1092
Ans.
3. HCF of two numbers is 82 and LCM is 244188. If one of the numbers is 1428, find
the other.
(Text Book Page No. 36)
Sol.
Let the required no. be x
x × 1428
=
82 × 244188
x
=
82 × 244188
1428
14022
=
Ans.
4. Find the least number which when divided by any one of 12, 15 or 10 leaves 5 as
remainder.
(Text Book Page No. 36)
Sol.
The least number which will be exactly divisible by 12, 15 and 10 is the LCM of these numbers. To get 5
as remainder, we mat add 5.
Finding LCM by division method :
2
2
3
5
LCM
\ To get 5 as remainder
Runway
12,
6,
3,
1,
1,
=
2×2×3×5
=
60
=
60 + 5 =
15,
15,
15,
5,
1,
65
54
10
5
5
5
1
Ans.
Math-O-Mania– 5
5. Find the greatest number which divides 155, 205 and 255 leaving 5 as remainder
in each case.
(Text Book Page No. 36)
Sol.
Subtracting the remainder 5 from every no :
155 – 5 = 150,
205 – 5 = 200,
255 – 5 = 250
Now, finding HCF of 150, 200 and 250
2 150,
3 75,
5 25,
5,
5
1
2
2
2
5
5
150 = 2 × 3 × 5 × 5
200
100
50
25
5
1
200 = 2 × 2 × 2 × 5 × 5
2
5
5
5
250
125
25
5
1
250 = 2 × 5 × 5 × 5
HCF = 2 × 5 × 5 = 50
Hence, the greatest number which divides 155, 205 and 255 leaving 5 as remainder in each case is
50.
Ans.
6. What is the greatest 4 digit number which is exactly divisible by 30 and 40 both ?
(Text Book Page No. 36)
Sol.
To a no. to be exactly divisible by 30 and 40 , we have to find the LCM of 30 and 40 and divide the no by
the LCM.
2
3
5
30
15
5
1
2
2
2
5
40
20
10
5
1
40 = 2 × 2 × 2 × 5
30 = 2 × 3 × 5
\ LCM
=
2×2×2×3×5 =
120
The greatest 4 -digit no. is 9999.
83
120 9999
– 960
399
– 360
39
Now we have to subtract the remainder obtained by dividing 9999 by 120 from the no 9999.
–
9999
39
9960
Hence, the required no is 9960.
Runway
Ans.
55
Math-O-Mania– 5
7. 5 bells ring at intervals of 10, 15, 20, 25, 35 seconds. If they ring together at 1
o'clock at what time again they will ring together ? (Text Book Page No. 36)
Sol.
The LCM of 10,15,20, 25, 35 is :
10,
5,
5,
5,
1,
1,
1,
2
2
3
5
5
7
LCM
=
2×2×3×5×5×7
Thus, 2100 seconds (or
=
15, 20, 25, 35,
15, 10, 25, 35,
15,
5, 25, 35,
5,
5, 25, 35,
1,
7,
1, 5,
1,
1,
7,
1,
1,
1,
1,
1,
2100
2100
= 35 (minutes) is the time after which the 5 bells will ring again together
60
= 1‘ o clock + 35 minutes
=
1 : 35
Ans.
8. Which is the smallest 5 digit number that is exactly divisible by 65 and 80 both ?
(Text Book Page No. 36)
Sol.
First, we find the LCM of 65 and 80
2
2
2
2
5
13
LCM
=
2 × 2 × 2 × 2 × 5 × 13 =
65,
65,
65,
65,
65,
13,
1
80
40
20
10
5
1
1
1040
Now, we have to multiply the LCM ( =1040) by 10 as 1040 × 9 = 9360 which is not a 5 digit no.
=
1040 × 10
=
10400
Hence, the smallest 5 digit number that is exactly divisible by 65 and 80 both is 10400.
Ans.
9. Find the largest number which divides 37, 51, 63 leaving remainders 5, 3, 7
respectively.
(Text Book Page No. 36)
Sol.
Subtracting the remainders from the numbers we get,
Runway
37 –
5
=
32
51 –
3
=
48
63 –
7
=
56
56
Math-O-Mania– 5
Now, we find out the HCF of 32, 48 and 56.
32
16
8
4
2
1
2
2
2
2
2
32 =
\ HCF
2
2
2
2
3
2×2×2×2×2
=
2×2×2
48 =
=
48
24
12
6
3
1
2
2
2
7
2×2×2×2×3
56 =
56
28
14
7
1
2×2×2×7
8
Thus, 8 is the required answer.
Ans.
10. Find the smallest number which when divided by 66, 99 as well as 88 leaves
remainder 6 in each case.
(Text Book Page No. 36)
Sol.
The smallest number which will be exactly divisible by 66, 99 and 88 is the LCM of these numbers.
To get 6 as remainder, we may add 6 to the LCM.
Finding LCM by division method :
2
2
2
3
3
11
LCM
=
66,
33,
33,
33,
11,
11,
1,
99,
99,
99,
99,
33,
11,
1,
2 × 2 × 2 × 3 × 3 × 11
=
792
\ To get 6 as remainder, the no is
=
792 + 6
=
798
88
44
22
11
11
11
1
Ans.
Try Me (Text Book : Page No 36)
Ans :
(Do yourself)
Mental Maths (Text Book : Page No 37)
Ans :
(Do yourself)
Fun Activity (Text Book : Page No 41)
Ans :
Runway
(Do yourself)
57
Math-O-Mania– 5
7.
Fractions
Exercise 7.1
1. State numerator and denominator in the following fractions : (Text Book Page No. 38)
Sol.
(a)
5
6
(b)
Numerator
1
3
2
9
(c)
=
5
Numerator
=
1
Numerator
Denominator =
6
Denominator =
3
Denominator = 9
(d)
91
19
(e)
Numerator
85
75
= 2
102
132
(f)
=
91
Numerator
=
85
Numerator
Denominator =
19
Denominator =
75
Denominator = 132
2. Pick out proper fractions from the following :
(a)
23
19
Ans. (b)
8
17
(b)
(c)
8
17
3
8
(c)
(d)
5
9
3
8
(f)
(d)
(Text Book Page No. 38)
5
9
(e)
28
19
11
5
Ans. (a)
11
5
(b)
(b)
18
7
18
7
(c)
(f)
16
17
8
15
(d)
15
19
(e)
17
77
92
16
(Text Book Page No. 39)
— Not mixed
(b)
16
25
— Not mixed
2
5
— Mixed
(d) 8
3
8
— Mixed
(e) 15
1
8
— Mixed
(f) 75
16
17
— Mixed
5. Check if the following pairs of fractions are like or unlike :
Runway
(f)
92
16
(c) 1
Ans. (a)
25
39
(Text Book Page No. 39)
4. Pick out mixed fractions from these :
Ans. (a)
(f)
25
39
3. Pick out improper fractions from the following :
(a)
= 102
3 , 2
7
3
Unlike
(b)
58
(Text Book Page No. 39)
3 , 1
4
4
Like
Math-O-Mania– 5
(c)
1 , 7
12 12
(d)
Like
(e)
1 , 5
15
17
Unlike
11. , 15
13 13
(f)
Like
19 , 17
20 20
Like
6. Pick out the unit fractions from following fractions :
Ans. (a)
(d)
1
8
Yes
(b)
1
20
(e)
Yes
2
9
No
(Text Book Page No. 39)
(c)
3
19
(f)
No
1
17
Yes
1
25
Yes
Exercise 7.2
1. Write the following improper fractions as mixed numbers : (Text Book Page No. 42)
Sol.
(a)
8
5
5
(b)
8
–5
3
1
17
4
4
17 4
– 16
1
17
1
= 4
4
4
\
8
5
(c)
35
6
(d)
6 35 5
– 30
5
35
5
= 5
6
6
16 81 5
– 80
1
81
1
\
= 5
16
16
131
5
(f)
\
(e)
\
Runway
131
5
= 1
3
5
\
5 131 26
– 10
31
– 30
1
1
= 26
5
81
16
201
30
30 201 6
– 180
21
\
59
201
30
= 26
21
30
Math-O-Mania– 5
2. Write the following mixed numbers as improper fractions : (Text Book Page No. 42)
Sol.
1
5
(a) 1
(b) 4
=
(5 × 1) + 1
5
=
5+1
5
=
6
5
3
10
(c) 7
70 + 3
10
=
=
73
10
=
8+1
2
(f)
63 + 3
7
=
=
2
5
50 + 2
5
15
=
(a)
(c)
Runway
1
8
1
8
= (15 × 8) + 3
8
= 120 + 3 =
8
66
7
(b)
=
1 × 2
8 × 2
1
8
=
1 × 3
8 × 3
1
8
=
1 × 4
8 × 4
5
6
5
6
2
16
3
=
24
4
=
32
=
(d)
=
5 × 2
6 × 2
5
6
=
5 × 3
6 × 3
5
6
=
5 × 4
6 × 4
52
5
3
8
3. Write 3 equivalent fractions for each of the following :
Sol.
9
2
(10 × 5) + 2
5
=
= (9 × 7) + 3
7
=
(2 × 4) + 1
2
=
3
7
(e) 9
=
(d) 10
(7 × 10) + 3
10
=
1
2
10
12
15
=
18
20
=
24
=
60
4
5
4
5
=
4 × 2
5 × 2
4
5
=
4 × 3
5 × 3
4
5
=
4 × 4
5 × 4
=
3 × 2
4 × 2
3
4
=
3 × 3
4 × 3
3
4
=
3 × 4
4 × 4
3
4
3
4
123
8
(Text Book Page No. 42)
8
10
12
=
15
16
=
20
=
6
8
9
=
12
12
=
16
=
Math-O-Mania– 5
(e)
2
7
2
7
(f)
=
2 × 2
7 × 2
2
7
=
2 × 3
7 × 3
2
7
=
2 × 4
7 × 4
4
14
6
=
21
8
=
28
=
1
11
1
11
=
1
11
=
1
11
=
1 × 2
2
=
11 × 2
22
1 × 3
3
=
11 × 3
33
1 × 4
4
=
11 × 4
44
4. Put <, > or = in the boxes :
(Text Book Page No. 42)
Ans. (a)
1
8
<
3
8
(b)
4
5
>
2
5
(c)
3
4
>
5
8
(d)
2
3
>
1
2
(e)
2
8
=
1
4
(f)
3
2
=
9
6
5. Reduce each fraction into smallest (lowest) terms :
Sol.
(a)
(c)
(e)
25
100
(b)
(Text Book Page No. 42)
18
24
=
5×5
(Prime factorization)
2×2×5×5
=
2×3×3
2×2×2×3
=
5×5
2×2×5×5
=
2×3×3
2×2×2×3
=
1
4
63
72
(d)
=
3×3×7
2×2×2×3×3
=
3×3×7
2×2×2×3×3
=
7
8
16
20
(f)
=
2×2×2×2
2×2×5
=
2×2×2×2
2×2×5
=
4
5
Runway
3
5
=
9
15
3
4
=
8
9
=
2
1
32
36
=
2×2×2×2×2
2×2×3×3
=
2×2×2×2×2
2×2×3×3
66
33
=
2 × 3 × 11
3 × 11
=
2 × 3 × 11
3 × 11
6. Fill in the missing numbers :
Ans. (a)
=
(Text Book Page No. 42)
(b)
5
9
=
61
50
90
(c)
1
2
=
5
10
Math-O-Mania– 5
(d)
3
4
15
=
20
(e)
5
8
=
40
64
(f)
7. Arrange the following in ascending order :
Ans. (a)
2 , 7 , 11
15 15 15
(b)
2 , 7 , 11
15
15
15
(d)
7 , 11 , 13
9
12 15
7 , 13 , 11
9
15
12
5 , 5 , 5
7
3
11
(c)
7 , 13 , 18
10 20 25
(f)
13 , 7 , 18
20 10 25
9 , 7 , 13 , 2
10 10 10
10
(b)
13 , 9 , 7 , 2
10 10 10 10
5 , 7 , 8
6
8
9
(Text Book Page No. 42)
1 ,
4
2
2 ,
4
3 ,
1
1
4
4
1 , (4 × 2) + 2 , 3 , (1 × 4) + 1
4
4
4
4
1 ,
10 ,
3 , 5
4
4
4
4
2 , 7 , 9 , 1
5
10 20 15
(d)
7 , 9 , 2 , 1
10 20
5
5
(e)
7 , 23 , 19
12 24 36
5 , 7 , 8
6
8
9
10 ,
4
2 ,
2
4
(c)
4
10
19 , 7 , 23
36 12 24
8. Arrange the following in descending order :
Ans. (a)
=
5
(Text Book Page No. 42)
5 , 5 , 5
11
7
3
(e)
2
5 ,
4
1 ,
1
4
3 ,
4
3 ,
4
1
4
1
4
7 , 15 , 3 , 9
8
4
24
16
15 , 7 , 3 , 9
8
16
4
24
5 , 13 , 5 , 3
12 36
6
8
(f)
5 , 5 , 3 , 13
12
6
8
36
2 , 7 , 13 , 7
9
9
27 18
7 , 13 , 7 , 2
18 27
9
9
Exercise 7.3
1. Add the following :
(a)
Runway
1
3
+
(Text Book Page No. 43)
1
5
(b)
62
2
7
+
2
3
Math-O-Mania– 5
Sol.
(a) Taking LCM of denominators
LCM of 3 and 5 is 15.
LCM of 7 and 3 is 21.
Making both the denominators as 15 :
Making both the denominators as 21 :
1 × 5
3 × 5
Þ
5
15
=
(c)
6
7
=
1 × 3
5 × 3
+
3
15
=
8
15
7
8
7
8
(d)
+
2 × 7
3 × 7
+
14
21
=
20
21
8
9
+
LCM of 7 and 8 is 56.
LCM of 8 and 9 is 72.
Making both the denominators as 56 :
Making both the denominators as 72 :
6 × 8
7 × 8
48
56
1
3
+
+
7 × 7
8 × 7
+
49
56
7 × 9
8 × 9
Þ
=
97
56
4
33
=
63
72
(f) 5
3
4
+
8 × 8
9 × 8
+
64
72
+
3
=
127
72
1
4
Taking LCM of denominators
Taking LCM of denominators
LCM of 3 and 33 is 33.
LCM of 4 and 4 is 4.
Making both the denominators as 33 :
Making both the denominators as 4 :
1 × 11
3 × 11
11
33
+
4 × 1
33 × 1
+
4
33
=
15
33
Þ
(5 × 4) + 3
4
+
(3 × 4) + 1
4
=
20 + 3
4
+
12 + 1
4
+
13
4
=
2
3
1
+
+
21
28
14
Taking LCM of denominators
(h)
LCM of 14, 21 and 28 is 84.
Runway
6
21
=
=
(g)
2 × 3
7 × 3
Þ
Taking LCM of denominators
=
Þ
+
+
Taking LCM of denominators
Þ
(e)
(b) Taking LCM of denominators
23
4
36
4
=
9
5
1
7
1
+
+
+
24
20
36
12
Taking LCM of denominators
LCM of 12, 24, 20 and 36 is 360.
63
Math-O-Mania– 5
Making both the denominators as 84 :
Making both the denominators as 360 :
Þ
1×6
14 × 6
+
2×4
21 × 4
+
3×3
28 × 3
Þ
=
6
84
+
8
84
+
9
84
=
30
360
=
6+8+9
84
23
84
=
30 + 75 + 18 + 70
360
=
1 × 30
5 × 15
1 × 18
7 × 10
+
+
+
12 × 30
24 × 15
20 × 18
36 × 10
+
75
360
2. Subtract the following :
Sol.
(a)
(b)
Making denominators of both the
fractions as 4.
1 × 2
2 × 2
2
4
=
(c)
Þ
=
3
5
Runway
1 × 1
4 × 1
–
1
4
=
+
193
360
1
3
–
5
4
LCM of 4 and 5 is 20.
3 × 5
4 × 5
Þ
=
1
4
15
20
=
(d)
7
8
–
–
1 × 4
5 × 4
–
4
20
=
11
20
1
5
LCM of 5 and 6 is 30
LCM of 8 and 5 is 40.
Making denominators of both the
fractions as 30.
Making denominators of both the
fractions as 40.
3 × 6
5 × 6
18
30
– 3
–
1 × 5
6 × 5
–
5
30
=
13
30
35
40
=
4
7
(f)
Þ
Making denominators of both the
fractions as 7.
Þ
–
7 × 5
8 × 5
Þ
LCM of 7and 7 is 7
(7 × 4) + 5
7
70
360
Making denominators of both the
fractions as 20.
1
6
–
(e) 4 5
7
Þ
–
18
360
(Text Book Page No. 43)
1
1
–
4
2
LCM of 2 and 4 is 4
Þ
+
(3 × 7) + 4
7
Þ
64
–
1 × 8
5 × 8
–
8
40
1
1
– 2
12
6
(6 × 3) + 1
–
6
=
27
40
3
18 + 1
6
19
6
(2 × 12) + 1
12
–
24 + 1
12
–
25
12
Math-O-Mania– 5
28 + 5
7
=
33
7
8
7
=
=
=
1
(g) 4
–
21 + 4
7
–
25
7
LCM of 6 and 12 is 12.
Making denominators of both the
fractions as 12.
=
1
7
5
8
– 2
1
2
=
38
12
=
13
12
(h) 3
1
4
Þ
(4 × 5) + 1
8
–
=
32 + 5
8
–
4+1
2
=
–
5
2
=
37
8
=
(2 × 2) + 1
2
Þ
25 × 1
12 × 1
–
25
12
=
– 3
(3 × 4) + 1
4
12 + 1
4
13
4
1
1
12
1
8
–
(3 × 8) + 1
8
–
24 + 1
8
–
25
8
LCM of 4 and 8 is 8.
Making denominators of both the
fractions as 8.
Making denominators of both the
fractions as 8.
Þ
37 × 1
8×1
–
5×4
2×4
Þ
=
37
8
–
20
8
=
1
8
=
17
8
=
2
3. Simplify the following :
1
12
=
(12 × 9) + 5
9
–
(2 × 12) + 1
12
+
(1 × 18) + 11
18
=
108 + 5
9
–
24 + 1
12
+
18 + 11
18
5
9
+ 1
13 × 2
4 × 2
26
8
–
–
25 × 1
8 × 1
25
8
1
8
(Text Book Page No. 44)
– 2
(a) 12
Runway
–
LCM of 8 and 2 is 8.
=
Sol.
19 × 2
6 × 2
11
18
65
Math-O-Mania– 5
=
113
9
25
12
–
29
18
+
The LCM of 9, 12 and 18 is 36.
Making denominators of all the fractions as 36 :
Þ
113 × 4
9×4
–
25 × 3
12 × 3
+
25 × 3
12 × 3
=
452
36
–
75
36
+
58
36
=
452 – 75 + 58
36
(b) 15
=
=
1
8
+ 4
26
13
(15 × 13) + 8
13
203
13
+
=
– 3
435
36
105
26
–
3
36
31
39
(4 × 26) + 1
26
+
= 12
–
(3 × 39) + 31
39
148
39
The LCM of 13, 26 and 39 is 78.
Making denominators of all the fractions as 78 :
203 × 6
13 × 6
Þ
1218
78
=
+
105 × 3
26 × 3
–
148 × 2
39 × 2
+
315
78
–
296
78
=
1218 + 315 – 296
78
(c)
6
=
=
2
7
– 3
1
6
(7 × 6) + 2
–
7
44
19
–
7
6
=
– 1
1237
78
= 15 67
78
4
21
(6 × 3) + 1
6
25
–
21
–
(1 × 21) + 4
21
The LCM of 7, 6 and 21 is 42.
Making denominators of all the fractions as 42 :
Þ
Runway
44 × 6
7×6
–
19 × 7
6×7
–
25 × 2
21 × 2
66
Math-O-Mania– 5
264
42
=
=
133
42
–
264 – 133 – 50
42
(d) 7 + 7
=
7
1
=
7
1
5
22
= 1
39
42
17
33
(22 × 7) + 5
22
159
22
+
81
42
=
– 3
+
50
42
–
–
(33 × 3) + 17
33
116
33
–
The LCM of 1, 22 and 33 is 66.
Making denominators of all the fractions as 66 :
Þ
7 × 66
1 × 66
+
159 × 3
22 × 3
–
116 × 2
33 × 2
=
462
66
+
477
66
–
232
66
=
462 + 477 – 232
66
(e) 5 + 18
4
15
=
5
1
+
=
5
1
+
– 5
–
= 10 47
66
14
25
(15 × 18) + 4
15
274
15
707
66
=
–
(25 × 5) + 14
25
139
25
The LCM of 1, 15 and 25 is 75.
Making denominators of all the fractions as 75 :
Þ
5 × 75
1 × 75
+
274 × 5
15 × 5
–
139 × 3
25 × 3
=
375
75
+
1370
75
–
417
75
=
(f)
Runway
375 + 1370 – 417
75
18
1
+
13
15
52
=
–
1328
75
= 17 53
75
5
39
67
Math-O-Mania– 5
=
(18 × 13) + 1
13
=
235
13
15
52
+
15
52
+
–
5
39
–
5
39
The LCM of 13, 52 and 39 is 156.
Making denominators of all the fractions as 156 :
Þ
235 × 12
+
13 × 12
2820
156
=
=
+
15 × 3
52 × 3
–
5×4
39 × 4
45
156
–
20
156
2820 + 45 – 20
156
=
2845
156
= 18
37
156
Exercise 7.4
1. Multiply the following :
Sol.
5
7
× 3
= 5
7
×
(a)
2
9
×
4
=
2
9
×
(b)
3
1
=
5×3
7×1
=
2×4
9×1
=
15
7
=
8
9
(c)
2
9
×
3
3
1
=
31
×
1
=
2
3
(e) 1
1
7
× 2
3
4
=
8
7
×
11
4
×
1
5
(d)
=
Runway
(Text Book Page No. 44)
2
9
=
2
93
=
×
1
5
4
1
2
2
1
×
1×2
5×1
2
=
5
(f)
3
=
68
1
2
× 2
7
2
×
1
4
9
4
Math-O-Mania– 5
=
82
×
7
=
22
7
(g) 5
6
7
11
41
7×9
2×4
63
=
8
1
4
× 2
(h) 1
4
5
5
14
×
=
4
5
=
1
14
85
×
14
× 6
=
41
7
=
41 × 85
7 × 14
=
51
×
42
=
3485
98
=
7
2
7
14
51
2. Find :
Sol.
(a)
1
1
of
3
6
1
1
×
=
3
6
=
(c)
(Text Book Page No. 44)
(b)
1
18
=
4
4
of
7
3
4
4
×
=
7
3
(d)
1
of 3
9
64 4
=
×
93
=
Runway
(f)
77
135
(g) 7
5
16
of
8
20
51
16 2 1
×
=
81
20 4
=
7
11
of
9
15
7
11
=
×
9
15
=
3
2
16
=
21
(e)
1
of 9
3
1
93
×
=
31
1
1
2
6
7
of
10
12
1
6
7
=
×
10
12 2
7
=
20
7
1
of 2
8
24
87 29 49
=
×
8
24
8
3
16
51 17
16 1
(h) 10
68
3
=
69
1421
64
Math-O-Mania– 5
Exercise 7.5
1. Divide the following :
Sol.
(a)
(c)
(e)
(g)
1
÷ 2
3
1
=
÷
3
1
=
×
3
1
=
6
1
÷ 4
6
1
=
÷
6
1
=
×
6
= 1
24
8
÷ 4
9
8
=
÷
9
28
=
×
9
2
=
9
8
÷ 8
9
8
=
÷
9
8
=
×
9
= 1
9
(i) 12 ÷
=
Runway
(Text Book Page No. 45)
(b)
2
1
1
2
(d)
4
1
1
4
(f)
4
1
1
4
(h)
8
1
1
8
4
5
12
÷
1
(j)
4
5
1
÷ 4
5
1
4
÷
=
5
1
1
1
×
=
5
4
1
=
20
5
÷ 3
9
5
3
=
÷
9
1
5
1
=
×
9
3
5
=
27
7
÷ 7
8
7
7
÷
=
8
1
7
1
×
=
8
7
1
=
8
9
÷ 8
10
9
8
=
÷
10
1
9
1
=
×
10
8
9
=
80
15 ÷
=
70
15
÷
1
5
4
5
4
Math-O-Mania– 5
=
12 3
×
1
5
41
=
= 15
(k) 16 ÷
16
÷
1
16 4
=
×
1
=
15 3
×
1
4
51
= 12
4
7
4
7
7
41
18
÷
1
18 2
×
1
=
=
= 28
9
5
18 ÷
(l)
=
9
5
5
91
10
Exercise 7.6
1. Veenu ate
2
4
of a cake and Anu ate
of the same cake. How much of the cake is
3
13
eaten up ?
(Text Book Page No. 46)
Sol.
Part of the cake eaten up by Veenu
=
Part of the cake eaten up by Anu
=
Total Part of the cake eaten up
=
=
Thus, quantity of the cake eaten up =
2. Teena had ` 50. She spent ` 4
2
3
4
13
2 + 4
3
13
26 + 12
39
38
39
=
38
39
Ans.
1
3
on books and ` 4
on toys. How many rupees
8
4
were left with her ?
Sol.
(Text Book Page No. 46)
Total amount of money that Teena had
=
` 50
Amount spent by Teena on books
=
` 4
Amount spent by Teena on toys
=
\ Amount left with Teena
=
=
Runway
1
8
3
` 4
4
=
=
33
–
8
400 – 33 – 38
8
50 –
71
33
8
19
`
4
`
19
4
=
329
8
=
` 41
1
8
Ans.
Math-O-Mania– 5
3. A fraction is added to
Sol.
Required fraction
=
=
=
4. Priya bought 9
1
3
to get
. Find the fraction.
3
7
3
7
(Text Book Page No. 46)
1
3
–
9–7
21
2
21
Ans.
1
m of cotton cloth at the rate of ` 10 per m. How much money is
2
Priya required to pay ?
Sol.
(Text Book Page No. 46)
1
m
2
Quantity of cloth bought
=
9
=
Rate of the cloth bought
=
` 10 / m
19
m
2
\ Total amount of money Priya is required to pay
=
5
10 ×
19
=
21
` 95
Ans.
5. In a cinema hall's parking, 90 cars can be parked at a time. During a night show,
5
of the parking lot was full. How many cars were there at that time ?
9
(Text Book Page No. 46)
Sol.
No. of cars that can be parked
=
Fraction or Part of the parking lot that was full
=
No. of cars present at that time
=
=
6. The product of two numbers is 6
90
5
9
5
× 90 10
91
50
Ans.
1
1
. If one number is 3 , find the other number.
4
2
(Text Book Page No. 46)
Sol.
Let the other no. be x.
Þ
Þ
Runway
x
×
3
x
×
1
2
1
4
=
6
7
2
=
25
4
x
=
25
×
42
21
7
=
72
25
14
Ans.
Math-O-Mania– 5
7. Each side of a square is 4
Sol.
2
m. Find its perimeter.
7
Side of the square
=
4
2
m
7
Perimeter of a square
=
4
×
=
4
× 30
7
=
17
=
(Text Book Page No. 46)
30
m
7
side
= 120 m
7
1
m
7
Ans.
8. The cost of 10 kg of sugar is ` 200 1 . What is the cost of 3 kg sugar ?
2
(Text Book Page No. 46)
Sol.
Cost of 10 kg of sugar
=
\ The cost of 3 kg of sugar =
=
9.
Sol.
` 200
1
2
401
× 3
2
`
=
1
401
×
× 3
10
2
=
` 60
10
1
1203
20
401
2
=
3
20
Ans.
3
of a number is 39. Find the number.
4
(Text Book Page No. 46)
Let the required no. be x.
3
4
×
x
=
39
x
=
39 ÷
x
=
3
4
13
4
39 ×
31
10. The cost of 1 l of milk is ` 18
=
52
Ans.
3
1
. Find the cost of 7 l of milk.
4
2
(Text Book Page No. 46)
Sol.
Cost of 1 l of milk
Cost of 7
=
1
15
l or
l of milk =
2
2
=
Runway
3
4
75
15
×
4
2
` 18
75
4
=
`
=
` 140
÷ 1
1125
8
73
5
8
Ans.
Math-O-Mania– 5
Mental Maths (Text Book : Page No 47)
(Do yourself)
Ans :
Fun Activity (Text Book : Page No 47)
(Do yourself)
Ans :
8.
Decimals
Exercise 8.1
1. How will you read the following decimal expressions ?
Sol.
(Text Book Page No. 50)
(a) 1.734
One point seven three four
(b) 2.77
Two point seven seven
(c) 9.312
Nine point three one two
(d) 10.250
Ten point two five zero
(e) 66.2189
Sixty six point two one eight nine
(f)
9.938
Nine point nine three eight
(g) 0.0012
Zero point zero zero one two
(h) 9.10804
Nine point one zero eight zero four
2. Write the following in expanded decimal form :
Sol.
(Text Book Page No. 50)
(a) 270.45
200 + 70 + 0 + 0.4 + 0.05
(b) 7.013
7 + 0.0 + 0.01 + 0.003
(c) 21.983
20 + 1 + 0.9 + 0.08 + 0.003
Runway
74
Math-O-Mania– 5
(d) 0.35
0.3 + 0.05
(e) 263.805
200 + 60 + 3 + 0.8 + 0.00 + 0.005
(f)
15.024
10 + 5 + 0.0 + 0.02 + 0.004
(g) 0.810
0.8 + 0.01 + 0.000
(h) 1.347
1 + 0.3 + 0.04 + 0.007
3. Write the following in expanded fractional form :
Sol.
(a) 4.326
(b) 8.127
= 4 + 0.3 + 0.02 + 0.006
= 4 +
= 8 + 0.1 + 0.02 + 0.007
3
2
6
+
+
10
100
1000
= 8 +
(c) 1.95
= 10 + 8 + 0.2 + 0.00 + 0.005 + 0.0003
9
5
+
10
100
= 10 + 8 +
(e) 13.560
(f)
= 10 + 3 + 0.5 + 0.06 + 0.000
= 10 + 3 +
2
5
3
+
+
10
1000
10000
0.93
= 0.9 + 0.03
5
6
+
10
100
=
(g) 0.09008
9
3
+
10
100
(h) 6.7803
= 0.0 + 0.09 + 0.000 + 0.0000 + 0.00008
=
1
2
7
+
+
10
100
1000
(d) 18.2053
= 1 + 0.9 + 0.05
= 1 +
(Text Book Page No. 50)
= 6 + 0.7 + 0.08 + 0.000 + 0.0003
9
8
+
100
100000
= 6 +
7
8
3
+
+
10
100 10000
4. Write the place value of the underlined digits in the following :
(Text Book Page No. 50)
Sol.
(a) 30.6 2 5
(b) 29. 1 4 3
2 ® hundredth
1 ® ten th
3 ® thousandth
Runway
75
Math-O-Mania– 5
(c) 1 8 .0 3 8
(d) 0.25 4
8 ® unit
4 ® thousandth
3 ® hundredth
5. Express the following fractions as decimals :
(a)
2
25
(b)
2×4
25 × 4
=
=
8
100
(Text Book Page No. 50)
5
10
=
0.5
= 0.08
4
5
(c)
=
(d)
4×2
5×2
=
8
10
=
5
10
=
35
100
= 0.5
92
10
(f)
= 9.2
(g)
1×5
2×5
=
= 0.8
(e)
1
2
17
1000
= 0.017
.1
100
(h)
= 0.001
7
20
=
7×5
20 × 5
= 0.35
6. Express the following decimals as fractions :
Sol.
(Text Book Page No. 50)
214
100
(b) 9.312
=
9312
1000
(c) 2.875 =
2875
1000
(d) 0.83
=
83
100
(e) 10.532 =
10352
1000
(f)
3.9
=
39
10
(g) 0.612
612
1000
(h) 12.385
=
12385
1000
(a) 2.14
Runway
=
=
76
Math-O-Mania– 5
Exercise 8.2
1. What numbers should be rounded off to :
(Text Book Page No. 51)
Ans. (a) 13.0
–
12.5, 12.6, 12.7, 12.8, 12.9, 13.1, 13.2, 13.3, 13.4
(b) 15.0
–
14.5, 14.6, 14.7, 14.8, 14.9, 15.1, 15.2, 15.3, 15.4
(c) 22.0
–
21.5, 21.6, 21.7, 21.8, 21.9, 22.1, 22.2, 22.3, 22.4
(d) 45.0
–
44.5, 44.6, 44.7, 44.8, 44.9, 45.1, 45.2, 45.3, 45.4
(e) 87.0
–
86.5, 86.6, 86.7, 86.8, 86.9, 87.1, 87.2, 87.3, 87.4
(f)
–
54.5, 54.6, 54.7, 54.8, 54.9, 55.1, 55.2, 55.3, 55.4
55.0
2. Round off the following decimals to the nearest tenths :
Ans. (a) 35.77
(b) 102.32
= 35.8
(Text Book Page No. 51)
(c) 107.87
= 102.3
(d) 88.88
= 107.9
(e) 44.44
= 88.9
(f)
= 44.4
207.42
= 207.4
3. Round off the following decimals to the nearest hundredths : (Text Book Page No. 51)
Ans. (a) 57.564
(b) 88.999
= 57.56
(c) 6.559
= 89.00
(d) 15.106
= 6.56
(e) 300.309
= 15.11
(f)
= 300.31
27.449
= 27.45
4. Round off the following decimals to the nearest tenth and hundredth in a table :
(Text Book Page No. 51)
Ans.
Tenth
Hundredth
Tenth
Hundredth
(a) 17.8
17.8
17.80
(b) 25.9
25.9
25.90
(c) 58.25
58.3
58.25
(d) 76.85
76.9
76.85
(e) 30.378
30.4
30.38
(f)
506.0
506.01
506.005
5. Put >, < or = in the boxes :
(Text Book Page No. 52)
Ans. (a) 3.693
<
3.963
(b) 8.031
<
8.301
(c) 2.002
<
2.200
(d) 3.6
>
0.36
(e) 6.007
>
5.007
(f)
<
4.21
Runway
77
4.12
Math-O-Mania– 5
6. Arrange the following in ascending order :
Ans. (a) 1.312, 8.501, 7.312, 2.542
(Text Book Page No. 52)
1.312, 2.542, 7.312, 8.501
(b) 3.876, 3.768, 3.678, 3.867
3.678, 3.768, 3.867, 8.876
(c) 0.63, 7.241, 8.325, 19.621
0.63, 7.241, 8.325, 19.621
(d) 6.1, 1.6, 1.62, 2.16
1.6, 1.62, 2.16, 6.1
7. Arrange the following in descending order :
Ans. (a) 37.635, 38.365, 37.356, 37.65
(Text Book Page No. 52)
38.365, 37.65, 37.635, 37.356
(b) 1.734, 1.374, 1.743, 1.437
1.743, 1.734, 1.437, 1.374
(c) 0.812, 0.802, 0.820, 0.128
0.820, 0.812, 0.802, 0.128
(d) 41.101, 41.001, 40.101, 40.011
41.101, 41.001, 40.101, 40.011
Exercise 8.3
1. Add the following :
Sol.
(Text Book Page No. 53)
(a) 4.3+0.637+439+0.0437
(b) 25+2.5+0.25+0.025
4.3000
0.6370
439.0000
+
0.0437
+
443.9807
25.000
2.500
0.250
0.025
27.775
(c) 9.7+10008+1.08+0.009
(d) 60+6.05+0.065+0.006
9.700
10008.000
1.080
+
0.009
+
10018.789
60.000
6.050
0.065
0.006
66.1 2 1
(e) 0.45+4.5+0.045+450
(f)
71.07+7.107+710.7+7.007
0.450
4.500
0.045
+ 450.000
71.070
7.107
710.700
+
7.007
454.995
795.884
2. Subtract the following :
Sol.
(Text Book Page No. 53)
(a) 42.04 – 8.46
42.04
–
8.46
(b) 53.13 – 19.76
53.13
– 19.76
33.58
Runway
33.37
78
Math-O-Mania– 5
(c) 12 – 0.00896
12.00000
–
0.00896
(d) 32.24 – 19.75
32.24
– 19.75
11.99104
12.49
(e) 45.27 – 38.424
–
(f)
45.270
38.424
14 – 0.8765
–
6.846
14.0000
0.8765
13.1235
3. Solve the following questions :
Sol.
(Text Book Page No. 53)
(a) 4.3 + 43 + 2.43 – 3.7
(b) 33 – 3.3 + 3.03 – 3.003
= (4.3 + 43 + 2.43) – (3.7)
+
= (33 + 3.03) – (3.3 + 3.003)
4.30
43.00
2.43
+
33.00
3.03
36.03
49.73
–
49.73
3.70
+
3.300
3.003
6.303
46.03
–
36.030
6.303
29.727
(c) 4.8 + 482 – 4.048
(d) 53.68 + 53 – 0.089 – 0.008
(4.8 + 482) – (4.048)
= (53.68 + 53) – (0.089 + 0.008)
4.8
+ 482.0
+
486.8
–
53.68
53.00
106.68
486.800
4.048
+
0.089
0.008
0.097
482.752
–
106.680
0.097
106.583
Runway
79
Math-O-Mania– 5
(e) 5.5 + 0.55 – 0.077 – 0.0011
= (5.5 + 0.55) – (0.077 + 0.0011)
+
5.50
0.55
6.05
+
0.0770
0.0011
0.0781
–
6.0500
0.0781
5.9719
4. Anita had ` 31.45. She purchased a pencil box for ` 3.78. What amount is left
with her?
(Text Book Page No. 53)
Sol.
Total money Anita had
=
` 31.45
Pencil box was purchased by her for =
` 3.78
Amount left with her
=
` 31.45 – ` 3.78
–
31.45
3.78
`
27.67
Ans.
5. Age of Seema is 7.5 years. She will be in class VI after 4.75 years. What will be her
age in class VI?
(Text Book Page No. 53)
Sol.
Age of Seema at present
=
7.5 years
She will be in class VI after 4.75 years
\ Seema‘s age in class VI
=
7.5 + 4.75
+
7.50
4.75
12.25
years
Ans.
6. The sum of two numbers is 2.564. If one of the number is 1.628, find the other.
(Text Book Page No. 53)
Sol.
The other number
=
2.564 – 1.628
–
2.564
1.628
0.936
Runway
80
Ans.
Math-O-Mania– 5
7. Neeta walked 8.3 km on Sunday, 8.9 km on Monday and 4.68 km on Tuesday.
What total distance is walked by her in these 3 days?
(Text Book Page No. 53)
Sol.
Distance travelled by Neeta in Sunday
=
8.3 km
Distance travelled by Neeta on Monday
=
8.9 km
Distance travelled by Neeta on Tuesday
=
4.68 km
Total distance walked by her in 3 days
8.30
8.90
4.68
+
km.
21.88
Ans.
8. Ajay gets ` 410.50 as his salary where as Vijay gets ` 316.25 only. By what
amount is Vijay's salary less than Ajay's?
(Text Book Page No. 53)
Sol.
Salary of Ajay
=
` 410.50
Salary of Vijay
=
` 316.25
\ Vijay‘s salary is less than Ajay‘s by
410.50
– 316.25
`
94.25
Ans.
9. A pair of leather shoes costs ` 925.75 and a pair of socks costs ` 110.75. What
total amount of money will you need to buy both?
(Text Book Page No. 53)
Sol.
Cost of pair of leather shoes =
` 925.75
Cost of a pair of socks
` 110.75
=
Total money needed to buy both
+
`
925.75
110.75
1036.50
Ans.
10. Kamal gets ` 576.75 from his mother as pocket money. His mother gets ` 1565.90
from Kamal's father. What amount of money it left with mother?
(Text Book Page No. 59)
Sol.
Amount of money Kamal‘s mother gets from his father =
` 1565.90
Amount of money Kamal gets from his mother
` 576.75
=
Amount of money left with her
1565.90
– 576.75
`
Runway
989.15
81
Ans.
Math-O-Mania– 5
Exercise 8.4
1. Without actual multiplication, find the value of :
Ans. (a) 7.32 × 10
=
73.2
(b) .698 × 100
=
69.8
(c) 5.324 × 100
=
532.4
(d) 8.3968 ×1000
=
8396.8
(e) 5.314 × 10
=
53.14
(f)
=
3.269
0.3269 × 10
2. Multiply the following :
Sol.
(Text Book Page No. 54)
(Text Book Page No. 54)
(a) 13.728 × 8.69
(b) 0.033 × 0.11
13.728
× 8.69
123552
82368×
109824××
119.29632
0.033
× 0.11
0033
0033×
0000××
0.00363
(c) 3.768 × 3.13
3.768
× 3.13
11304
3768×
11304××
11.79384
(d) 5.83 × 32.7
5.83
× 32.7
4081
1166×
1749××
190.641
(e) 0.225 × 20.5
(f)
0.225
× 20.5
1125
0 000×
04 50××
04.61 2 5
0.627 × 7.2
0.627
× 7.2
1254
4389×
4.5144
3. A vegetable seller sold 18 kg vegetables at the rate of ` 7.5 per kg. Calculate how
much money he gets ?
(Text Book Page No. 54)
Sol.
Price of 1 kg vegetables
=
` 7.5
\ Price of 18 kg vegetables
=
` 7.5 × 18
7.5
× 18
600
75×
135.0
\ The vegetable seller gets ` 135.
Runway
Ans.
82
Math-O-Mania– 5
4. The bus fare for one person from Meerut to Delhi is ` 28.65. What would be the
fare for 13 passengers ?
(Text Book Page No. 54)
Sol.
Fare for one person from Meerut to Delhi =
` 28.65
Fare for 13 person from Meerut to Delhi
` 28.65 × 13
=
28.65
× 13
8595
2865×
372.45
Thus, the fare for 13 passengers would be ` 372.45.
Ans.
5. The charges of one registered letter is ` 2.75. What will be the charges of 9 such
letters ?
(Text Book Page No. 54)
Sol.
Charge of 1 registered letter =
` 2.75
Charge of 9 such letters
` 2.75 × 9
=
2.75
×9
24.75
`
Ans.
6. One litre of kerosene costs ` 35.25. What will be the cost of 100 l of kerosene ?
(Text Book Page No. 54)
Sol.
Cost of 1 l kerosene
=
` 35.25
\ Cost of 100 l kerosene
=
` 35.25 × 100
`
35.25
× 100
0000
0000×
3525××
3525.00
Ans.
7. 2.25 m of cloth is needed to stitch a shirt. If 25 shirts are to be stitched, how much
cloth is needed ?
(Text Book Page No. 54)
Sol.
To stitch 1 shirt 2.25 m of cloth is needed.
\ To stitch 25 shirts quantity of cloth needed =
2.25
× 25
1125
450×
56.25
Runway
83
25 × 2.25 m
m
Ans.
Math-O-Mania– 5
Exercise 8.5
1. Divide the following :
Sol.
(Text Book Page No. 55)
(a) 2.24 ÷ 1.6
(b) 4.744 ÷ 0.16
Multiplying both the divisor and
Multiplying both the divisor and
the divided by 100 we get
the divided by 1000 we get
Þ
2.24
2.24 × 100
=
1.6
1.6 × 100
=
Þ 4.744 = 4.744 × 1000
0.16
0.16 × 1000
224
160
=
1.4
160 224
– 160
640
– 640
×
Hence, the quotient is 1.4.
4744
160
29.65
160 4744
– 320
1544
– 1440
1040
– 960
800
– 800
×
Ans.
Hence, the quotient is 29.65
(c) 0.408 ÷ 0.17
(d) 0.985 ÷ 8
Multiplying both the divisor and
Multiplying both the divisor and
the divided by 1000 we get
the divided by 1000 we get
Þ
0.408
0.408 × 1000
=
0.17
0.17 × 1000
=
Ans.
Þ
408
170
985
8000
0.123
8000 9850
– 8000
18500
– 16000
25000
– 24000
1000
=
2.4
170 408
– 340
680
– 680
×
Hence, the quotient is 2.4.
0.985
0.985 × 1000
=
8
8 × 1000
Ans.
Hence, the quotient is 0.123 (approx.) Ans.
Runway
84
Math-O-Mania– 5
(e) 580.47 ÷ 4
145.11
4 580.47
–4
18
– 16
20
– 20
04
–4
07
–4
3
Hence, the quotient is 145.11 (approx.)
(f)
21.02
21 441.42
– 42
21
– 21
042
42
×
Hence, the quotient is 21.02.
(h) 625.75 ÷ 0.25
0.0163
32 0.522
– 32
202
– 192
100
– 96
4
Multiplying both the divisor and
the divided by 1000 we get
Þ
Ans.
625.75 = 625.75 × 100 62575
=
0.25 × 100
0.25
25
2503
25 62575
– 50
125
– 125
075
– 75
×
Hence, the quotient is 2503.
(i)
Ans.
9.92 ÷ 21
0.4723
21 9.92
– 84
152
– 147
50
– 42
80
– 63
17
Hence, the quotient is 0.4723 (approx.)
Runway
Ans.
Ans.
(g) 0.522 ÷ 32
Hence, the quotient is 0.0163 (approx.)
441.42 ÷ 21
Ans.
85
Math-O-Mania– 5
2. Without actual division, find the quotients :
Ans. (a) 7.25 ÷ 10
(Text Book Page No. 55)
(b) 28.97 ÷ 100
(c) 2.395 ÷ 100
0.2897
0.02395
0.725
(d) 1.732 ÷ 1000
(e) 0.4744 ÷ 100
0.001732
0.004744
(f)
283.2 ÷ 1000
0.2832
(g) 2.4 ÷ 100
(h) 2438.14 ÷ 10
0.024
243.814
(i)
0.3 ÷ 10
0.03
Mental Maths (Text Book : Page No 56)
(Do yourself)
Ans :
Fun Activity (Text Book : Page No 56)
(Do yourself)
Ans :
9.
The Unitary Method
Exercise 9
1. The height of 19 books is 57.57 inches. Find the height of one book.
(Text Book Page No. 58)
Sol.
Height of 19 books
=
57.57 inches
\ Height of 1 book
=
57.57
19
=
3.03 inches
Ans.
2. Rajat paid ` 7.50 for 5 inland letters. Find the cost of one inland letter.
(Text Book Page No. 58)
Sol.
Cost of 5 inland letters
=
` 7.50
\ Cost of 1 inland letter
=
7.50
5
=
` 1.50
Ans.
3. There were 7500 students in 15 classes. If all the classes have same strength, what
is the strength of each class ?
(Text Book Page No. 58)
Sol.
Strength of 15 classes
=
7500 students
Each class has same strength
\ Strength of each class
=
7500
15
=
500 students
Ans.
4. 132 steel bars were used for 11 windows. How many bars are required for one
window ?
(Text Book Page No. 58)
Runway
86
Math-O-Mania– 5
Sol.
For 11 windows no. of steel Bars used =
For 1 widows no. of steel Bars used
=
132
132
11
=
12 bars
Ans.
5. 25 bottles of cold drinks cost ` 212.00. What is the cost of one bottle ?
(Text Book Page No. 58)
Sol.
Cost of 25 bottle of cold drinks
=
` 212
\ Cost of 1 bottle of cold drink
=
212
25
=
` 8.48
Ans.
6. The cost of 14 metres curtain cloth is ` 962.50. Find the cost of one metre cloth.
(Text Book Page No. 58)
Sol.
Cost of 14 m curtain cloth
=
` 962.50
\ Cost of 1 m curtain cloth
=
962.5
14
=
` 68.75
Ans.
7. 19 badminton rackets were bought for ` 2375. Find out the cost of one racket.
(Text Book Page No. 58)
Sol.
Cost of 19 badminton rackets
=
` 2375
\ Cost of 1 badminton racket
=
2375
19
=
` 125
8. If the cost of 12 pens is ` 67.20, find the cost of 26 pens.
Sol.
Cost of 12 pens
=
` 67.20
\ Cost of 1 pen
=
67.2
12
=
67.2
× 26
12
Cost of 26 pens
=
Ans.
(Text Book Page No. 58)
` 145.6
Ans.
9. 24 rubbers were bought for ` 84. What would be the cost of 15 rubbers ?
(Text Book Page No. 58)
Sol.
Cost of 24 rubbers
=
\ Cost of 1 rubber
=
Cost of 15 rubbers
` 84
84
24
84
24
=
× 15
=
` 52.50
Ans.
10. To store 138 l milk, we require 6 cans. How many cans are required to store 253 l
milk ?
Sol.
(Text Book Page No. 58)
To store 138 l milk, no. of cans required=
To store 1 l milk, no. of cans required
Runway
=
6
6
138
87
Math-O-Mania– 5
To store 253 l milk, no. of cans required =
6 × 253
138
=
11 cans
Ans.
11. Uniform of 18 students cost ` 4059. What would be the cost of uniform for 23
students ?
(Text Book Page No. 58)
Sol.
Cost of uniform for 18 students
=
` 4059
\ Cost of uniform of 1 student
=
4059
18
4059 × 23
18
Cost of uniform for 23 students
=
=
` 5186.50
Ans.
12. An aeroplane flies 480 km in one hour. How far does it fly in 30 minutes ?
(Text Book Page No. 58)
Sol.
1 Hour =
60 minutes
In 60 minutes the aeroplane flies
=
In 30 minutes the aeroplane flies
=
480 km
480 × 30
60
=
240 km
Ans.
13. 12 dozen of mangoes cost ` 124.80. Find the cost of 120 mangoes.
(Text Book Page No. 58)
Sol.
1 dozen mangoes
=
12 mangoes
\ 12 dozen mangoes
=
12 × 12
=
` 124.80
Cost of 144 mangoes
\ Cost of 1 mango
Cost of 120 mangoes
=
144 mangoes
=
` 104
= 124.80
144
= 124.80 × 120
144
Ans.
14. A bus travels 18 km in 24 minutes. How long would it go in 32 minutes ?
(Text Book Page No. 58)
Sol.
In 24 minutes the bus travels
=
18 km
In 1 minute the bus travels
=
18 km
24
\ In 32 minutes the bus travels
=
18 × 32
24
=
24 km
Ans.
15. 270 buns were prepared with 48.6 kg of flour. How much flour is required to
prepare 378 buns ?
(Text Book Page No. 58)
Sol.
270 buns have been prepared with
=
48.6 kg flour
1 bun can be prepared by
=
48.6 kg flour
270
48.6 × 378
270
\ 378 buns can be prepared by flour =
Runway
88
=
68.04 kg
Ans.
Math-O-Mania– 5
16. Rashmi painted 90 doors with 14.4 l of paint. How many l of paint is needed to
paint 132 doors ?
(Text Book Page No. 58)
Sol.
90 doors can be painted with paint
=
1 door can be painted with paint
=
\ 132 doors can be painted with paint
=
14.4 l
14.4
l
90
14.4
× 132
90
=
21.12 l
17. If 50 beads weigh 5 g, find the weight of 185 beads.
Sol.
Weight of 50 beads
=
Weight of 1 bead
=
\ Weight of 185 beads
=
Ans.
(Text Book Page No. 58)
5g
5 g
50
5
× 185
50
=
18.50 g
Ans.
18. If 8 men built a huge wall in 2 days, how many days are required to built the same
wall by 16 men ?
(Text Book Page No. 58)
Sol.
8 men build a wall in
=
2 days
1 man build a wall in
=
2 × 8 days
\ 16 men build a wall in
=
2×8
16
=
1 day
Ans.
19. A road was built by 45 workers in 90 days. In how many days could 60 workers
complete the same road ?
(Text Book Page No. 58)
Sol.
45 workers built a road
=
90 days
1 worker can build a road in
=
90 × 45 days
\ 60 worker could build that road in
=
90 × 45
60
=
67.5 days
Ans.
20. Food material is stored to feed 12 persons for 10 days. How long would it feed 15
persons?
(Text Book Page No. 58)
Sol.
Food can be fed to 12 persons for
=
10 days
Food can be fed to 1 person for
=
10 × 12 days
\ Food can be fed to 15 persons for
=
10 × 12
15
=
8 days
Ans.
Mental Maths (Text Book : Page No 59)
Ans : (Do yourself)
Try Me (Text Book : Page No 59)
Ans : (Do yourself)
Runway
89
Math-O-Mania– 5
10.
Simplification
Exercise 10
Simplify the following :
1.
Sol.
8×4–5+3
=
32 – 5 + 3
=
35 – 5 = 30
4.
5
Sol.
=
(Text Book Page No. 62)
2.
Sol.
30 ÷ 6 + 10 – 2 × 5
=
5 + 10 – 10
=
=
15 + (3 – 2)
15 – 10
=
15 + 1
5
=
16
5.
Sol.
51
3
7
5
+
÷
–
10
5
2
2
7 × 2 – 3
5
2
5
7
3
–
5
5
6.
Sol.
= 51 + 14 – 3
10
5
= 65 – 3
10
5
= 65 – 6
10
= 59
=5 9
10
10
7.
Sol.
9.
Sol.
Sol.
7.12 ÷ 0.02 × 0.04 – 2.24
=
356 × 0.04 – 2.24
=
14.240 – 2.24
=
12
6 + [6 + {6 + 6(6 + 6 + 6)}]
=
6 + [6 + { 6 + 6 (18)}]
=
6 + [ 6 + {6 + 108}]
=
6 + [6 + 114]
=
6 + 120
=
126
10 – 2 + 2 × 6
=
10 – 2 + 12
=
22 – 2
=
20
10 × 10 + [400 ÷ {100 – (50 – 3 × 10)}]
=
10 × 10 + [400 ÷ {100 – (50 – 30)}]
=
10 × 10 + [400 ÷ {100 – 20}]
=
10 × 10 + 5
=
100 + 5
=
105
Runway
90
5
3
1
1
1
+
÷
–
of 1
8
4
2
2
2
8.
1
Sol.
=
3
5
3
1
3
+
÷
–
×
2
8
4
2
2
=
3
5
4
3
+
×
–
2
8
3
4
=
3
5
3
+
–
2
6
4
=
9+5
–
6
=
14
3
–
6
4
=
28 – 9
12
=
19
12
10 – 8 ÷ 4 + 2 × 6
=
15 + (3 – 5 – 3)
=
1
1
1
3
+3
÷2 –
10
2
2
5
= 51 +
10
51
=
+
10
3.
3
4
= 1 7
12
Math-O-Mania– 5
10.
Sol.
81 + [159 – 2 { 7 × 8 + (13 – 2 × 5)}]
=
81 + [159 – 2 { 7 × 8 + 13 – 10}]
=
[{25 – (18 – 4)} of 10] – 80
81 + [159 – 2 {69 – 10}]
=
[{25 – 14} of 10 ] – 80
=
81 + [159 – 2 {59}]
=
[11 × 10] – 80
=
81 + [159 – 118]
=
110 – 80
=
81 + 41
=
30
=
122
Sol.
=
7
6
–
15
8
=
7
6
1
–
×
15
8
3
=
7
1
–
15
4
=
28 – 15
60
=
13
60
5
1
+ 1
8
4
15.
=
=
=
=
Sol.
18.
Sol.
Sol.
13.
12.
17.
[{25 – (18 – 4)} of 10] – 80
=
7
7
1
–
–
15
8
8
Sol.
11.
1
of
3
Sol.
1
of
3
14.
Sol.
of
5
5
+
8
4
1
2
×
1
2
Sol.
2.4 – 0.28
=
2.12
8.3 + 3.36 – 0.2
=
11.66 – 0.2
=
11.46
Runway
5.9 – 3
=
2.9
[{63 – (19 + 7)} ÷ 5] + 5
=
[{63 – 26} ÷ 5] + 5
=
[37 ÷ 5] + 5
=
37
+ 5
5
=
37 + 25
5
=
62
5
=
2
5
3
5
4
+
–
8
8
25
125 + 32 – 75
200
82
200
41
100
9
–
5
19.
Sol.
= 12
1
3
5
4
+
×
–
5
8
8
5
=
8.3 + 8.4 × 0.4 – 0.2
=
=
=
2.4 – (1.3 – 0.6) of .4
=
5.4 – 3 + 0.05
=
15
16
2.4 – (0.7) × .4
=
16.
5 + 10
1
×
8
2
15
1
×
8
2
=
5.4 – 0.6 ÷ 0.2 + 0.5
=
1
1
+
5
8
9
–
5
8+5
40
= 9 – 13
5
40
= 72 – 13
40
91
= 59
40
Math-O-Mania– 5
20.
Sol.
4.2 + {(5.6 – 2.4) ÷ 1.6} × 2.5 – 1.5
=
4.2 + {3.2 ÷ 1.6} × 2.5 – 1.5
=
4.2 + 2 × 2.5 – 1.5
=
4.2 + 5.0 – 1.5
=
9.2 – 1.5
=
7.7
Mental Maths (Text Book : Page No 63)
Ans : (Do yourself)
Fun Activity (Text Book : Page No 63)
Ans : (Do yourself)
11.
Shopping Bills
Exercise 11
1. Mohan bought 12 kg of rice at the rate of ` 15 per kg, 2 kg of sugar at the rate of
` 14 per kg and 1 kg flour at the rate of ` 9.50 per kg from Sudha Provision Store,
New Nehru Place Ghaziabad. Prepare a bill for him.
(Text Book Page No. 66)
Sol.
SUDHA PROVISION STORE
New Nehru Place, Ghaziabad
Bill No. 0999
Date 01 / 07 / 2011
To,
S. No.
Mr. Mohan
Particulars
1.
Rice
2.
3.
Qty.
Rate
Amount
`
P.
12 kg
15.00
180 00
Sugar
2 kg
14.00
28 00
Flour
1 kg
9.50
9 50
Total
217 50
For SUDHA PROVISION STORE
(Auth. Signatory)
Ans.
2. Mamta bought the following items from “Manglam Fruit shop, Dehradun”. Make
a bill for her.
(Text Book Page No. 66)
(a) 10 oranges at the rate of `1.30 each.
(b) 2 dozen bananas at the rate of `15 per dozen.
Runway
92
Math-O-Mania– 5
(c) 5 kg of mangoes at the rate of ` 23 per kg.
Sol.
MANGLAM FRUIT SHOP
Dehradun
Bill No. 0123
Date 01 / 07 / 2011
To,
Mamta
S. No.
Particulars
Per
Amount
`
P.
10 pc.
1.30 pc.
13 00
Qty.
Rate
1.
Orange
2.
Bananas
2 dzn
15.00 dzn
30 00
3.
Mangoes
5 kg
23.00
115 00
kg
Total
158 00
For MANGLAM FRUIT SHOP
(Auth. Signatory)
Ans.
3. Richa bought the following items for domestic use from “Chandran Mart, Delhi”
Make a bill for her.
(Text Book Page No. 67)
(a) 6 toilet soaps at the rate of ` 116 per dozen.
(b) 1 toothpaste at the rate of ` 38.50 each.
(c) 6 detergent soaps at the rate of ` 8 each.
(d) 2 packs of cleaning powder, at the rate of ` 21 each.
Sol.
CHANDRAN MART
Delhi
Bill No. 0445
Date 01 / 07 / 2011
To,
S. No.
1.
2.
3.
4.
Richa
Particulars
Soaps
Toothpaste
Detergent Soap
Cleaning Powder
Total
Qty.
Rate
Per
6
1
6
2
116.00
38.50
8.00
21.00
dzn
pc
pc
pc
pcs.
pc.
pcs.
pcs.
Amount
`
58
38
48
42
P.
00
50
00
00
186 50
For CHANDRAN MART
(Auth. Signatory)
Runway
93
Math-O-Mania– 5
4. Mr Sohan arranged a picnic for his class. He purchased following items from
“Kapoor Bakers, Kanpur”. Make a bill for him.
(Text Book Page No. 67)
(a) 60 packets of biscuits at the rate of ` 5 each.
(b) 60 packets of popcorn at the rate of ` 3.20 each.
(c) 15 packets fruit cake at the rate of ` 15 per packet.
Sol.
KAPOOR BAKERS
Kanpur
Bill No. 0889
Date 01 / 07 / 2011
To,
Sohan
S. No.
1.
2.
3.
Particulars
Biscuits
Popcorn
Fruit Cakes
Qty.
Rate
Per
60 pcs.
60 pcs.
15 pcs.
5.00 pc.
3.20 pc.
15.00 pc
Amount
`
300
192
225
Total
P.
00
00
00
717 00
For KAPOOR BAKERS
(Auth. Signatory)
Ans.
5.
Answer the following questions on the basis of bill :
(Text Book Page No. 67)
VENUS MATCHING STORE
No.-99
Quantity
2
2
3
4
1
Tel : 406445
100, N.D. Market, Meerut
To, Mrs Reeta Tyagi, Patel Nagar, Muzaffarnagar.
Particulars
Date : 15- 07- 10
Rate
Sarees
Blouse Piece (80 cm)
Saree Falls
Suits
Gown Length of 3 m
` 350 each
` 50 each
` 12 each
` 250 each
` 36 Per m
Total
`
Amount
P
700
100
36
1000
108
00
00
00
00
00
1944
00
Signature
Sol.
(a) How many sarees are purchased ?
2
(b) What is the address and phone number of the shop ?
Venus Matching Store, 100, N.D. Market, Meerut, Phone no. 406445
Runway
94
Math-O-Mania– 5
(c) Write the date of purchase.
15-07-2010
` 12
(d) What is the cost of 1 saree fall ?
` 1944
(e) What is the total amount of the bill ?
6. Which of the following statements are true ?
Sol.
(Text Book Page No. 67)
(a) A bill is a useless document for the customer.
False
(b) Shopkeeper prepares a bill.
True
(c) All types of bills look alike.
True
(d) Total cost of various items included in the bill can be found out from the bill.
True
(e) Name of the store from which things are bought does not appear on the bill.
False
Mental Maths (Text Book : Page No 68)
(Do yourself)
Ans :
Fun Activity (Text Book : Page No 68)
(Do yourself)
Ans :
12.
Percentage
Exercise 12.1
1. Complete the following :
1
= 50%
2
5
(e)
= 20%
25
Ans. (a)
(b)
(f)
(Text Book Page No. 71)
1
= 25%
4
15
= 93.75%
16
1
= 12.5%
8
7
(g)
= 58.33%
12
2
= 66.6%
3
19
(h)
= 95%
20
(c)
(d)
2. Change into per cent :
Sol.
(a) 4
1
2
=
9
2
(b) 4
3
7
31
7
=
(c) 3
1
4
13
4
=
=
9 × 100
2 × 100
=
31 × 100
7 × 100
=
13 × 100
4 × 100
=
50
9
1
× 100 ×
21
100
=
31
1
× 100 ×
7
100
=
13
1
× 100 ×
41
100
= 450%
(d) 4
=
Runway
(Text Book Page No. 71)
5
8
25
= 442.85%
=
37
8
37 × 100
= 462.5%
8 × 100
(e) 10
=
15
16
=
175
16
175 × 100
= 1093.75%
16 × 100
95
= 325%
(f)
2
=
9
20
=
49
20
49 × 100
= 245%
20 × 100
Math-O-Mania– 5
(g) 14
3
5
=
73
5
11
25
(h) 12
=
311
25
73 × 100
= 5 × 100
311 × 100
= 25 × 100
= 1460%
= 1244%
3. Change into fractions :
Sol.
(a) 6%
(b) 20%
3
6
100 50
= 3
50
=
(d) 95%
19
= 95
100 20
= 9
20
(g) 63
(Text Book Page No. 71)
3
%
4
= 255 %
4
255
=
÷ 100
4
255
1
×
4
100
51
255
=
=
400
80
=
(c) 75%
20 1
=
3
= 75
100 4
= 3
4
100 5
= 1
5
(e) 0.25%
(f)
91
= 7.28 = 728
100
10000 1250
91
=
1250
1
= 0.25 = 25
100
10000 400
1
=
400
(h) 43
7.28%
2
%
3
= 131 %
3
131
=
÷ 100
3
=
131
1
×
3
100
=
131
300
4. Change into decimals :
Sol.
(a) 2.5%
=
2.5
25
=
100
1000
= 0.025
(d) 72.2%
=
72.2
722
=
100
1000
= 0.722
Runway
(Text Book Page No. 71)
(b) 8.5%
=
(c) 24.4%
8.5
85
=
100
1000
=
= 0.085
(e) 6
2
%
3
=
24.4
244
=
100
1000
= 0.244
20
%
3
(f)
37
1
75
% =
%
2
2
=
20
1
×
3
100
=
75
1
×
2
100
=
2
30
=
3
8
96
= 0.066
= 0.375
Math-O-Mania– 5
(g) 0.785%
=
3
588
% =
%
5
5
(h) 117
0.785
7.85
=
100
1000
= 0.00785
=
588
1
×
5
100
=
588
500
= 1.176
5. Change into per cent :
Sol.
(a) 0.4
=
(Text Book Page No. 71)
(b) 0.5
40
100
= 40 ×
1
100
= 50 ×
25
100
1
100
30
100
= 30 ×
1
100
= 30%
(f)
75
100
=
(g) 0.08
0.18
=
18
100
1
= 75 × 100
= 18 ×
= 75%
= 18%
1
100
(h) 0.0075
8
100
= 8 ×
1
100
(e) 0.75
= 25%
=
=
= 50%
(d) 0.25
= 25 ×
50
100
=
= 40%
=
(c) 0.3
=
1
100
0.75
100
= 0.75 ×
= 8%
1
100
= 0.75%
Exercise 12.2
1. Find the value of the following :
Sol.
(a) 20% of 80
Runway
(Text Book Page No. 72)
(b) 5% of 40
= 20 × 80
100
=
5
× 40
100
= 16
= 2
(c) 90% of 90
= 90 × 90
100
= 81
97
Math-O-Mania– 5
(d) 2
=
=
=
1
% of 5 km
2
(e) 37
1
% of 36 kg
2
5
% of 5 km
2
5
1
×
× 5
2
100
75
% of 36 kg
2
75
1
×
× 36
=
2
100
25
200
=
=
1
8
(f)
=
3
× 36
8
= 3 × 4.5 kg
= 0.125 × 1000
= 13.5 kg
1
% of 45 cm
3
40
% of 45 cm
3
40
1
×
× 5
=
3
100
=
= 0.125 km
13
=
2
× 45
15
= 6 cm
= 125 m
(g) 6
1
% of 6500 mm
5
31
% of 6500 mm
5
31
1
×
× 6500
=
5
100
=
=
31
× 65
5
= 31 × 13 mm
(h) 5
1
% of 7500 mm
3
(i)
48
1
% of 800 m
2
16
% of 7500 mm
3
16
1
×
× 7500
=
3
100
97
% of 800 m
2
97
1
×
× 800
=
2
100
= 16 × 25
= 97 × 4
= 400 mm
= 388 m
=
=
= 403 mm
2. Solve the following :
Sol.
(Text Book Page No. 72)
(a) What percentage of 72 is 8 ?
=
8
× 100
72
= 800
72
= 11.1%
Ans.
(b) What percentage of 200 is 20 ?
20
× 100
200
2000
=
200
=
= 10%
Runway
Ans.
98
Math-O-Mania– 5
(c) What percentage of 500 g is 1 kg ?
=
1000
× 100
500
=
100000
500
(
= 200%
1 kg = 1000 gm)
Ans.
(d) What percentage of ` 650 is ` 130 ?
=
130
× 100
650
=
13000
650
= 20%
Ans.
(e) What percentage of 525 km is 25 km ?
=
25
× 100
525
=
2500
525
= 4.76%
Ans.
3. Find the number whose :
Sol.
(Text Book Page No. 72)
(a) 8% is 16
Let the number be, x.
Given 8% of x
=
16
=
16
=
16 × 100
8
=
200
=
15
=
15
x
=
15 × 100
5
x
=
300
8
× x
100
x
Ans.
(b) 5% is 15
Let the number be, x.
Given 5% of x
5
× x
100
Runway
Ans.
99
Math-O-Mania– 5
(c) 12% is 3
Let the number be, x.
Given 12% of x
=
3
12
× x
100
x
=
3
=
x
=
3 × 100
12
25
=
10
10
× x
100
x
=
10
=
x
=
10 × 100
10
100
=
45
=
45
=
=
45 × 100
30
150
=
5
=
5
=
5 × 100
15
100
3
Ans.
(d) 10% is 10
Let the number be, x.
Given 10% of x
Ans.
(e) 30% is 45
Let the number be, x.
Given 30% of x
30
× x
100
x
(f)
Ans.
15% is 5
Let the number be, x.
Given 15% of x
15
× x
100
x
=
=
33 1
3
Ans.
4. The cost of furniture and curtains is ` 15,000. If the cost of curtains is 75%, find
the cost of curtains.
(Text Book Page No. 72)
Runway
100
Math-O-Mania– 5
Sol.
Total cost of furniture and curtains
=
` 15,000
Cost of curtain
=
75% of total cost
=
75 × 15000
100
=
` 11250
Ans.
5. Rajan’s father earns ` 3,600 per month. He spends its 12% for house rent. How
much does he spends for house rent?
(Text Book Page No. 72)
Sol.
Salary of Rajan‘s father
=
Percentage of salary spent for house rent =
` 3600
12%
Amount that Rajan‘s father spends as house rent
=
12
× 3600
100
=
` 432
Ans.
6. There are 3575 students in our school. 56% of them are boys. Find the number of
boys in the school.
(Text Book Page No. 72)
Sol.
Total no. of students
=
3575
56% of students
=
56% of 3575
=
56
× 3575
100
=
2002 boys
Ans.
7. In half-yearly exams, Ashok secured 86% marks out of 300 marks. Find out the
marks obtained by him.
(Text Book Page No. 72)
Sol.
86% of 300
=
86
× 300
100
=
258
Thus, marks obtained by Ashok in half-yearly exams = 258
8.
Sol.
1
5
part of a glass is empty. What percentage is filled?
Part of the glass that is empty
=
1 = 1 × 100
5
5 × 100
=
1 × 100 × 1
100
5
1
20 ×
=
100
=
Runway
Ans.
101
(Text Book Page No. 72)
20%
Ans.
Math-O-Mania– 5
9. In a jungle, there are 750 animals. 150 animals of them are herbivores. Find the
percentage of herbivores animals.
(Text Book Page No. 72)
Sol.
Total no. of animals
=
750
No. of animals that are herbivores
=
\ Percentage of herbivores animals
=
150
150 × 100
750
=
20%
Ans.
10. Ruchi has ` 1500 with her. She spend ` 555 for a saree. What percentage of
money she still has ?
(Text Book Page No. 72)
Sol.
Total amount of money that Ruchi has
=
` 1500
Money spent by her for a saree
=
` 555
\ Money still left with her
=
` 1500 – ` 555
=
` 945
=
945
× 100
1500
=
63%
Percentage of money she still has
Ans.
Mental Maths (Text Book : Page No 73)
(Do yourself)
Ans :
Fun Activity (Text Book : Page No 73)
(Do yourself)
Ans :
13.
Average
Exercise 13
1. Find the average of following groups of numbers :
Sol.
(Text Book Page No. 75)
(a) 15, 16, 18, 17, 22 and 20
15 + 16 + 18 + 17 + 22 + 20
6
108
=
=
6
=
18
Ans.
3.44
Ans.
(b) 2.5, 3.0, 4.5, 3.8, 3.4
2.5 + 3.0 + 4.5 + 3.8 + 3.4
5
17.2
=
=
5
=
Runway
102
Math-O-Mania– 5
(c) 7 ,
15 17
1
,
and 6
2
2
2
= The numbers are 7,
15 17 13
,
,
2
2
2
= Adding up these numbers and then dividing by 4.
7+
=
15
17 13
+
+
2
2
2
4
=
14 + 15 + 17 + 13
1
×
4
2
=
59
8
=
7.375
Ans.
(d) 320, 230, 540, 400, 325.50, 415.50, 385.00
320 + 230 + 540 + 400 + 325.5 + 415.5 + 385.00
7
2616
=
= 373.71
7
=
Ans.
(e) 9.5, 8.5, 10.75, 7.95, 6.75, 11.35, 12
9.5 + 8.5 + 10.75 + 7.95 + 6.75 + 11.35 + 12
7
66.8
=
= 9.543
7
=
Ans.
2. Temperature recordings in a city during a week are as follows: 32.5ºC on
Monday, 36ºC on Tuesday, 36.5ºC on Wednesday, 37ºC on Thursday, 35.5ºC on
Friday, 35.5ºC on Saturday and 34ºC on Sunday. Find the average daily
temperature.
(Text Book Page No. 75)
Sol.
Sum of daily temperature recordings
=
(32.5 + 36 + 36.5 + 37 + 35.5 + 35.5 + 34) °C
=
247°C
No. of days
=
7
Average daily temperature
=
247
°C
7
=
35.28 °C
Ans.
3. Sheela purchased 5 greeting cards for new year, costing ` 10.50, ` 6.50, ` 30, `
8.25 and ` 9.20. Find the average cost of each card.
(Text Book Page No. 75)
Sol.
Sum of costs of greetings cards
Runway
=
` (10.50 + 6.50 + 30 + 8.25 + 9.20 )
=
` 64.45
103
Math-O-Mania– 5
No. of cards
=
5
Average cost
=
`
=
` 12.89
64.45
5
Ans.
4. A Rickshaw puller collected ` 162, ` 264, ` 185, ` 295, ` 173 and ` 350 in 6 days.
Find his average daily income.
(Text Book Page No. 75)
Sol.
=
` (162 + 264 + 185 + 295 + 173 + 350)
=
` 1429
No. of days
=
6
Average daily income
=
`
=
` 238.16
Sum of daily incomes
1429
6
Ans.
5. A businessman earns in five consecutive months ` 14,056, ` 20,500, ` 12,600,
(Text Book Page No. 75)
` 18,505 and ` 15,600. Find his average monthly income.
Sol.
Total earning of the businessman =
` (14,056 + 20,500 + 12,600 + 18,505 + 15,600)
=
` 81,261
No. of days
=
5
Average monthly income
=
`
=
` 16,252.20
81261
5
Ans.
6. The two families consumed wheat in 4 weeks as follows :
WEEK
FAMILY (A)
FAMILY (B)
1st week
3.2 kg
2.9 kg
2nd week
2.8 kg
3.2 kg
3rd week
4.2 kg
4.4 kg
4th week
3.6 kg
3.5 kg
Which family requires more ?
Sol.
(Text Book Page No. 76)
Total consumption of wheat by family (A) during 4 weeks
=
(3.2 + 2.8 + 4.2 + 3.6) kg =
13.8 kg
Total consumption of wheat by family (B) during 4 weeks
=
Runway
(2.9 + 3.2 + 4.4 + 3.5) kg =
104
14 kg
Math-O-Mania– 5
As the consumption of wheat by family (B) during 4 weeks is more therefore the requirement of
family (A) is more.
Ans.
7. The average of 5 numbers is 50. 4 numbers are 53, 48, 64 and 44. Find the fifth
number.
(Text Book Page No. 76)
Sol.
We know average
=
Let the fifth number be x.
\ Average
=
Sum of total quantities
Total number of quantities
53 + 48 + 64 + 44 + x
5
209 + x
50
=
250
=
5
x + 209
x
=
250 – 209
=
41
Therefore the fifth number is 41.
Ans.
8. Average weight of 3 books is 500 gm. What is the total weight of 3 books ?
(Text Book Page No. 76)
Sol.
Average weight
=
500
Total weight
=
=
Total weight
[
3
No. of books = 3]
Total weight
3
1500 gm
Thus, the total weight of 3 books is 1500 gm.
Ans.
9. Find out the average of all even numbers between 9 and 25. (Text Book Page No. 76)
Sol.
The even numbers between 9 and 25 are 10, 12, 14, 16, 18, 20, 22, 24
Average
=
=
Sum of total quantities
Total number of quantities
10 + 12 + 14 + 16 + 18 + 20 + 22 + 24
8
=
136
8
=
17
Ans.
10. Find out the average of first 5 multiples of 3 greater then 6.
Runway
105
(Text Book Page No. 76)
Math-O-Mania– 5
Sol.
The first 5 multiples of 3 greater than 6 are 9, 12, 15, 18, 21
Their average
Sum of multiples
=
5
9 + 12 + 15 + 18 + 21
=
5
=
75
5
=
15
Ans.
11. Find out the average of first 10 prime numbers.
Sol.
(Text Book Page No. 76)
The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Their average
=
Average
=
Sum of prime nos.
No. of prime nos.
2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29
10
=
129
10
=
12.9
Ans.
12. The average of five numbers is 14. Four numbers are as 12, 13, 14 and 16. Find the
fifth number.
(Text Book Page No. 76)
Sol.
Average
=
14 =
Sum
Sum
.
No. of quantities
Sum
5
=
14 × 5
=
70
12 + 13 + 14 + 16 + x
=
70
x + 55
=
70
x
=
70 – 55
=
15
Let the fifth no be x.
Thus, the fifth no is 15.
Runway
Ans.
106
Math-O-Mania– 5
13. Data of rainfall in different cities is given below :
CITY
MONTHS
JUNE
JULY
AUGUST
SEPTEMBER
Kolkata
30.2 cm
31.7 cm
32.3 cm
28.3 cm
Srinagar
3.6 cm
5.9 cm
6.2 cm
6.3 cm
Ahmedabad
9.3 cm
31.1 cm
20.2 cm
18.2 cm
Madurai
2.8 cm
4.9 cm
10.7 cm
9.8 cm
Calculate the average rainfall in each city.
Sol.
Average rainfall for Kolkata
Average rainfall for Srinagar
=
=
Average rainfall for Ahmedabad =
=
Average rainfall for Madurai
30.2 + 31.7 + 32.3 + 28.3
=
=
=
=
(Text Book Page No. 76)
4
30.63 cm
3.6 + 5.9 + 6.2 + 6.3
4
5.5 cm
9.3 + 31.1 + 20.2 + 18.2
4
19.70 cm
2.8 + 4.9 + 10.7 + 9.8
4
7.05 cm
Ans.
14. Two factories A and B have 8 and 10 members in administrative staff. Their
salaries are given below in the table. Find, which of the two factories pay better
average salary ?
(Text Book Page No. 76)
FACTORY (A)
FACTORY (B)
Sol.
(i)
`
4500
(ii)
`
4000
(iii)
`
3800
(iv)
`
3100
(v)
`
5000
(vi)
`
5000
(vii)
`
4000
(viii)
`
3600
(ix)
(x)
`
6000
`
5300
`
3800
`
3600
`
3400
`
3300
`
3600
`
3000
`
5500
`
3900
Average salary for factory (A)
=
=
Average salary for factory (B)
=
=
Runway
4500 + 4000 + 3800 + 3100 + 5000 + 5000 + 4000 + 3600
8
33000
8
=
` 4125
6000+5300+3800+3600+3400+3300+3600+3000+5500+3900
10
41400
10
=
107
` 4140
Ans.
Math-O-Mania– 5
Mental Maths (Text Book : Page No 77)
(Do yourself)
Ans :
Puzzle (Text Book : Page No 77)
(Do yourself)
Ans :
Fun Activity (Text Book : Page No 77)
(Do yourself))
Ans :
14.
Profit and Loss
Exercise 14.1
1. Complete the following table : (Put ‘X’ in the invalid box)
Sol.
S.No.
C.P.
S.P.
PROFIT
(i)
` 620
` 580
X
` 40
(ii)
` 650
` 700
` 50
X
(iii)
` 100
` 88
X
` 12
(iv)
` 82
` 72
X
` 10
(v)
` 4896
` 4798
X
` 98
(Text Book Page No. 79)
LOSS
2. A shopkeeper purchased a shirt for ` 250 and sold it for ` 225. Find his profit or
(Text Book Page No. 79)
loss.
Sol.
=
` 250
The selling Price (S.P.) of the shirt =
` 225
The cost Price (C.P.) of the shirt
In above case C.P. is more tha S.P., therefore there is loss.
Loss
=
C.P. – S.P.
=
250 – 225
=
25
Hence, there is loss of ` 25.
Ans.
3. Chandan bought a land for ` 120000 and sold it on a loss of ` 4500. At what price
(Text Book Page No. 79)
the land was sold.
Sol.
\
Loss
=
C.P. – S.P. (as we know)
S.P.
=
C.P. – loss
S.P.
=
120000 – 4500
Thus, the land was sold for ` 115500.
Runway
=
` 115500
Ans.
108
Math-O-Mania– 5
4. A milkman had 10 l of milk. 2 l milk spilt up unfortunately. He sold remaining as
` 20 per l. If he suffered a loss of ` 50. What was the C.P. of 10 l of milk ?
(Text Book Page No. 79)
Sol.
Total quantity of milk that the milkman had =
10 l
Quantity of milk that split
=
2l
Remaining milk
=
10 – 2
=
8l
S.P.
=
20 × 8
=
` 160
Loss
=
` 50
G.P.
=
S.P. + Loss
=
160 + 50
=
` 210
The remaining milk was sold for ` 20 per l.
\
We know
Thus the C.P. of the 10 l of milk = ` 210
Ans.
5. A man bought 10 lampshades. He sold them at ` 75 each, thereby making a profit
of ` 70 in total. What was the C.P. of one lampshade ?
(Text Book Page No. 79)
Sol.
S.P. of 1 lampshade
=
` 75
\ S.P. of 10 lampshades
=
75 × 10
=
` 70
=
S.P. – Profit
=
750 – 70
C.P. of 10 lampshades
=
` 680
\ C.P. of 1 lampshade
=
680 ÷ 10
Total Profit
C.P.
=
` 750
=
` 680
=
` 68
Ans.
6. Shikha bought 3 dozen eggs for ` 72. 7 eggs were broken and 5 eggs were found
rotten. She sold the remaining eggs at ` 3 each. Find her profit or loss.
(Text Book Page No. 79)
Sol.
1 dozen
=
12
3 dozen
=
12 × 3
=
36
=
24
=
` 72
36 eggs were brought for ` 72.
7 eggs were broken and 5 eggs were found rotten.
\ Remaining eggs
=
36 – 7 – 5
Remaining eggs were sold for ` 3 each
S.P. of 1 egg
=
3
S.P. of 24 eggs
=
24 × 3
=
` 72
C.P.
Runway
109
Math-O-Mania– 5
S.P.
=
` 72
As C.P. and S.P. of the eggs are same.
\ There are no profit and loss.
Ans.
7. Sarika bought a saree for ` 375. She sold it to Anuja at a profit of ` 25. Anuja’s
friend Rekha purchased the saree and gave Anuja a profit of ` 10 only. What was
the C.P. of saree for Rekha.
(Text Book Page No. 79)
Sol.
C.P. of saree for sarika
=
` 375
C.P. of saree for Anuja
=
375 + 25
=
` 400
=
400 + 10
=
` 410
C.P. of saree for Rekha
[ Profit = 25]
[ Profit = 10]
Ans.
8. By selling an old car, a man loses ` 12500. How much did he pay for the car if it
was sold for ` 75200 and ` 1500 were spent on its repairs. (Text Book Page No. 79)
Sol.
S.P. of the car
=
` 75200
Loss
=
` 12500
C.P.
=
S.P. + Loss
=
75200 + 12500
=
` 87700
Money spent on the repairs after purchasing the car
\ Money paid by the man for the car
=
` 1500
=
87700 – 1500
=
` 86200
Ans.
9. Naveen bought 10 old calculators at ` 60 each. He spent ` 15 on the packing and
battery repairs of each. If he wants to make a profit of ` 100 on them, what should
be the selling price.
(Text Book Page No. 79)
Sol.
C.P. of 1 calculator
=
` 60
Money spent on 1 calculator for packing and battery repairs
\
C.P.
=
` 15
=
60 + 15
=
` 75
Naveen wants to make a profit of ` 100 on the sale of the 10 calculators
\ Profit that Naveen has to make as the sale of 1 calculator
S.P.
=
100 ÷ 10
=
Profit + C.P.
=
10 + 75
Thus, the selling price of 1 calculator ` 85.
=
` 85
Ans.
10. Harmeet Singh bought 15 cartons of unripe mangoes from an orchard at ` 250
Runway
110
Math-O-Mania– 5
each. 5 cartons were sold hand to hand by him on a profit of ` 25 on each. 7
cartons were sold by him on profit of ` 30 each. Rest 3 cartons of mangoes got
rotten and were thus thrown away. Find the gain or loss of Harmeet Singh in
rupees.
(Text Book Page No. 80)
Sol.
C.P. of 1 carton of unripe mangoes
=
250
250 × 15
=
` 3750
=
5 × 25
=
125 [ Profit = ` 25 per carton]
=
C.P. + Profit
=
(250 × 5) + 125
=
1250 + 125
=
` 1375
Profit for the next 7 cartons sold
=
7 × 30
=
210 [ Profit = Rs 30 per carton]
S.P. for the next 7 cartons
=
C.P. + Profit
=
(250 × 7) + 210 =
\ C.P. of 15 cartons of unripe mangoes =
Profit for the first 5 cartons sold hand to hand
\ S.P. for the first 5 carton
` 1960
The rest 3 cartons of mangoes got rotten and were thus thrown away.
\ S.P. of cartons sold
=
1375 + 1960
=
C.P . – S.P.
=
3750 – 3335
=
` 415
=
` 3335
As the G.P. is more than S.P. so there is Loss
Loss
Ans.
Exercise 14.2
1. Renu bought a carpet for ` 12000.00 and sold it for ` 12500.00. Find her profit
(Text Book Page No. 81)
per cent.
Sol.
\
C.P. of carpet
=
` 12000
S.P. of carpet
=
` 12500
Profit
=
12500 – 12000
=
500
Profit percentage
=
=
=
=
Runway
Profit
× 100
C.P.
500
× 100
12000
25
6
4.16%
111
Ans.
Math-O-Mania– 5
2.
Vinay bought a sofa for ` 2400 and sold it for ` 2800. Find his profit percent.
(Text Book Page No. 81)
Sol.
C.P. of sofa
=
` 2400
S.P. of sofa
=
` 2800
Profit
=
S.P. – C.P.
=
2800 – 2400
Profit Per cent
=
Profit
× 100
C.P.
400
× 100
2400
50
3
=
16.66%
=
=
=
400
Ans.
3. A shopkeeper bought tea at ` 120 per kg. He sold it at the cost of ` 135 per kg. Find
his profit or loss per cent.
(Text Book Page No. 81)
Sol.
C.P. of tea
=
` 120 / kg
S.P. of tea
=
` 135 / kg
Profit
=
S.P. – C.P.
=
135 – 120
Profit Per cent
=
Profit
× 100
C.P.
15
× 100
120
1500
120
=
12.5%
=
=
=
15
Ans.
4. A man bought 100 bananas for ` 75 and sold them at the cost of ` 0.60 each. Find
his profit or loss per cent.
(Text Book Page No. 81)
Sol.
C.P. of 100 bananas
=
` 75
S.P. of 100 bananas
=
` 0.60 each
S.P. of 100 bananas
=
0.60 × 100
=
C.P. – S.P.
=
75 – 60
=
` 60
=
15
As C.P. is more than S.P., \ There is loss
Loss
Loss per cent
Runway
=
Loss
× 100
C.P.
112
Math-O-Mania– 5
=
=
=
15
× 100
75
100
5
20%
Ans.
5. 10 l of milk was bought for ` 85. For what price should it be sold to gain 25% ?
(Text Book Page No. 81)
Gain %
Sol.
25
Gain
\
Profit
=
× 100
C.P.
Gain
=
× 100
85
25 × 85
=
=
100
S.P. – C.P.
=
Profit
S.P.
=
C.P. + Profit
=
85 + 21.2
S.P. of 10 l milk
=
S.P. of 1 l milk
=
106.2
106.2
10
` 10.62
=
=
` 21.2
106.2
So to gain 25% the milk should be sold at ` 10.62 per litre.
Ans.
6. 40 packets of namkeen were bought for ` 60. To gain 20% for how much should
they be sold ?
(Text Book Page No. 81)
Profit %
Sol.
20%
Profit
× 100
C.P.
Profit
=
× 100
60
=
20 × 60
100
=
Profit
Profit
=
` 12
S.P. – C.P.
=
Profit
S.P.
=
Profit + C.P.
=
12 + 60
=
72
C.P. of 40 packets
=
` 72
72
= ` 1.80
40
\ To gain 20% the namkeen packets must be sold for ` 1.80 per packet.
\
Runway
C.P. of 1 packet
=
113
Ans.
Math-O-Mania– 5
7. Firoz paid ` 472 for a toy car. He spent ` 28 on its transportation. If it is sold for
` 490. Find his profit or loss per cent.
(Text Book Page No. 81)
Sol.
Original cost of the toy car
=
472
Money spent by Firoz on its transportation
=
28
Thus, total money spent by Firoz on the try car =
472 + 28
\ C.P. of the toy car
=
` 500
S.P. of the toy car
=
` 490
=
C.P. – S.P.
=
500 – 490
=
` 500
=
` 10
As C.P. is more than S.P. therefore there is loss.
Loss
Thus we have to calculate loss %
Loss%
Loss
× 100
C.P.
10
=
× 100
500
=
=
2%
Ans.
8. Fruit seller purchased 10 kg mangoes for ` 20 per kg and 15 kg bananas for ` 12
per kg. He packed them equally in 5 baskets and sold 1 basket for ` 85 each. Find
his profit or loss%.
(Text Book Page No. 81)
Sol.
Cost of 1 kg mangoes
=
20
\ Cost of 10 kg mangoes
=
10 × 20
Cost of 1 kg bananas
=
` 12
\ Cost of 15 kg bananas
=
12 × 15
C.P.
=
Cost of Mangoes + Cost of Bananas
=
200 + 180
S.P. of fruits (Bananas and Mangoes)
=
` 85 for each basket
No. of baskets
=
5
\ S.P. of fruits
=
5 × 85
=
` 200
=
Rs 180
=
` 380
=
` 425
=
` 45
As S.P. is more than C.P. therefore there is profit.
Profit
Profit %
Runway
=
S.P. – C.P.
=
425 – 380
=
Profit × 100
C.P.
=
45 × 100 =
380
114
11.84%
Ans.
Math-O-Mania– 5
9. A computer company manufactures 20 computers daily. The manufacturing cost
of each computer is ` 25680. Packing charges is ` 520 per computer. The
company sells them at a profit of 10%. What should be the S.P. of one computer ?
(Text Book Page No. 81)
Sol.
Manufacturing cost of each computer
=
\ Manufacturing cost of 20 computers =
` 25680
25680 × 20
=
` 513600
Packing charges per computer
=
` 520
\ Packing charge for 20 computers
=
520 × 20
=
` 10400
=
513600 + 10400
Thus C.P. of 20 computers
=
` 524000
=
52400
=
576400
This company sells the computers at a profit of 10%
Profit percentage
Profit
× 100
C.P.
Profit
=
× 100
524000
=
10
Profit
=
S.P.
=
10 × 524000
100
C.P. + Profit
\ S.P. of 20 computers
=
576400
Thus S.P. of 1 computer
=
576400
20
` 28820
=
Ans.
10. Fill the following table (Put ‘X’ in invalid box) :
Ans.
Profit%
(Text Book Page No. 81)
S.No.
C.P.
S.P.
(i)
` 2500
` 2000
(ii)
` 19
` 21
10.52%
×
(iii)
` 2550
` 2805
10%
×
(iv)
` 10000
` 9500
×
5%
(v)
` 4100
` 4018
×
2%
×
Loss %
20%
Mental Maths (Text Book : Page No 82)
Ans :
(Do yourself)
Fun Activity (Text Book : Page No 82)
Ans :
Runway
(Do yourself)
115
Math-O-Mania– 5
15.
Simple Interest
Exercise 15.1
1. Complete the following table :
Ans.
(Text Book Page No. 85)
S.No.
Principal
Rate
Time
Amount
SI
(i)
` 600
2.5%
2 years
` 630
` 30
(ii)
` 1200
4%
1 year
` 1248
` 48
(iii)
` 1000
9%
4 years
13600
3600
(iv)
` 7600
6
1 year
` 8094
` 494
(v)
` 1350
8%
5 years
` 1890
540
1
%
2
2. Find the interest on ` 1200 for 4 years at the rate of 6 1 % per annum.
2
Sol.
Simple Interest
=
Principle × Rate × Time
100
Principle (P)
=
` 1200
Rate (R)
=
Time
=
6 1 %
2
4 years
SI
=
P×R×T
100
=
` 312
(Text Book Page No. 85)
= 13 %
2
=
1200 × 13 × 4
100 × 2
Ans.
3. Nisha took a loan of ` 25000 from a bank for 6 years at the rate of 5% p.a. What
amount she has to pay after 6 years ?
(Text Book Page No. 85)
Sol.
Simple Interest
=
SI
=
P×R×T
100
` 7500
Amount
=
Principle + Interest
=
25000 + 7500
=
=
25000 × 5 × 6
100
` 32500
Hence, after 6 years Nisha has to pay an amount of ` 32500.
Ans.
4. Nikhil borrowed ` 10,000 from his friend and paid ` 3200 as interest after 4
years ? Find the rate of interest.
(Text Book Page No. 85)
Sol.
Simple Interest
Runway
=
Principle × Rate × Time
100
116
Math-O-Mania– 5
3200
=
Rate
=
=
10000 × R × 4
100
100 × 3200
10000 × 4
8%
Ans.
5. Ramlal borrowed ` 6000 from the landlord Rajendra Babu to purchase a cow. He
paid ` 7800 back to Rajendra Babu. If the rate of interest was 10% p.a. Find after
how much time the debt was cleared.
(Text Book Page No. 85)
Sol.
Amount
=
Principle + Interest
7800
=
6000 + Interest
Interest
=
` 1800
SI
=
P×R×T
100
1800
=
6000 × 10 × T
100
T
=
=
1800 × 100
6000 × 10
3 years
Hence, the debt was cleared after 3 years.
Ans.
6. Raman kept a certain principle in a bank for 10 years and received ` 300 as
interest. Rehman kept the same principle for 7 years. If the rate of interest was
3% p.a. for both. Find the amount received by Rehman after 7 years ?
(Text Book Page No. 85)
Sol.
Simple Interest
=
P×R×T
100
300
=
P × 3 × 10
100
300 × 100
= ` 1000
3 × 10
Rehman also kept the same principle amount, so for Rehman Interest on a principle amount of
for a period of 7 years should be calculated as follows :
P×R×T
1000 × 3 × 7
Interest
=
=
100
100
P amount for Raman
Amount
=
=
` 210
=
Principle + Interest
=
1000 + 210
=
` 1210
So, the amount received by Rehman after 7 years = ` 1210.
Runway
117
` 1000
Ans.
Math-O-Mania– 5
7. Calculate the interest on ` 1000 in 5 years if the interest is ` 300 in 6 years at the
same rate.
(Text Book Page No. 85)
Sol.
Interest
=
P×R×T
100
R
=
Interest × 100
P×T
=
300 × 100
1000 × 6
=
5%
Now, it is required to calculate interest at the same rate for a period of 5 years.
SI
=
SI
=
=
P×R×T
100
1000 × 5 × 5
100
` 250
Ans.
8. If the simple interest for 8 years on a certain sum is ` 96. Find the SI for 6 years on
the same sum at the same rate.
(Text Book Page No. 85)
Sol.
SI for a cetain sum for 8 years is ` 96.
\ SI for a the same sum for 6 years at the same rate will be
=
=
96 × 6
8
` 72
Ans.
9. Calculate the time in which ` 5000 will amount to ` 5500 at 5% p.a. simple
interest.
(Text Book Page No. 85)
Sol.
Amount
=
` 5500
Principle
=
` 5000
\ SI
=
Amount – Principle
=
5500 – 5000
=
` 500
We know,
Runway
SI
=
500
=
T
=
P×R×T
100
5000 × 5 × T
100
500 × 100
5000 × 5
=
118
2 years
Ans.
Math-O-Mania– 5
10. Sohan borrowed ` 2500 from Seema at 5% p.a. simple interest for 4 years and
lent this money to Anuj at 6% p.a. for 4 years. Calculate the profit earned by
Sohan at the end of 4 years.
(Text Book Page No. 85)
Sol.
S.I. for Sohan
P×R×T
100
2500 × 5 × 4
=
100
=
=
S.I. for Anuj
` 500
P×R×T
100
2500
×6×4
=
100
=
=
` 600
Profit earned by Sohan after 4 years
=
600 – 500
=
` 100
Ans.
Exercise 15.2
1. Find the simple interest of :
Sol.
(Text Book Page No. 86)
(a) ` 3600 for 3 years 4 months at 8% per annum.
Principle
=
` 3600
Time
=
3 years 4 months
=
3+
=
40 years
12
Rate
=
8% p.a.
SI
=
P×R×T
100
4
12
=
36 + 4
12
= 3600 × 8 × 40
100 × 12
=
` 960
Ans.
(b) ` 1250 for 146 days at 9% per annum.
Runway
Principle
=
` 1250
Rate
=
9% p.a.
119
Math-O-Mania– 5
Time
SI
=
146 days
=
146
years
365
=
P×R×T
100
= 1250 × 9 × 146
100 × 365
=
(c) ` 5000 for 1
` 45
Ans.
1
1
years at 7 % per annum.
4
2
Principle
=
` 5000
Time
=
1
Rate
=
SI
=
P×R×T
100
=
5000 × 15 × 5
2 × 4 × 100
=
` 468.75
1
years
4
1
7
p.a.
2
Principle
3
% per annum.
4
= ` 3000
Times
=
73 days
=
73 years
365
3
3 % p.a.
4
=
5
years
4
= 15 p.a.
2
Ans.
(d) ` 3000 for 73 days at 3
Rate
=
SI
=
= 15 % p.a.
4
P×R×T
100
= 3000 × 15 × 73
100 × 4 × 365
=
` 22.50
Ans.
2. Teena borrowed ` 5000 from Meena at 8% per annum interest. She returned the
money after 146 days. What interest did she pay ? Find the amount she paid.
(Text Book Page No. 86)
Runway
120
Math-O-Mania– 5
Sol.
Principle
=
` 5000
Rate
=
8% p.a.
Time
=
146 days
SI
=
P×R×T
100
=
146
365
= 5000 × 8 × 146
100 × 365
Hence Teena paid interest = ` 160.
Amount
=
Principle + Interest
=
5000 + 160
=
` 5160
So, Teena paid amount = ` 5160.
Ans.
3. Naresh deposited ` 7000 in a bank at a rate of 6% p.a. Find the amount he gets
back after 8 months.
(Text Book Page No. 86)
Sol.
Principle
=
` 7000
Rate
=
6% p.a.
Time
=
8 months
=
8 years
12
P×R×T
100
=
=
7000 × 6 × 2
100 × 3
=
=
Principle + Interest
=
7000 + 280
=
` 7280
Interest
Amount
=
2 years
3
` 280
\ After 8 months, the amount that Naresh gets back = ` 7280.
Ans.
4. ` 1600 amount to ` 1648 at 15% rate of simple interest. Calculate the number of
(Text Book Page No. 86)
days for which the money was invested.
Sol.
Amount
=
` 1648
Principle
=
`. 1600
Interest
=
1648 – 1600
Interest
=
P×R×T
100
48
=
1600 × 15 × T
100
Runway
=
121
` 48
Math-O-Mania– 5
T
=
48 × 100
1600 × 15
1 year
=
365 days
0.2 year
=
365 × 0.2
=
73 days
=
0.2 years
So the money was invested for 73 days.
Ans.
5. An interest of ` 28.50 was earned on a certain sum of money after 9 months when
the rate on interest was 15%. Had the rate been 7
interest in rupees ?
Sol.
=
?
Rate
=
15% p.a.
Time
=
9 months
=
9
years
12
SI
=
P×R×T
100
28.50
=
P × 3 × 15
4 × 100
28.50 × 4 × 100
3 × 15
1
15
Now the interest is 7 % or
%
2
2
As we know,
Interest
= P×R×T
100
6.
Sol.
% p.a. what would be the
(Text Book Page No. 86)
Principle
P
1
2
=
=
253.33 × 15 × 3
100 × 2 × 4
=
` 14.25
=
3 years
4
=
`. 253.33
Ans.
Thus, had the rate been 7 1 % p.a. the interest rupees would be ` 14.25.
2
1
` 5200 were deposited in a bank at a rate of 16 % p.a. for 219 days. Calculate
2
(Text Book Page No. 86)
the amount got back.
Principle
=
Rate
=
Time
=
Runway
` 5200
16 1 % p.a.
2
219 days
122
=
33 % p.a.
2
=
219 years
365
Math-O-Mania– 5
Interest
Amount
=
P×R×T
100
=
5200 × 33 × 219
100 × 2 × 365
=
` 514.80
=
Principle + Interest
=
5200 + 514.80
=
` 5714.80
Ans.
Mental Maths (Text Book : Page No 87)
Ans : (Do yourself)
Fun Activity (Text Book : Page No 87)
Ans : (Do yourself)
16.
Distance, Time and Speed
Exercise 16.1
1. An athlete runs a 1500 m race in 3 minutes. What is his speed in m/min ?
(Text Book Page No. 89)
Sol.
Speed
=
Distance
Time
=
1500
3
=
500 m/min.
Ans.
2. Amita walked a distance of 300 m at the speed of 3 m/s. Find the time taken by
her.
(Text Book Page No. 89)
Sol.
Time
=
=
=
Distance
Speed
300
3
100 sec
Ans.
3. The speed of a bus is 60 km per hour. It runs for 4 hours. Find out the distance
covered by it.
(Text Book Page No. 89)
Sol.
Speed
=
60 km/h
Time
=
4 hrs
Distance
=
Speed × Time
=
60 × 4
Runway
=
123
240 km
Ans.
Math-O-Mania– 5
4. A train runs at the speed of 90 km/h. How much time is required to cover a
distance of 405 km ?
(Text Book Page No. 89)
Sol.
Distance
=
405 km
Speed
=
90 km/h
Time
=
405
90
=
4.5 hrs
Ans.
5. The speed of a motorcycle is 45 km/h. It travelled continuously for 2 hrs. What is
the distance covered by it ?
(Text Book Page No. 89)
Sol.
Speed
=
45 km/h
Time
=
2 hrs
Distance
=
Speed × Time
=
45 × 2
=
90 km
Ans.
6. Rupali walks to school from her house. She walks at 3 km/h, and reaches the
school in
Sol.
1
2
hr. How far is her school from house ?
Speed
=
3 km/h
Time
=
1
hrs.
2
Distance
=
Speed × Time
=
3×
=
(Text Book Page No. 89)
1
2
1.5 km
So the school of Rupali is at a distance of 1.5 km from her house.
Ans.
7. An aeroplane flies at a speed 600 km/h. How much time it requires to cover 1500
km ?
(Text Book Page No. 89)
Sol.
Distance
=
1500 km
Speed
=
600 km/h
Time
=
=
=
Runway
Distance
Speed
1500
600
2.5 hrs
Ans.
124
Math-O-Mania– 5
8. A train has to cover a distance of 200 km in 3 hrs. What speed it should run with ?
(Text Book Page No. 89)
Sol.
Distance
=
200 km
Time
=
3 hrs
Speed
=
=
=
Distance
Time
200
3
66.67 km/h
Ans.
9. A bird flew for 63 seconds at the speed of 9 m/s. How far could it fly ?
(Text Book Page No. 89)
Sol.
Time
=
63 sec
Speed
=
9 m/sec
Distance
=
Speed × Time
=
63 × 9
=
567 m
So the bird could fly for 567 m.
Ans.
Exercise 16.2
1. Convert in km/h :
Sol.
(Text Book Page No. 90)
(a) 5 m/s
= 5 ×
18
5
=
18 km/h
18
5
=
90 km/h
18
5
=
72 km/h
=
45 km/h
18
5
=
108 km/h
18
5
=
115.2 km/h
(b) 25 m/s
= 25 ×
(c) 20 m/s
= 20 ×
(d) 12.5 m/s
= 12.5 ×
18
5
(e) 30 m/s
= 30 ×
(f)
32 m/s
= 32 ×
Runway
125
Math-O-Mania– 5
2. Convert to m/s :
Sol.
(Text Book Page No. 90)
(a) 9 km/h
=
9 ×
5
18
=
2.5 m/s
5
18
=
5 m/s
5
18
=
10 m/s
5
18
=
15 m/s
5
18
=
17.5 m/s
5
18
=
22.5 m/s
(b) 18 km/h
= 18 ×
(c) 36 km/h
= 36 ×
(d) 54 km/h
= 54 ×
(e) 63 km/h
= 63 ×
(f)
81 km/h
= 81 ×
3. A car covered a distance of 70 km in 5 hrs. Find out the speed in m/s.
(Text Book Page No. 90)
Sol.
Distance
=
70 km
Time
=
5 hrs
Speed
=
Distance
Time
=
=
14 km/h
5
14 × 18
=
3.88 m/s
=
70
5
=
35
9
Thus, the car has the speed = 3.88 m/s
Ans.
4. A horse can run 30 m in 2 seconds. Calculate the time in hrs it will take to reach 27
km far.
(Text Book Page No. 90)
Sol.
In 2 seconds the horse can run 30 m.
30
Hence, in one second the horse can run 2
=
15 m
Thus, speed of the horse
=
15 m/s
=
15 ×
=
54 km/h
Runway
126
18
5
Math-O-Mania– 5
Distance to cover
=
27 km
Speed
=
54 km/h
Time
=
27
54
=
0.5 hr
=
1
2
Thus, the time the horse will take to reach 27 km will be 0.5 hr.
Ans.
5. A cyclist rode 180 m in 24 seconds. Find its time to cover 27 kms in hrs.
(Text Book Page No. 90)
Sol.
In 24 seconds the cyclist rode 180 m
=
15
m
2
15
m/s
2
15
18
×
2
5
27 km/h
Distance to cover
=
27 km
Time
=
Distance
Speed
=
1 hr
\ In 1 second the cyclist can ride =
Thus speed of the cyclist
180
=
24
=
=
×
=
27
27
6. A bird flies 100 metres in 20 seconds. Find her speed in :
(a)
Sol.
(b)
Sol.
Ans.
(Text Book Page No. 90)
m/s
Distance
=
100m
Time
=
20 sec
Speed
=
=
Distance
Time
5 m/s
=
5 m/s
=
5×
=
18 km/h
=
100
20
Ans.
km/h
Speed of the bird
18
5
Ans.
7. Jenny covers a distance of 12.8 km in 2 hours 8 minutes on bicycle. Find her speed
Runway
127
Math-O-Mania– 5
in km/h.
Sol.
(Text Book Page No. 90)
Distance
Time
=
12.8 km
=
128
km
10
=
2 hours 8 minutes
=
2 + 8
60
=
128
hrs
60
=
Speed
=
32 hrs
15
Distance
Time
=
128
10
=
6 km/h
1 hr
= 60 min
1
1 min =
hr
60
8
8 min =
hrs
60
hrs
128
10
=
32
15
×
15
32
Thus the speed of Jenny is 6 km/h.
Ans.
Mental Maths (Text Book : Page No 91)
(Do yourself)
Ans :
Do You Konw (Text Book : Page No 91)
(Do yourself)
Ans :
17.
Temperature
Exercise 17
1. Fill up the boxes :
Sol.
(Text Book Page No. 94)
(a) 32ºF
= 0 ºC
(b) 0ºC
=
32 °F
(c) 212ºF
= 100 ºC
(d) 100ºC =
212 ºF
(e) F = 9 C + 32
5
Runway
(f)
128
C = 5 (F – 32 )
9
Math-O-Mania– 5
2. Change into Celsius scale :
Sol.
(Text Book Page No. 94)
(a) 40ºF
=
(b) 86ºF
(40 – 32) ×
5
°C
9
=
5
9
=
8×
=
4.44 °C
(c) 156ºF
=
5
°C
9
5
9
=
54 ×
=
30 °C
=
5
9
=
124 ×
=
68.8 °C
(e) 102ºF
(f)
(102 – 32) ×
5
°C
9
70 ×
=
38.88 °C
(185 – 32) ×
=
153 ×
=
85 °C
(g) 200ºF
5
9
(175 – 32) ×
143 ×
=
79.44 °C
(h) 37ºF
(200 – 32) ×
5
°C
9
=
(37 – 32) ×
=
168 ×
5
9
=
5×
=
93.33 °C
=
2.77 °C
Runway
5
°C
9
5
9
(Text Book Page No. 94)
(a) 15ºC
=
5
°C
9
5
9
=
3. Change into Fahrenheit scale :
Sol.
5
°C
9
175ºF
=
5
9
=
=
5
°C
9
(d) 185ºF
(156 – 32) ×
=
(86 – 32) ×
(b) 20ºC
15 ×
9
+ 32 °F
5
=
20 ×
9
+ 32 °F
5
=
27 + 32
=
36 + 32
=
59 °F
=
68 °F
129
Math-O-Mania– 5
(c) 40ºC
=
(d) 75.5ºC
40 ×
9
+ 32 °F
5
=
=
72 + 32
=
135.9 + 32
=
104 °F
=
167.9 °F
(e) 90ºC
=
(f)
90 ×
25ºC
9
+ 32 °F
5
=
25 ×
162 + 32
=
45 + 32
=
194 °F
=
77 °F
(h) 60ºC
9
+ 32 °F
5
=
40.3 ×
=
362.7
+ 32 °F
5
=
72.54 + 32
=
104.54 °F
=
60 ×
9
+ 32 °F
5
=
108 + 32
=
140 °F
4. Simplify (give answers in ºF) :
Sol.
9
+ 32 °F
5
=
(g) 40.3ºC
(a)
9
+ 32 °F
5
75.5 ×
(Text Book Page No. 94)
52ºC + 14ºF
We have to convert both in the same scale to add the temperature.
Hence,
52 °C
=
=
=
\
Runway
52°C + 14° F
52 ×
9
+ 32 °F
5
468
+ 32
5
468 + 160
5
=
125.6 °F
=
125.6 °F + 14 °F
=
139.6 °F
130
=
625
5
Ans.
Math-O-Mania– 5
(b)
Sol.
40ºC – 20ºF
We have to convert both in the same scale to find the resultant temperature.
Hence,
\
(c)
40 °C
40°C – 20 ° F
40 ×
=
(72 + 32) °F
=
104 °F – 20 °F
=
84 °F
=
2×
2 °C
=
=
\
Sol.
=
104 °F
Ans.
25ºF + 2ºC
Sol.
(d)
9
+ 32 °F
5
=
25°C + 2° F
9
+ 32 °F
5
18
+ 32
5
18 + 160
5
=
35.6 °F
=
25 + 35.6
=
60.6 °F
=
178
5
Ans.
95ºF – 3ºC
Hence,
3 °C
=
=
=
95 – 37.4
=
57.6 °F
=
95°F – 3° C
9
+ 32 °F
5
27 + 32
5
27 + 160
5
37.4 °F
=
\
3×
=
187
5
Ans.
5. On Tuesday, the minimum temperature at Mumbai was 21.2ºC. On Wednesday,
the minimum went down by 1.9ºC. On Thursday, the minimum went up by 2.8ºC.
Find the minimum on Thursday.
(Text Book Page No. 94)
Sol.
Minimum temperature of Mumbai on Tuesday
=
21.2 °C
On Wednesday temperature went down by
=
1.9 °C
\ Minimum temperature on Wednesday
=
21.2 – 1.9
On Thursday, the minimum temperature went up by
=
2.8 °C
Thus the minimum temperature on Thursday
=
19.3 + 2.8
=
22.1 °C
Runway
131
=
19.3 °C
Ans.
Math-O-Mania– 5
6. On a certain day, the maximum temperature at Shimla was 32.7ºC. Next day, the
maximum went up by 2.6ºC. On third day, the maximum came down by 3.8ºC.
(Text Book Page No. 94)
Find the maximum on third day.
Sol.
Maximum Temperature of Shimla on a certain day
=
32.7°C
On the next day the maximum temperature went up by
=
2.6 °C
Thus, the maximum temperature of Shimla on the next day =
=
32.7 + 2.6
35.3 °C
On the third day the maximum temperature come down by =
3.8 °C
Thus, on the third day the maximum temperature of Shimla =
35.3 – 3.8
=
31.5 °C
Ans.
7. In a city, the maximum and minimum temperatures on a certain day were 34.4ºC
and 16.4ºC. The next day, the maximum went up by 1.8ºC and minimum went
down by 1.4ºC. Find the maximum and minimum temperatures on the next day.
(Text Book Page No. 94)
Sol.
On a certain day the maximum temperature of a city was 34.4 °C.
The next day, the maximum temperature of that city went up by 1.8 °C
\ The maximum temperature of the city on the next day
=
34.4 + 1.8
=
36.2 °C
On the first day, the minimum temperature of that city was 16.4 °C
On the next day, the minimum temperature of that city went down by 1.4 °C
\ The minimum temperature of that city on the next day
=
16.4 – 1.4
=
15°C
Thus, the maximum and minimum temperature of the city on the next day were 36.2°C and 15°C
respectively.
Ans.
Mental Maths (Text Book : Page No 95)
Ans :
(Do yourself)
Fun Activity (Text Book : Page No 95)
Ans :
Runway
(Do yourself)
132
Math-O-Mania– 5
18.
Introduction to Algebra
Exercise 18.1
1. Express the following in algebraic form :
(Text Book Page No. 97)
Ans. (a) 3 is added to x.
3+x
(b) x is increased by 21.
x + 21
(c) 9 is taken away from a.
a–9
(d) a is decreased by 3.
a–3
(e) c is taken away from the sum of a and b.
(a + b) – c
(f)
p times q is added to 7 times r.
pq + 7r
(g) 15 times m is divided by 2 times n.
15m/2n
(h) The product of 3 and x is subtracted from the sum of y and 5.
(y + 5) – 3x
(i)
The difference '3 minus y' is divided by the product of 3 and x.
(3 – y) / 3x
(j)
Half of x is taken away from one-third of y.
y
– x
3
2
2. Express these in words :
(Text Book Page No. 97)
(b) x + y
Ans. (a) b + y
Sum of b and y
Sum of x and y
(c) x – 8
(d) 12 – x
Subtract 8 from x
(e) m × n
(f)
6 divided by x
m multiply by n
Runway
Subtract x from 12
6
x
133
Math-O-Mania– 5
(g)
2
of y
100
(h)
2% of y
(i)
5
y
100
5% of y
x
y
+
9
4
(j)
One fourth of x added to one-ninth of y
1
(x+y)
7
One seventh of the sum of x and y
Exercise 18.2
1. Find the coefficients of :
(Text Book Page No. 98)
Ans. (a) x in 8x + 9y + 18xy
2
2
(b) yz in 2xy + 3yz + 9zx
2
8
3
(c) x y in 18abx + 19a b + 20ab – 17x y
2
2
2
2
(d) b in 10a + 7b
– 17
7
2. Say following terms in each pair are like or unlike :
Ans. (a) 3a, 4a
(b) 15y, 10x
(c) 2d,2b
Unlike
Unlike
Like
(d) 3xy , 5y x
2
2
(Text Book Page No. 98)
(e) x , 3x
3
Like
3
(f)
Like
4ab, 2bc
Unlike
3. Find the sum of the following :
Sol.
(a) 5x, 14x
5x + 14x = 19x
(d) 3x, 2x, 9x
3x + 2x + 9x = 14 x
(Text Book Page No. 99)
(b) x, 7x
(c) 2abc, 20abc
x + 7x = 8x
(e) 4ab, 5ab, 6ab
2abc + 20abc = 22 abc
(f)
(a) 16x, 3x
2
(Text Book Page No. 99)
(c) 9x, 2x
16x – 3x
4ab – 2ab
9x – 2x
= 13 x
= 2ab
= 7x
(d) 10ab, 10ab
Runway
(b) 4ab, 2ab
2
6x2 + 7y2x = xy (6x + 7y)
4ab + 5ab + 6ab = 15ab
4. Subtract the second from the first term :
Sol.
6x , 7y x
(e) 3x3, x3
10ab – 10ab
3x – x
=0
= 2x
3
(f)
3
15abx, 5abx
15abx – 5 abx
= 10 abx
3
134
Math-O-Mania– 5
5. Simplify :
Sol.
(a) x + x + x + x + x
= 5x
(d) 2x + x + x – x
= 3x
(Text Book Page No. 99)
(b) a + a + a + b + b
(c) p + p + a + p + a
= 3a + 2b
= 3p + 2a
(e) 8xy +2xy – 4xy
(f)
= 6x y
2
2
2
b + 2b – c – c
2
= 3b – 2c
2
2
Exercise 18.3
1. Multiply the following algebraic expressions :
Sol.
(a) x × 5
(Text Book Page No. 99)
(c) xy × 6
(b) 6 × a
= (x) × (5)
= (6) × (a)
= (x y) × (6)
= 5x
= 6a
= 6x y
(d) 9 × ab
(e) p × 7
(f)
2q × 4p
= (9) × (ab)
= (p) × (7)
= (2×4) × (q×p)
= 9ab
= 7p
= 8qp
(g) 5 × p × q
(h) 10 × a × b
= (5) × (p×q)
= (10) × (a×b)
= 5 pq
= 10ab
2. Divide the following algebraic expressions :
Sol.
(a) xy ÷ 10
=
x
xy
10
(d) 9a ÷ 3
(b) 10 ÷ ab
=
(Text Book Page No. 99)
(c) m ÷ 8
10
ab
(e) 18b ÷ 6
=
(f)
m
8
14x ÷ 7
14x
7
14
=
x
7
=
9a
3
9
=
a
3
= 18b
6
18
=
b
6
=
= 3a
= 3b
= 2x
(g) 36xy ÷ 9
(h) 5xyz ÷ 5
= 36xy
9
36
=
xy
9
= 5xyz
5
= 5 x yz
5
= 4xy
= 1 × x yz
= x yz
Runway
135
Math-O-Mania– 5
Exercise 18.4
1. If x = 4, find the value of :
Sol.
(a) x + 5
(Text Book Page No. 99)
(b) x + 14
(c) 5 + x
= 4+5
= 4 + 14
= 5+4
= 9
= 18
= 9
(d) 3x + 6
(e) 7 + 8x
(f)
x–4
= (3 × 4) + 6
= 7 + (8 × 4)
= 4–4
= 12 + 6
= 7 + 32
= 0
= 18
= 39
(g) 10 – x
(h) 3x – 7
= 10 – 4
= (3 × 4) – 7
=6
= 12 – 7
= 5
2. If a = 5, find the value of :
Sol.
(a) 3a
(Text Book Page No. 99)
(b) 4a
(c) 7a
= 3×5
= 4×5
= 7×5
= 15
= 20
= 35
(d) a × 5
(e) 3a – 14
(f)
a×6
= 5×5
= (3 × 5) – 14
= 5×6
= 25
= 15 – 14
= 30
= 1
(g) 15a ÷ 3
(h) 14a ÷ 7
= (15 × 5) ÷ 3
= (14 × 5) ÷ 7
= 75 ÷ 3
= 70 ÷ 7
= 25
= 10
3. Find the value of the following, if a = 3 and b = 2 :
Sol.
(a) a + 2b
(b) 6a – 2b
(Text Book Page No. 99)
(c) 3a + 4b – 6
= 3 + (2 × 2)
= (6 × 3) – (2 × 2)
= (3 × 3) + (4 × 2) – 6
= 3+4
= 18 – 4
= 9+8–6
= 7
= 14
= 17 – 6
= 11
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Math-O-Mania– 5
(d) 4b – 2a
(e) 2a × 4b
(f)
6a × 5b
= (4 × 2) – (2 × 3)
= (2 × 3) × (4 × 2)
= (6 × 3) × (5 × 2)
= 8–6
= 6×8
= 18 × 10
= 2
= 48
= 180
(g) a × 4b
(h) 2b ÷ a
= 3 × (4 × 2)
= (2 × 2) ÷ 3
= 3×8
= 4÷3
= 24
=
4
3
Lab Activity (Text Book : Page No 100)
Ans : (Do yourself)
Fun Activity (Text Book : Page No 100)
Ans : (Do yourself)
19.
Geometry
Exercise 19.1
1. Fill in the blanks :
(Text Book Page No. 102)
Ans. (a) The length of a line segment is definite.
(b) The length of a line is indefinite.
(c) The lines which do not intersect are parallel lines.
(d) Two or more lines passing through common point are concurrent lines.
(e) Perpendicular lines make an angle of 90°.
2. Identify the following :
Ans. (a)
(Text Book Page No. 102)
(b)
Ray
(d)
Point
(e)
Line
Runway
(c)
Parallel Lines
(f)
Line segment
137
Intersecting Lines
Math-O-Mania– 5
(g)
(h)
Concurrent Lines
Perpendicular Lines
Exercise 19.2
1. Classify the following angles as acute, obtuse, right, straight, complete or reflex,
(Text Book Page No. 105)
etc. :
Ans. (a) 75º
(b) 125º
acute
obtuse
(e) 200º
(f)
reflex
(i)
(c) 180º
straight
164º
(g) 22.5º
obtuse
360º
(j)
complete
acute
False
True
(c) Vertically opposite angles are supplementary.
False
(d) Adjacent angles are always equal in degrees.
False
(e) Lines that are not parallel always intersect.
3. Write the compliments of following angles :
True
(Text Book Page No. 105)
50º
Let the complement of the angle be x.
\
Sol.
right
(Text Book Page No. 105)
(b) Perpendicular lines make a right angle.
(b)
(h) 90º
acute
Ans. (a) Parallel lines have an angle of 90º between them.
Sol.
reflex
59º
2. Write True or False :
(a)
(d) 210º
50 + x
=
90
x
=
90 – 50
=
40°
65º
Let the complement of the angle be x.
\
Runway
65 + x
=
90
x
=
90 – 65
=
25°
138
Math-O-Mania– 5
(c)
Sol.
78º
Let the complement of 78° be x.
\
(d)
Sol.
78 + x
=
90
x
=
90 – 78
=
12°
42º
Let the complement of 42° be x.
\
42 + x
=
90
x
=
90 – 42
=
48°
4. Write the supplements of following angles :
(a)
Sol.
86º
Let the supplement of the angle be x.
\
(b)
Sol.
x + 86
=
180
x
=
180 – 86
=
94°
121º
Let the supplement of the angle be x.
\
x + 121 =
x
(c)
Sol.
Sol.
180
=
180 – 121
=
59°
80º
Let the supplement of the angle be x.
\
(d)
(Text Book Page No. 105)
x + 80
=
180
x
=
180 – 80
=
100°
154º
Let the supplement of the angle be x.
\
x + 154 =
x
Runway
180
=
180 – 154
=
26°
139
Math-O-Mania– 5
5. Name the pairs of adjacent angles in following figures :
A
(a)
B
(Text Book Page No. 105)
C
O
Ans. ÐAOB & ÐBOC; ÐAOC & ÐAOB ; ÐAOC & ÐBOC
(b)
R
Q
S
P
Ans. ÐQPR & ÐSPR; ÐRPS & ÐSPQ; ÐQPR & ÐQPS
M
N
L
O
P
Ans. ÐMLN & ÐNLO; ÐNLO & ÐOLP; ÐMLP & ÐMLN; ÐNLP & ÐOLP, etc.
6. Find the angles marked aº, bº, cº in following figures :
(Text Book Page No. 105)
(i)
aº
155º
bº
cº
Sol.
Let the angle given equal to 155° be marked as d°.
As we can see, the angles Ðd and Ðb are vertically opposite angles.
Thus, Ðd and Ðb are equal in magnitude
b° =
d°
b
155°
=
Also, we can see from the above figure that the four angles Ða, Ðb, Ðc, Ðd from a complete angle
which taken a complete turn around a point, so the sum of the 4 angles equals 360°
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140
Math-O-Mania– 5
\
a+b+c+d
=
360
a + 155 + c + 155
=
260
a+c
=
360 – 155 – 155
a+c
=
50
We can see from the given figure that Ða and Ðc are equal as they are vertically opposite angles.
Thus,
a
But
\
=
c
a+c
=
50
a+a
=
50
2a
=
50
a
=
25
c
=
25°
[a = c]
Thus, the three angles are a = 25°, b = 155°, c = 25°.
Ans.
(ii)
bº
aº
10º cº
120º
110º
Sol.
Ða and the angle given as equal to 120° are equal in magnitude as they are vertically opposite angles
[as can be seen in the figure.]
\
a
=
120°
Similarly Ðb is equal in magnitude to the angle given as equal to 10° as these both angles are vertically
opposite angles.
\
b
=
10°
Ða and Ðc are supplementary angles as they are an a straight line. Thus their sum is equal to 180°.
Þ
\
a+c
=
180
120 + c =
180
c
=
180 – 120
=
60°
Ans.
7. Construct these angles with the help of protractor :
(Text Book Page No. 106)
(a) 125º
(b) 75º
(c) 146º
(d) 110º
(e) 56º
(f)
(g) 135º
(h) 45º
95º
Ans. Do your self
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Math-O-Mania– 5
Exercise 19.3
1. In which of the following cases a triangle with given angles is possible :
(Text Book Page No. 108)
(a)
Sol.
30º, 60º, 45º
According to the angle sum property of a D, the sum of the three angles of a D is always 180°.
Thus for a triangle to exist, the sum of its angles must be 180°.
In this case the 3 angles are 30°, 60°, 45°
30 + 60 + 45
=
135°
Which is not equal to 180°. Hence the D with the given angles is not possible.
(b)
Sol.
Ans.
120º, 90º, 50º
Sum of the angles
=
120 + 90 + 50
=
260°
Which is not equal to 180°. Hence the D with given angles cannot be constructed.
(c)
Sol.
78º, 82º, 20º
Sum of the angles
=
78 + 82 + 20
=
180°
So the D can be constructed with the three given angles.
(d)
Sol.
Ans.
90º, 90º, 0º
Sum of the angles
=
90 + 90 + 0
=
180°
The sum of the angles equals 180° but in this case the third angle equals 0° which suggests that the
angle does not have value. For a D to exist.
All the three angles must have some value. Thus the D is not possible with the given angles.
(e)
Sol.
Ans.
110º, 65º, 5º
Sum of the angles
=
110 + 65 + 5
=
180°
As all the angles in this case have some values and their sum equals 180°, so the D can be
constructed.
Ans.
(f)
Sol.
100º, 70º, 20º
Sum of the angles
Runway
=
100 + 70 + 20
=
190°
142
Math-O-Mania– 5
Clearly, the sum of the angles is not equal to 180°. So the D cannot be constructed with these angles.
Ans.
2. Write true or false :
(Text Book Page No. 108)
Ans. (a) A triangle can have 2 right angles.
False
(b) A triangle can have all angles acute.
True
(c) A triangle whose 2 sides are equal is scalene.
False
(d) A triangle has all 3 sides equal in equilateral.
True
(e) Sum of all angles of a triangle is always 180º.
True
3. In a DABC, ÐA = 60º, ÐB = 110º, find ÐC.
Sol.
In DABC,
ÐA + ÐB + ÐC
=
180°
60 + 110 + ÐC
=
180
=
180 – 110 – 60
=
10°
ÐC
(Text Book Page No. 108)
(as we know)
Ans.
4. In an isosceles triangle ABC, AB = AC. If ÐB = 65º, find ÐC and ÐA.
(Text Book Page No. 109)
Sol.
A
B
In
C
DABC, AB = AC (given)
ÐB
=
ÐC
ÐB
=
65° (given)
ÐC
=
65°
ÐA + ÐB + ÐC
=
180°
ÐA + 65 + 65
=
180
=
180 – 65 – 65
=
50°
i.e.,
\
\
ÐA
[ DABC as isosceles D]
Ans.
5. In a triangle ÐA =ÐB = ÐC. Is it equilateral or isosceles ?
Sol.
ÐA
=
ÐB
=
(Text Book Page No. 109)
ÐC (given)
As all the three angles are equal in magnitude, thus the D been formed by these angles should be
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143
Math-O-Mania– 5
equilateral.
Ans.
6. Measure the angles and sides of these triangles and tell what types of triangles are
these :
(Text Book Page No. 109)
(i)
On the basis of angles
(a)
(ii) On the basis of sides.
(b)
(c)
(d)
(e)
Ans. Do yourself
7. Construct a DABC in which BC = 4 cm, AC = 5 cm and AB = 6 cm.
(Text Book Page No. 109)
Ans. Do yourself
8. Construct a DPQR in which PQ = 3 cm, ÐP = 110º, ÐQ = 40º.
(Text Book Page No. 109)
Ans. Do yourself
9. Construct a DXYZ in which XY = 7 cm, YZ = 9 cm, ÐY = 100º.
(Text Book Page No. 109)
Ans. Do yourself
10. Construct an equilateral triangle with side = 6.5 cm.
(Text Book Page No. 110)
Ans. Do yourself
Exercise 19.4
1. Watch the figure and fill in the blanks accordingly :
A
C
(Text Book Page No. 122)
B
.U
.X
D
O
.Y
.V
E
F
Ans. (a) O is the centre of the circle.
(b) CD, BE are the diameters of the circle.
(c) OE, OB, OC, OD are the radii of the circle.
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Math-O-Mania– 5
(d) AB and EF are the chords of the circle.
(e) Point A, B, C, D, E and F are on the circle.
(f)
Exterior point of the circle are U and V.
(g) Points interior to the circle are X and Y.
(h) CO + OD = diameter CD.
2. Draw circles with the help of a compass taking the following radii :
(Text Book Page No. 110)
(a) 5 cm
(b) 4.2 cm
(c)
3.5 cm
Ans. Do yourself
3. Find the diameter of a circle whose radius is :
(a)
Sol.
(b)
Sol.
(Text Book Page No. 110)
8 cm
Radius
=
8 cm
Diameter
=
2 × Radius
Radius
=
4 cm
Diameter
=
2 × Radius
=
2 × 8
=
16 cm
=
2×4
=
8 cm
=
2 × 5.5
=
11 cm
4 cm
(c)
5.5 cm
Sol.
Radius
=
5.5 cm
Diameter
=
2 × Radius
4. Find the radius of a circle whose diameter is :
(a)
10 cm
Radius
Sol.
(b)
=
Diameter
2
=
10
2
=
5 cm
=
Diameter
2
=
12
2
=
6 cm
=
Diameter
2
=
9.6
2
=
4.8 cm
12 cm
Radius
Sol.
(c)
(Text Book Page No. 110)
9.6 cm
Sol.
Radius
Lab Activity (Text Book : Page No 111)
Ans : (Do yourself)
Fun Activity (Text Book : Page No 111)
Ans : (Do yourself)
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145
Math-O-Mania– 5
20.
Mensuration
Exercise 20.1
1. Find the perimeter of a rectangle having :
(a)
Sol.
Length = 60 m
Breadth = 50 m
Perimeter of a rectangle = 2 [ l + b]
l
=
60 m
b
=
50 m
\ Perimeter of the rectangle
(b)
Sol.
(c)
Sol.
(d)
Sol.
(e)
Sol.
(f)
Sol.
Length = 7.5 cm
=
2 × 110 m
=
220 m
=
2 [ l + b]
=
2 [ 7.5 + 2.5 ]c m
=
2 × 10 cm
=
20 cm
=
41.2 m
=
22 m
=
58 m
=
510 m
Breadth = 12 m
Perimeter
Length = 6m
2 [ 60 + 50 ] m
Breadth = 2.5 cm
Perimeter
Length = 8.6 m
=
=
2 [ l + b]
=
2 [ 8.6 + 12 ] m
=
2 × 20.6 m
=
2 [ l + b]
=
2[6+5]m
=
2 × 11 m
=
2 [ l + b]
=
2 [ 17 + 12 ] m
=
2 × 29 m
Breadth = 5 m
Perimeter
Length = 17 m Breadth = 12 m
Perimeter
Length = 180 m
Perimeter
Breadth = 75 m
=
2 [ l + b]
=
2 [ 180 + 75 ] m
=
2 × 255 m
2. Find the side of a square whose perimeter is :
(a)
(Text Book Page No. 113)
(Text Book Page No. 113)
64 m
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146
Math-O-Mania– 5
Sol.
(b)
Sol.
Side of a square
=
Sol.
Sol.
=
Perimeter
4
400 m
=
1600
4
Perimeter
4
140 m
=
560
4
Perimeter
4
4.94 cm
=
19.76
4
Perimeter
4
4.70 cm
=
18.80
4
Perimeter
4
17.17 cm
=
68.68
4
1600 m
Side of a square
560 m
Side of a square
=
19.76 cm
Side of a square
=
=
(e)
Sol.
18.80 cm
Side of a square
=
=
(f)
Sol.
64
4
16 m
=
(d)
=
=
=
(c)
Perimeter
4
68.68 cm
Side of a square
=
=
3. A rectangular field is 825 m long and 225 m broad. Find its perimeter.
(Text Book Page No. 113)
Sol.
As the field is rectangular, the perimeter of the filed =
2[l+b]
=
2 (825 + 225) m
=
2 × 1050 m
=
2100 m
Ans.
4. A square courtyard is 8 m in length. Find the length of wire required to fence
(Text Book Page No. 113)
around it.
Sol.
Length of the square courtyard
=
8m
\ Side
=
8m
Perimeter
=
4 × side
=
4×8
=
32 m
Hence, the length of wire required to fence around the square courtyard = 32 m
Runway
147
Ans.
Math-O-Mania– 5
5. Perimeter of a square is 12 km. Find its side in metres.
Sol.
Perimeter of a square
=
12 km
Perimeter
4
12
=
=
4
But it is required to find the side‘s length in metres.
Side
(Text Book Page No. 113)
=
1 km
=
1000 m
3 km
=
3 × 1000 m
=
3000 m
3 km
Thus, the side of the square = 3000 m.
Ans.
6. A square play ground has a side of 550 m. How long would I walk if I wish to take
two rounds of it ?
(Text Book Page No. 113)
Sol.
Side of the square playground in metres
=
550 m
\ Perimeter of the square playground
=
4 × 550
=
2200 m
If it is required to take two rounds of that square playground then the length of the ground that is
walked
= 2 × 2200 m
=
4400 m
Ans.
7. Length of a rectangular garden is 220 m. If its perimeter is 740 m. Find the
breadth of the garden.
(Text Book Page No. 113)
Sol.
Length
=
220 m
Perimeter
=
740 m
Breadth
=
=
=
Perimeter
– Length
2
740 – 220
=
2
150 m
370 – 220
Thus the breadth of the garden = 150 m
Ans.
Exercise 20.2
1. Find the area of the rectangle having the length and breadth as follows :
(Text Book Page No. 113)
(a)
Length = 8 cm
Runway
Breadth = 3 cm
148
Math-O-Mania– 5
Sol.
(b)
Sol.
(c)
Sol.
(d)
Sol.
Area of the rectangle
Length = 15.5 m
Sol.
(f)
Sol.
Area of the rectangle
Length = 3.5 cm
=
(8 cm × 3 cm)
=
24 cm2
=
l×b
=
( 15.5 m × 7.5 m)
=
116.25 m2
Breadth = 1.5 cm
Area of the rectangle
Length = 6 m
=
l×b
=
( 3.5 cm × 1.5 cm)
=
5.25 cm2
Breadth = 75 cm
Length
=
6m
Breadth
=
75 cm
=
0.75 m
Length = 6 m
=
75
m
100
=
l×b
=
( 6 m × 0.75 m)
=
4.5 m2
Breadth = 3 m
Area of the rectangle
Length = 8 m
=
l×b
=
( 6 m × 3 m)
=
18 m2
Breadth = 80 cm
Length
=
8m
Breadth
=
80 cm
=
0.8 m
Area of the rectangle
Runway
l×b
Breadth = 7.5 m
Area of the rectangle
(e)
=
=
80
m
100
=
l×b
=
( 8 m × 0.8 m)
=
6.4 m 2
149
Math-O-Mania– 5
2. Fill the following spaces for a rectangle :
S.No.
Ans.
Length
Breadth
Area
Perimeter
38 m
(a)
10 m
9m
90 m2
(b)
63 m
5m
315 m2
136 m
(c)
50 m
60 m
3000 m2
220 m
(d)
3.6 cm
12 cm
43.20 cm2
31.2 cm
3. Find the area of the square whose side is given below :
(a)
Sol.
(b)
Sol.
(c)
Sol.
(d)
Sol.
(e)
Sol.
(f)
Sol.
(Text Book Page No. 115)
(Text Book Page No. 115)
6 cm
Area of the square
=
side × side
=
(6 cm × 6 cm)
=
36 cm2
=
side × side
=
(2.5 cm × 2.5 cm)
=
6.25 cm2
=
side × side
=
(30.5 cm × 30.5 cm)
=
930.25 cm2
=
side × side
=
(3.5 cm × 3.5 cm)
=
12.25 cm2
=
side × side
=
(18.5 m × 18.5 m)
=
342.25 m2
=
side × side
=
(15 m × 15 m) sqm
=
225 m2
2.5 cm
Area of the square
30.5 cm
Area of the square
3.5 cm
Area of the square
18.5 m
Area of the square
15 m
Area of the square
Runway
150
Math-O-Mania– 5
4. Perimeter of a square field is 52 m. Find its area.
Sol.
Perimeter of the square field
=
52 m
\ Side of the square field
=
52
4
=
side × side
=
(13 m × 13 m )
=
169 m2
Area
(Text Book Page No. 115)
=
13 m
Ans.
5. The measure of all four sides of a rectangular tablecloth is 214 cm. If its length is
65 cm, find the area it can cover.
(Text Book Page No. 115)
Sol.
Measure of all four sides of a rectangular table cloth
\
Perimeter
=
214 cm
Length
=
65 cm
=
214
– 65
2
=
107 – 65
=
l × b
=
65 × 42
=
2730 cm2
\ Breadth
Area of the rectangular table cloth
=
214 cm
=
42 cm
Ans.
Exercise 20.3
1. Find out the area of following figures :
D
(a)
2 cm
(Text Book Page No. 116)
C
4 cm
6 cm
E
2 cm
F
A
Sol.
8 cm
B
We divide the following figure into parts.
D
2 cm
C
4 cm
6 cm
E
2 cm
X
2 cm
F
A
Runway
8 cm
151
B
Math-O-Mania– 5
Area of AF X B
Area of XEDC
Total area of figure AFEDCB
F
1m
G
A
=
8×2
=
16 cm2
=
XE × ED
=
2×4
=
8 cm2
=
(AFXB + XEDC)
=
24 cm2
Ans.
E
B
1m
C
8m
D
We divide the following figure into parts.
3m
3m
A
Area of ABGH
Area of CDEF
Total area of figure ABCDEFGH
(c)
8m
A
C
3m
F
1m
G
H
B
1m
C
8m
=
8×3
=
24 m2
=
5×3
=
15 m2
=
(ABGH + CDEF)
=
(24 + 15) m2
=
39 m2
E
5m
Sol.
3m
AB × FA
5m
H
3m
(b)
=
D
Ans.
B
D
F
K
Runway
2m
G
2m
10 m
E
H
I
6m
J
2m
L
152
Math-O-Mania– 5
We divide the following figure into parts.
A
8m
B
C
D 2m E
F
Area of rectangle ACFB
Area of rectangle EDHI
Area of rectangle GKLJ
Total area of figure ABFEIJLKGHDC
(d)
A
C
5m
8m
6m
J
2m
L
=
AC × AB
=
2×8
=
ED × EI
=
2 × 10
=
GK × KL
=
2×6
=
(ACFB + EDHI + GKLJ)
=
16 + 20 + 12
=
48 m2
A
8m
=
16 m2
=
20 m2
=
12 m2
Ans.
E
D
5m
6m
F
H
I
K
L
J
Dividing the given figure into parts.
E
14 m
H
4m
G
K
Runway
5m
F
6m
D
6m
5m
6m
C
B
4m
Sol.
I
B
14 m
G
2m
H
K
2m
G
2m
10 m
Sol.
I
J
L
153
Math-O-Mania– 5
Area of ABED
Area of EDHI
Area of HILK
Area of EIJF
Area of CDGH
Total area of figure ABEFJILKHGCD
=
AD × AB
=
4×8
=
ED × DH
=
8×6
=
IH × IL
=
8×4
=
EF × FJ
=
5 × 6
=
CD × DH
=
5×6
=
(ABED + EDHI + HILK + EIJF + CDGH)
=
(32 + 48 + 32 + 30 + 30)
=
172 m2
=
32 m2
=
48 m2
=
32 m2
=
30 m2
=
30 m2
Ans.
2. Find out the perimeters of following figures :
(a)
2 cm
(Text Book Page No. 116)
2 cm
2 cm
2 cm
3 cm
4 cm
4 cm
7 cm
Sol.
Perimeter =
4+2+2+3+2+2+4+7
(b)
4 cm
6 cm
3 cm
7 cm
3 cm
7 cm
Ans.
= 48 cm
Ans.
= 68 cm
Ans.
4 cm
2 cm
= 26 cm
2 cm
4 cm
Sol.
Perimeter =
6 cm
4+4+2+6+3+7+7+3+6+2+4
8 cm
(c)
8 cm
2 cm 3 cm
3 cm 2 cm
4 cm
4 cm
4 cm
4 cm
2 cm 3 cm
3 cm 2 cm
8 cm
Sol.
Perimeter =
Runway
8 cm
8+8+2+3+4+4+3+2+8+8+2+3+4+4+3+2
154
Math-O-Mania– 5
12 cm
(d)
3 cm
3 cm
3 cm
3 cm
6 cm
6 cm
6 cm
6 cm
3 cm
3 cm
3 cm
3 cm
12 cm
Sol.
Perimeter =
=
12 + 3 + 3 + 6 + 6 + 3 + 3 + 12 + 3 + 3 + 6 + 6 + 3 + 3
72 cm
Ans.
Exercise 20.4
1. Find the volume of a solid wood with length 40 cm, breadth 9 cm and height 7 cm.
(Text Book Page No. 117)
Sol.
Length
=
40 cm
Breadth
=
9 cm
Height
=
7 cm
Volume of box
=
l×b × h
=
(40 × 9 × 7) cm3
=
2520 cm3
Ans.
2. The volume of a solid box is 1600 cm3. If the length and breadth are 10 cm and 8
cm, find the height.
(Text Book Page No. 117)
Sol.
Volume of the solid box =
1600
=
Length × Breadth × Height
10 × 8 × Height
Height =
1600
10 × 8
=
20 cm
Ans.
3. A toffee is 8 cm long, 3 cm broad and 2 cm in height. Find its volume.
(Text Book Page No. 117)
Sol.
Length
=
8 cm
Breadth
=
3 cm
Height
=
2 cm
Volume of the toffee
=
Length × breadth × height
=
(8 × 3 × 2) cm3
=
48 cm3
Runway
Ans.
155
Math-O-Mania– 5
4. Which has a greater volume, a cube having edge 6 cm or a cuboid with
dimensions : l = 7.5 cm, b = 6 cm, h = 2 cm ?
(Text Book Page No. 117)
Sol.
Volume of the cube
=
side × side × side
Side
=
6 cm
\ Volume
=
(6 × 6 × 6) cm3
=
216 cm3
=
length × breadth × height
=
(7.5 × 6 × 2) cm3
=
90 cm3
Volume of the cuboid
Thus the cube has more volume.
Ans.
5. A tank is 0.20 m long, 1.5 m broad and 0.3 m high. Find its volume.
(Text Book Page No. 117)
Sol.
Length
=
0.20 m
Breadth
=
1.5 m
Height
=
0.3 m
Volume of the tank
=
length × breadth × height
=
(0.20 × 1.5 × 0.3) m3
=
0.09 m3
Ans.
6. The length, breadth and height of classroom are 5 m, 4.5 m and 3 m. Find the
volume of the air present in the room.
(Text Book Page No. 117)
Sol.
Length
=
5m
Breadth
=
4.5 m
Height
=
3m
Volume of the classroom
=
length × breadth × height
=
(5 × 4.5 × 3) m3
=
67.5 m3
So the volume of the air present in room = 67.5 m3.
7. Fill up the blanks :
(A)
Ans.
(Text Book Page No. 117)
In case of cube :
Side
Volume
(b) 10 cm
778.688 cm3
1000 cm3
(c) 18.5 m
6331.625 m3
(a) 9.2 cm
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Math-O-Mania– 5
(B)
In case of cuboid :
Length
Breadth
Height
(a) 3 m
2m
1m
Volume
6 m3
(b) 7 m
5m
3m
105 m3
(c) 40 cm
30 cm
20 cm
24000 cm3
Mental Maths (Text Book : Page No 118)
Ans : (Do yourself)
Lab Activity (Text Book : Page No 119)
Ans : (Do yourself)
Fun Activity (Text Book : Page No 119)
Ans : (Do yourself)
21.
1.
Data Handling
Exercise 21
The following table gives the birthday of 50 children in a class :
Days
Number of
Children
Monday
Tuesday Wednesday Thursday
7
5
Friday
Saturday
Sunday
4
13
9
10
2
Represent the above information by a pictograph.
(Text Book Page No. 123)
Ans. Do yourself
2. The following table shows the family budget of Mr Manoj :
Item
Food
Clothing
Education
House rent
Other
expenses
Savings
Cost
(in `)
3500
1000
2000
1500
500
2500
Represent the above information by a bar graph.
(Text Book Page No. 123)
Ans. Do yourself
3. The following table gives the number of children in various classes in school :
Runway
Class
I
II
III
IV
V
VI
VII
VIII
Number of
Students
100
95
115
90
85
105
95
110
157
Math-O-Mania– 5
Represent the above information by a pictograph and answer the following
questions :
(Text Book Page No. 123)
Ans. (a) How many students are in class V ?
85
(b) In which class, 105 students are studying ?
VI
(c) In which two classes same number of students study ?
II and VII
(d) How many more students study in class III than V ?
30
(e) How many total students study in the school ?
795
4. The following table gives the number of children taking part in different games in
a school :
Games
Cricket
Football
Chess
Badminton
Table
Tennis
Tennis
No. of
Children
120
175
25
90
80
45
Draw a bar graph to represent this data.
(Text Book Page No. 123)
Ans. Do yourself
Population in millions
5. The following bar graph represents the population of a city in different years :
50
45
40
35
30
25
20
15
10
5
0
2002
2003
2004
2005
Years
2006
2007
2008
Read the bar graph and answer the following questions :
2009
(Text Book Page No. 124)
Ans. (a) What was the population of the city in the year 2006 ?
25 millions
(b) How much population is increased in the year 2005 from the year 2004 ?
5 millions
(c) What is the difference in population in 2003 and 2008 ?
25 millions
(d) Is the population of the city increasing regularly ?
Yes
6. The following table shows the favourite fruits of 25 students of class V :
(Text Book Page No. 124)
Geeta
Kamal
Runway
Banana
Banana
Shyam
Suresh
Apple
Banana
158
Geeta
Tarun
Orange
Banana
Julie
Ranjan
Guava
Apple
Math-O-Mania– 5
Manoj
Komal
Raman
Kareena
Meera
(i)
Apple
Orange
Banana
Guava
Apple
Seema
Nitu
Suman
Anant
Orange
Guava
Apple
Orange
Rajesh
Priya
Naresh
Ashu
Apple
Banana
Banana
Grapes
Kavita
Mayur
Gopal
Ashok
Grapes
Orange
Apple
Banana
Represent the above data in tabular form using tally marks.
Ans. Do yourself
(ii)
Answer the following questions :
(a) How many students like banana ?
8 students
(b) How many students like grapes ?
2 students
(c) Which is the favourite fruit of most students ?
Banana
(d) Which fruit is liked by the least students ?
Grapes
(e) How many more students like apples than grapes ?
5 students
7. The following pictograph represent the number of animals in the forest. If one
picture stands for 10 animals, then answer the questions that follow :
(Text Book Page No. 124)
Tigers
Foxes
Deers
Elephants
Monkeys
(a) Which animal is least in number ?
Tiger
(b) Which animal is maximum in number ?
Monkey
(c) How many elephants are there in the forest ?
30
(d) Which animal is 50 in number in the forest ?
Deer
(e) Is their any animal which is present in more number than monkeys ?
No
Lab Activity (Text Book : Page No 125)
Ans : (Do yourself)
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Math-O-Mania– 5
22.
Patterns
Exercise 22.1
1. Study each pattern, find the rule and fill in the missing numbers :
(Text Book Page No. 127)
Ans. (a)
1
3
5
7
9
11
13
15
17
(b)
4
11
18
25
32
39
46
53
60
(c)
3
6
9
12
15
18
21
24
27
(d)
2
6
18
54
162
486
1458
4374
13122
(e)
95
87
79
71
63
55
47
39
31
2. Complete the following patterns :
Ans. (a)
(Text Book Page No. 127)
(b)
46
5
10
127
3
59
14
34
16
63
7
50
23
31
22
9
(b)
6
7
5
8
68
1
40
28
3. (a)
(c)
255
4
8
35
72
(c)
5
32
128
53
23
79
45
42
9
7
56
75
72
21
11
6
168
75
7
9
41
15
2
6
22
3
1
72
90
349
274
Exercise 22.2
1. Observe the following pattern and extend it by two steps :
Sol.
Runway
1
×
9
+
2
=
11
12
×
9
+
3
=
111
123
×
9
+
4
=
1111
1234
×
9
+
5
=
11111
12345
×
9
+
6
=
111111
160
(Text Book Page No. 128)
Math-O-Mania– 5
2. Study the following pattern and continue it :
Sol.
(Text Book Page No. 128)
1
×
8
+
1
=
9
12
×
8
+
2
=
98
123
×
8
+
3
=
987
1234
×
8
+
4
=
9876
12345
×
8
+
5
=
98765
3. Observe the following pattern and fill in the blanks :
Sol.
1
=
1
(1 × 1)
1+3
=
4
(2 × 2)
1+ 3 +5
=
9
(3 × 3)
1+3 +5 +7
=
16
(4 × 4 )
1+3+5+7+9
= 25
(5 × 5)
1 + 3 + 5 + 7 + 9 + 11 + 13
= 49
(7 × 7 )
(Text Book Page No. 128)
4. Observe the pattern, find the rule an extend it by two steps : (Text Book Page No. 128)
Sol.
(2 × 2) – (1 × 1)
=
3
(2 + 1)
(3 × 3) – (2 × 2)
=
5
(3 + 2)
(4 × 4) – (3 × 3)
=
7
(4 + 3)
(5 × 5) – (4 × 4)
=
9
(5 + 4)
(6 × 6) – (5 × 5)
=
11
(6 + 5)
5. Study the pattern and fill in the blanks :
Sol.
(Text Book Page No. 129)
1+2+3+4
=
10
(2 +
3) ×
2
=
10
2+3+4+5
=
14
(3 +
4) ×
2
=
14
3+4+5+6
=
18
(4 +
5) ×
2
=
18
4+5+6+7
= 22
(5 +
6) ×
2
=
22
25 + 26 + 27 + 28
= 106
(26 +
27) ×
2
= 106
226 + 227 + 228 + 229
= 910
(227 + 228) ×
2
= 910
6. Find the rule in the following pattern and write next step :
Sol.
Runway
11 × 11
= 121
1×1
1+1
1
21 × 21
= 441
2×2
2+2
1
31 × 31
= 961
3×3
3+3
1
41 × 41
= 1681
4×4
4+4
1
161
(Text Book Page No. 129)
Math-O-Mania– 5
7. Observe the following pattern and fill in the blanks :
Sol.
111
÷
3
= 37
222
÷
6
= 37
333
÷
9
= 37
444
÷
12
= 37
555
÷
15
= 37
888
÷ 24
= 37
(Text Book Page No. 129)
Exercise 22.3
Study each pattern and continue the sequence :
Sol.
(Text Book Page No. 129)
(a)
(b)
(c)
(d)
(e)
(f)
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Math-O-Mania– 5