Download maths in minutes - Tarun Publications

S.D. Sharma
Contents
1. Review
3
2. Roman Numerals
8
3. 8 and 9–Digit Numbers
8
4. The Four Basic Operations
12
5. Estimation
19
6. Multiples and Factors
24
7. Highest Common Factor (H.C.F) and
Lowest Common Multiple (L.C.M)
29
8. Fractions
36
9. Fundamental Operations On Fractions
46
10. Decimals
74
11. Average
97
12. Measurment
100
13. Time
107
14. Bills
124
15. Fundamental Concepts of Geometry
125
16. Angles
126
17. Triangles
129
18. Circles
131
19. Perimeter and Area
135
20. Volume
139
21. Data Handling
142
2
Review
Exercise 1.1
1.
5–V
100 – C
50 – L
1000 – M
500 – D
5. (a) 613250 = 600000 + 10000 + 3000 + 200 + 50 + 0
(b) 2122230 = 2000000 + 100000 + 20000 + 2000 + 200 + 30 + 0
6. (a) 6,00,000 + 60,000 + 6,000 + 600 + 60 + 6 = 666666
(b) 30,00,000 + 50,000 + 4,000 + 500 + 50 + 9 = 3054559
7. (a) The place value of 8 in 5,68,402 = 8,000.
Solve parts (b) to (d) as part (a).
8. Place values of 8 in number 83,892 are :
83,892
800
80,000
80,000 and 800
So, the difference = 80,000 – 800 = 79,200
9. Place value of 6 in 3,26,328 is 6,000
Face value of 6 in 3,26,328 is 6
Difference = 6,000 – 6 = 5,994
10. (a) 6,38,251 > 2,80,358
(b) 4,60,300 < 4,60,303
(c) 9,08,638 > 9,00,309
(d) 7,53,290 = 7,53,290
11. (a) The successor of 3,24,250 is 3,24251
(b) The successor of 6,05,032 is 6,05,033
(c) The successor of 8,51,936 is 8,51,937
12. (a) The predecessor of 9,61,286 is 9,61,285
(b) The predecessor of 6,30,650 is 6,30,649
(c) The predecessor of 8,50,931 is 8,50,930
15. To write smallest number, the smallest digit is written extremely left and
then other digits in ascending order.
3
But 0 is never written at first; it is written on second number. So smallest
number = 102358
Thus, in order to write greatest number from given digits, the greatest
digit is written extremely left and then other digits in descending order.
So, the greatest number = 853210
16. (a)
6
+
6
2
3
2
4
3
0
0
5
1
3
8
7
1
5
0
7
7
7
8
6
(b)
+
3
4
2
1
0
8
0
3
9
3
2
2
2
1
1
0
6
3
8
3
7
9
5
6
3
7
3
2
0
5
6
9
17. Number of soaps produced by a factory = 21,639
Number of soaps produced by another factory = 35,281
Total number of soaps produced by the two factories
= 21,639 + 35,281 = 56920
18. Greatest five-digit number = 99999
Smallest six-digit number
= 100000
On adding the two numbers, we get
9 9
9
9 9
+ 1 0 0
0
0 0
1 9 9
9
9 9
19. (a)
3
– 2
3
9
4
5
3
1
0
7
2
8
9
9
(b)
9
– 6
2
0
8
1
8
9
8
0
3
7
20. Total income of the family = ` 3,20,659 per year
Mother earns = ` 1,93,280
Contribution of father = Total income – money earned by mother
= ` 3,20,659 – ` 1,93,280
= ` 1,27,379
= ` 20,000 – ` 18,659
= ` 1,305
20,320; 20,340; 20,360 _____; _____; _____
Here the difference between two consecutive numbers
is 20. So the next three numbers will be 20,380; 20,400; 20,420
21. Required number
22. (a)
(b)
17,525; 17,625, 17,725; _____; _____; _____
Here the difference between two consecutive numbers
is 100. So next three numbers will be 17,825; 17,925; 18,025
4
Exercise 1.2
1. (a)
3
5
8
(c) 9
6
5
5
3
×
6
1
6
6
2
3
5
1
6
8
9
6
5
6
(b)
6
3
2
2
7
×
6
7
2
5
9
3
2
9
2
×
4
1
8
1
3
8
3
2
4
9
4
8
8
8
3
6
3
4
9
7
9
2
1 0
9
1
2
9
9
4
2. When a number is multiplied by 10 the product may be obtained by
puting a zero extreme right side the number. In order to multiply a number
with 100, two zeros are put on the number. Thus three and four zeros are
put in ease of 1000 and 10000 respectively and so on.
(a) 258 × 10
= 2580
(b) 456 × 40
= 456 × 4 × 10
= 18240
(c) 7301 × 100
= 730100
(d) 7632 × 300
= 7632 × 3 × 100 = 2289600
(e) 26429 × 1000 = 26429000
(f) 32108 × 400 = 32108 × 4 × 100
= 128432000
3. ∵ Cost of one DVD player
= ` 3247
∴ Cost of 129 DVD players = ` 3247 × 129
= ` 418863
5
4. (a) 58356 ÷ 27
2161
27 ) 58356
–54
(b) 58638 ÷ 33
1776
33 ) 58638
–33
43
–27
256
–231
165
–162
253
–231
36
–27
228
–198
9
30
quotient = 2161
remainder = 9
quotient = 1776
remainder = 30
(c) Solve as above.
5. ∵ Cost of one chair = ` 83680
Cost of 16 chairs = ` 83680 ÷ 16 = ` 5230
6.
(a) 50
1 × 50 = 50, 2 × 25 = 50, 5 × 10 = 50
Factors of 50 are : 1, 2, 5, 10, 25 and 50.
(b) 55
1 × 55 = 55, 5 × 11 = 55
Factors of 55 = 1, 5, 11 and 55.
(c) 60
1 × 60 = 60, 2 × 30 = 60, 3 × 20 = 60, 4 × 15 = 60,
5 × 12 = 60, 6 × 10 = 60
Factors of 60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
Exercise 1.3
1. (a) 21 km 11 hm 9 dam 7 m
= 21000 m + 1100 m + 90 m + 7 m = 22197 m
(b) 62 hm 26 dam 11 m
= 6200 m + 260 m + 11 m = 6471 m
(c) 10 kg 8 hg 3 dag 5 g
= 10,000 g + 800 g + 30 g + 5 g = 10,835 g
(d) 6 l 6 dl 5 cl 9 ml
= 6,000 ml + 600 ml + 50 ml + 9 ml = 6659 ml
6
2. (a) 675 cm
(b) 6126 mm
(c)
8659 g
(d)
6359 ml
3. (a)
=
=
=
=
=
=
=
600 cm + 75 cm = 6 m 75 cm
6000 mm + 100 mm + 20 mm + 6 mm
6 m 12 cm 6 mm
8000 g + 659 g
8 kg 59 g
6000 ml + 359 ml
6 l 359 ml
BC = 16 cm, AB = 22 cm
Perimeter
(b)
(c)
(d)
= 2 × (AB + BC)
= 2 × (16 + 22)
= 2 × 38 = 76 cm
AB = 19 cm, BC = 10 cm, AC = 11 cm
∴ Perimeter
= AB + BC + AC
= 19 + 10 + 11 = 40 cm
AB = 13 cm
Since ABCD is a square and side a = 13 cm
Perimeter of square = 4 × AB
= 4 × 13 = 52 cm
AB = 5 cm, BC = 7 cm, CD = 3 cm, DE = 8 cm, EA = 6 cm
Perimeter = AB + BC + CD + DE + EA
= 5 + 7 + 3 + 8 + 6 = 29 cm
Exercise 1.4
1. (a)
2 hours after 11 a.m. = 11 + 2 = 13
= 1 p.m.
(b) 4 hours after 9 p.m. = 9 + 4 = 13
= 1 a.m.
(c) 1 hour 50 minutes after 10 : 20 a.m.
= 10 : 20
+ 1 : 50
12 : 10
= 12 : 10 p.m.
(d) 30 minutes before 12 : 50 p.m. = 12 : 50 – 30 minutes = 12 : 20 p.m.
2. Anuj left school at 1 : 40 p.m. He reached home at 2 : 50 p.m. Time taken by
him to reach home = 2 : 50 – 1 : 40
= 1 h 10 minutes
3. (a) 10 days 20 hours = 10 × 24 + 20 hours
= 240 + 20 = 260 hours
7