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1. Real Numbers
Exercise 1.1
One Mark Questions
1. Write all the possible remainders that may be left if a positive integer p
is divided by 3.
Sol. The possible remainders may be 0, 1 or 2.
2. For two given positive integers p and q such that p = m × q + n, where m
and n are unique integers, write the relation between q and n.
Sol. According to Euclid’s division lemma, 0 ≤ n < q.
3. What is the general form of every positive odd integer in terms of some
integer p ?
Sol. 2p + 1
4. A positive integer n when divided by 9, gives 7 as remainder. What will
be the remainder when (3n – 1) is divided by 9 ?
Sol. n = 9m + 7, where m is an integer.
Now, 3n – 1 = 3 [9m + 7] – 1 = 27m + 21 – 1 = 27m + 20
= 27m + 18 + 2 = 9 (3m + 2) + 2
So, when 3n – 1 is divided by 9, the remainder is 2.
Two Marks Questions
5. Show that any positive odd integer is of the form 4m + 1 or 4m + 3,
where m is some integer. (2014-GG-RO-103; 2013-6N1LCY2, KW5N3TF, 036LAAB, BP63NCD)
Sol. Let p be any odd positive integer and q be 4. By Euclid’s division lemma, there
exists m and r such that
p = 4m + r, where 0 ≤ r < 4
⇒ p may be equal to 4m, 4m + 1, 4m + 2 or 4m + 3 according as r is 0, 1, 2 or 3
respectively.
But, p is an odd integer. Therefore, p cannot be equal to 4m or 4m + 2.
So, p = 4m + 1 or p = 4m + 3
Therefore, any odd integer is of the form 4m + 1 or 4m + 3.
6. Use Euclid division algorithm to find if the following pair of numbers is
co-prime : 121, 573.
(2015-0OUM5KY, 210HJIN, P9FEYXB)
Sol. We have :
573 = 121 × 4 + 89
121 = 89 × 1 + 32
89 = 32 × 2 + 25
32 = 25 × 1 + 7
25 = 7 × 3 + 4
7=4×1+3
4=3×1+1
3=1×3+0
So, HCF of 573 and 121 is 1.
Hence, 121 and 573 are co-prime.
(1)
7. State Euclid division lemma. If Euclid lemma is used for a < b as
a = bq + r, then which of a, b, q or r is necessarily zero. (2015-SBYU6F6, VK07P2M)
Sol. In this case, q should be zero, because a < b. For example, when 9 is divided by
13 (9 < 13), the quotient must be zero.
8. Use Euclid’s division algorithm to find HCF of 550 and 155. (2015-VK07P2M)
Sol. We have :
550 = 155 × 3 + 85
155 = 85 × 1 + 70
85 = 70 × 1 + 15
70 = 15 × 4 + 10
15 = 10 × 1 + 5
10 = 5 × 2 + 0
RELIABLE
So, HCF of 550 and 155 is 5.
9. Show that of the numbers n,ENTERPRISES
n + 2 and n + 4, only one of them is divisible
(2015-N3F53UQ, PPPTE8F; 2014-ROTRAI; 2013-2Y6FRII; 2012-40, 45, 55, 63, 69)
by 3.
Sol. Let n be any positive integer. Then,
n = 3q or 3q + 1 or 3q + 2
If n = 3q, then n = 3q is divisible by 3, n + 2 = 3q + 2 is not divisible by 3 and also
n + 4 = 3q + 4 = 3 (q + 1) + 1 is not divisible by 3.
If n = 3q + 1, then n = 3q + 1 is not divisible by 3, n + 2 = 3q + 1 + 2 = 3(q + 1) is
divisible by 3 and n + 2 = 3q + 1 + 4 = 3 (q + 1) + 2 is not divisible by 3.
If n = 3q + 2, then n = 3q + 2 is not divisible by 3, n + 2 = 3q + 2 + 2 = 3(q + 1) + 1
is not divisible by 3 and n + 4 = 3q + 2 + 4 = 3 (q + 2) is divisible by 3.
Thus, only one, out of n, n + 2 and n + 4, is divisible by 3.
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Three Marks Questions
FR
AD
10. Find HCF of numbers 134791,
6341 and 6339 by Euclid’s division
DOWNLO
E
algorithm.
(2014-51XEUPE)
E
Sol. We first find the HCF of 134791 and 6341 by Euclid’s division algorithm. We
have :
RELIABLE
134791 = 6341 × 21 + 1630
ENTERPRISES
6341 = 1630 × 3 + 1451
1630 = 1451 × 1 + 179
1451 = 179 × 8 + 19
179 = 19 × 9 + 8
19 = 8 × 2 + 3
8 =3×2+2
3 =2×1+1
2 =1×2+0
So, HCF of 134791 and 6341 is 1.
Now, for HCF of 1 and 6339, we have :
6339 = 1 × 6339 + 0
So, 1 is the HCF of 1 and 6339.
Hence, 1 is the HCF of the three given numbers.
Copyrighted Material
11. Prove that the square of any positive integer cannot be of the form
6m + 2 or 6m + 5 for any integer m.
(2015-WC7M2LC, 4CAELXQ, 4A8SK09)
Sol. Let a be a positive integer.
So, by Euclid’s division lemma, we can write
a = 6q + r, where r = 0, 1, 2, 3, 4 or 5 (when a is divided by 6)
When r = 0, a = 6q
⇒
a2 = (6q)2 = 6 × 6q2, which is of the form 6m, where m = 6q2.
When r = 1, a = 6q + 1
⇒
a2 = (6q + 1)2 = 36q2 + 12q + 1
= 6 (6q2 + 2q) + 1, which is of the form 6m + 1, where m = 6q2 + 2q.
When r = 2, we have :
a = 6q + 2
⇒
a2 = (6q + 2)2 = 36q2 +RELIABLE
24q + 4
2
ENTERPRISES
= 6 (6q + 4q) + 4, which is of the form 6m + 4, where m = 6q2 + 4q.
When r = 3, a = 6q + 3
⇒
a2 = (6q + 3)2 = 36q2 + 36q + 9
= 6 (6q2 + 6q + 1) + 3, which is of the form 6m + 3,
where m = 6q2 + 6q + 1.
When r = 4, a = 6q + 4
⇒
a2 = (6q + 4)2 = 36q2 + 48q + 16 = 36q2 + 48q + 12 + 4
= 6 (6q2 + 8q + 2) + 4, which is of the form 6m + 4,
where m = 6q2 + 8q + 2.
When r = 5, a = 6q + 5
⇒
a2 = (6q + 5)2 = 36q2 + 60q + 25 = 36q2 + 60q + 24 + 1
= 6 (6q2 + 10q + 4) + 1, which is of the form 6m + 1,
where m = 6q2 + 10q + 4.
From the above, it can be seen that the
square of a positive integer can be of the
OWN
form 6m or 6m + 1 or 6m + 3 or 6m + E4 D
only. LO
E
Hence, it cannot be of the form 6m + 2 or 6m + 5.
12. Use Euclid’s division lemma to show that the square of any positive
integer is either of the form 3m orRELIABLE
3m + 1 for some integer m.
AD
FR
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ENTERPRISES
(2014-0MPO8HFB; 2013-MAVDEL; 2012-21, 35, 46, 64, 68; 2011-560015, 560026, 560036;
2010-1040110-A1, 1040119-A1, 1040117-B1, B2, 1040119-C1, C2)
Sol. Let a be a positive integer. Then, it can be expressed as 3q or 3q + 1 or 3q + 2.
Now, we have to show that the square of each of these can be written in the form
3m or 3m + 1.
If a = 3q, we have: a2 = (3q)2 = 3 × 3q2
= 3m, where m = 3q2.
If a = 3q + 1, then a2 = (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1
= 3m + 1, where m = 3q2 + 2q.
2
If a = 3q + 2, then a = (3q + 2)2
= 9q2 + 12q + 4
= 3(3q2 + 4q + 1) + 1
= 3m + 1, where m = 3q2 + 4q + 1.
Thus, a2 is either of the form 3m or 3m + 1.
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13. Use Euclid’s division lemma to show that the cube of any positive integer
is of the form 9m, 9m + 1 or 9m + 8.
(2015-SU8ZO6N, VK07P2M; 2014-DAVNHEM; 2012-49, 51, 56, 69; 2011-560012, 560013,
560018, 560025, 560028, 560029, 560030; 2010-1040109-A1, 1040119-B1, 1040123-C1)
Sol. Let a be any positive integer. Then it is of the form 3p, 3p + 1 or 3p + 2. Now,
we have to show that the cube of each of these can be expressed in the form 9m, 9m + 1
or 9m + 8.
If a = 3p, then
a3 = (3p)3 = 9(3p3) = 9m, where m = 3p3.
If a = 3p + 1, then a3 = (3p + 1)3 = (3p)3 + 3(3p)2.1 + 3(3p).12 + 13
= 27p3 + 27p2 + 9p + 1
= 9(3p3 + 3p2 + p) + 1
= 9m + 1, where m = 3p3 + 3p2 + p.
RELIABLE
If a = 3p + 2, then a3 = (3p + 2)3
ENTERPRISES
= (3p)3 + 3(3p)2.2 + 3(3p).22 + 23
= 27p3 + 54p2 + 36p + 8
= 9(3p3 + 6p2 + 4p) + 8
= 9m + 8, where m = 3p3 + 6p2 + 4p.
Thus, a3 is either of the form 9m, 9m + 1 or 9m + 8 for some integer m.
14. Use Euclid’s division lemma to show that any positive odd integer is of
the form 6q + 1, 6q + 3 or 6q + 5, where q is some integer.
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(2015-YX1GOMQ, R8FAAGS, P9KKWON; 2014-8OA1MNR; 2013-DGS, Set B, A9EUJ82;
2012-48, 60; 2011-560021; 2010-1040117-C1, 1040125-A1)
FR
AD
Sol. Let a be any positive integer and b = 6. Then, by Euclid’s division lemma,
a = 6m + r, 0 ≤ r < 6 and so the possible remainders are 0, 1, 2, 3, 4 and 5.
That is, a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is the
quotient.
DOWNLO
If a = 6q or 6q + 2 or 6q + 4, then
EE a is an even integer.
And, if a = 6q + 1 or 6q + 3 or 6q + 5, then a is an odd integer.
Also, an integer can either be even
or odd.
RELIABLE
∴ Any positive odd integer is of the
form
6q + 1 or 6q + 3 or 6q + 5, where q is some
ENTERPRISES
integer.
15. Find the HCF of numbers 72 and 96 by Euclid’s division algorithm and
express it in the form 96m + 72n, where m and n are integers. (2015-OMJMARQ)
Sol. Here, 96 > 72. So, by Euclid’s division algorithm, we have :
96 = 72 × 1 + 24
72 = 24 × 3 + 0
So, HCF is 24.
This HCF 24 is of the form 96 × 1 + 72(–1), i.e., of the form 96m + 72n, where m = 1
and n = –1.
Practice Exercise
Two Marks Question
1. Show that square of any positive odd integer is of the form 8k + 1, where k is an
integer.
(2015-DE77B0O, V9RXEIN; 2014-A9USG3O; 2013-D7VSV63)
Copyrighted Material
Three Marks Questions
2. Show that any positive even integer is of the form 6m, 6m + 2 or 6m + 4, where
m is some integer.
(2015-6OJAT3P; 2014-BIDKAR; 2012-57; 2010-1040117-C2)
3. Prove that n2 + n is divisible by 2 for any positive integer n.
(2014-51XEUPE)
4. Show that any positive odd integer is of the form 8q + 1 or 8q + 3 or 8q + 5 or
8q + 7.
(2012-43; 2011-560023)
2
5. Show that n leaves the remainder 1 when divided by 8, where n is an odd
positive integer.
(2014-92QIRIA)
6. Show that every positive even integer is of the form 2q and that every positive
odd integer is of the form 2q + 1 for some integer q.
(2015-IQ4KULN)
7. Use Euclid’s algorithm to find the HCF of 408 and 1032.
(2015-P9FEYXB)
RELIABLE
Answers
ENTERPRISES
7. 24
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Exercise 1.2
One Mark Questions
1. The LCM of 12 and 24 is :
(a) 12
(b) 72
Sol. We have : 12 = 22 × 3
and
24 = 23 × 3
So, LCM of 12 and 24 is 23 × 3 = 24(c)
(2014-0MPO8HFB)
(c) 24
(d) 144
FR
4
AD
2. If LCM (77, 99) = 693, then HCF (77, 99) is :
(2014-DAVNHEM)
(a) 11
(b) 7
(c) 9
(d) 22
77 × 99
77 × 99
= OWN = 11 (a)
Sol. HCF (77, 99) =
D
LCM(77, 99)E 693 LO
E
3. What is the exponent of 2 in the prime factorisation of 336 ?
RELIABLE
(2014-DAVNHEM)
Sol. We have : 336 = 2 × 3 × 7
ENTERPRISES
So, exponent of 2 is 4.
4. Which of the following numbers is a prime number ?
(2014-MPSHEM)
(a) 105
(b) 109
(c) 117
(d) 119
Sol. We have :
105 = 7 × 3 × 5, 109 = 1 × 109, 117 = 3 × 3 × 13 and 119 = 7 × 17.
So, (b) 109 is a prime number.
Two Marks Questions
5. Check whether 8n can end with the digit 0 for any natural number n.
(2015-0MJMARQ, F04CK4V, YR77QHS)
Sol. If the number 8n, for any natural number n, is to end with digit 0, then it
would be divisible by 2 and 5. That is, the prime factorisation of 8n would contain the
n
primes 2 and 5. This is not possible because 8 = (23)n = (2 × 2 × 2)n.
Copyrighted Material
Practice Exercise
Two Marks Questions
1. Check whether 15n can end with the digit zero (0) for any natural number n.
(2015-P9KKWON; 2012-63; 2011-560027, 560034; 2010-1040119-B1)
2. Find the LCM of 90 and 225 by Prime Factorisation method.
(2015-5QEFK0S)
3. Find the smallest number divisible by numbers from 2 to 9 (both inclusive).
(2015-5U7RUFD, 6WIM19E, FN6I2G3)
4. Show that the numbers 231 and 396 are not co-prime. (2015-ASPC3C9; 2012-74)
5. Find the least positive integer which when diminished by 5 is exactly divisible
by 36 and 54.
(2015-CU1G7T6, 55LGL7Q; 2014-3EK66RB)
6. State fundamental theorem of arithmetic. Explain it for the number 1176.
o
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RELIABLE
(2015-7ZGXP6Y, P9KKWON)
(2015-53NSA59)
7. Find HCF of the numbers 31, ENTERPRISES
310 and 3100.
8. Find HCF and LCM of 12, 63 and 99 using prime factorisation method.
(2015-BXMWD0D, JYWB6DQ)
Three Marks Questions
9. Find HCF of 306 and 657. Also, find their LCM using their HCF. (2015-DE77B0O)
10. If the HCF of 144 and 180 is expressed in the form 13m – 3, find the value of m.
(2015-DIVHEM, YHM6P3R)
11. Show that 12n cannot end with digits 0 or 5 for any natural number n.
(2015-6WIM19E, 0MJMARQ)
3
12. Prove that for any positive integer n, n – n is divisible by 6.
(2015-OVMHRSI, PRX7O9M; 2014-9SY5YW6; 2011-560022)
13. Write 32875 as a product of prime factors. Is this factorisation unique ?
(2015-0VIVP2T, 4O33TZ9, 158JKB3; 2014-JGNVVPT)
FR
AD
W L
14. Explain whether the number E3D×O5 N
× 13
O × 46 + 23 is a prime number or a
E
composite number.
(2015-N3F53UQ; 2014-WSOLZ33)
15. Three alarm clocks ring at intervals of 4, 12 and 20 minutes respectively. If they
RELIABLE
start ringing together, after how
much time will they next ring together ?
ENTERPRISES
(2015-DQ9KL4J, SD9P51Z, HMJI3HE)
16. During a sale, colour pencils were being sold in packs of 24 each and crayons in
packs of 32 each. If you want full packs of both and the same number of pencils
and crayons, how many minimum number of each would you need to buy ?
(2015-HAV0HQY)
17. Find the greatest number of six digits exactly divisible by 18, 24 and 36.
(2015-ZN6473Z)
Four Mark Questions
18. An army group of 308 members is to march behind an army band of 24 members
in a parade. The two groups are to march in the same number of columns. What
is the maximum number of columns in which they can march ? (2015-0OUM5KY)
19. Show that n2 – 1 is divisible by 8, if n is an odd positive integer. (2015-5LLSNDW)
Copyrighted Material
VALUE BASED QUESTIONS
Q. 1. There are 156, 208 and 260 students in groups A, B and C respectively.
Buses are to be hired to take them to a deserted land near a colony, where
people used to throw garbage. These students decided to develop a pond there.
(a) Find the minimum number of buses to be hired if students of each group
are accommodated separately and same number of students should be
accommodated in each bus.
(b) Which value is depicted by the students ?
Sol. (a) For minimum number of buses, maximum number of students should be
accommodated in each bus.
So, we find the HCF of 156, 208 and 260.
RELIABLE
Now, 156 = 22 × 3 × 13
ENTERPRISES
208 = 24 × 13
260 = 22 × 5 × 13
So,
HCF = 22 × 13 = 52
Hence, required minimum number of buses
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156 208 260
+
+
52
52
52
= 3 + 4 + 5 = 12
(b) Value : To keep the places in our neighbourhood neat and clean.
Q. 2. Priya and Ravish planned to participate in a cycle race to be organised
for national integration. They decided to practice on circular path around a
sports field. Priya takes 18 minutes to complete one round, while Ravish takes
12 minutes for the same. Suppose they both start at the same time and go in the
DOWNLO
same direction.
EE
After how many minutes will they meet again at the starting point ?
What is the value depicted by Priya and Ravish ?
Sol. For LCM of 18 and 12
ENTERPRISES
18 = 2 × 32
12 = 22 × 3
So, LCM = 22 × 32 = 36
Thus, they will meet again at the starting point after 36 minutes.
Value : Priya and Ravish believe in National Integration, without any gender bias.
Q. 3. In Sports Day activities of a school, three cyclists start together and
can cycle 48 km, 60 km and 72 km a day round the field. The field is circular,
whose circumference is 360 km.
(a) After how many rounds they will meet again ?
(b) Sports activities are important in day to day life, why? Write two reasons.
AD
FR
=
RELIABLE
(2014-ZZDRO103)
Sol. (a) LCM of 48 km, 60 km and 72 km.
48 = 24 × 3
Copyrighted Material
Practice Paper – 1 (Solved)
MATHEMATICS
Time Allowed : 3 hours
CLASS–X
Maximum Marks : 90
General Instructions :
(i) All questions are compulsory.
(ii) The question paper consists of 31 questions divided into four sections A, B, C and D.
Section–A comprises of 4 questions of 1 mark each; Section–B comprises of 6 questions
of 2 marks each; Section–C comprises of 10 questions of 3 marks each and Section–D
comprises of 11 questions of 4 marks each.
(iii) There is no overall choice in this question paper.
(iv) Use of calculator is not permitted.
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RELIABLE
SECTION
:A
ENTERPRISES
Question numbers 1 to 4 carry one mark each.
1. If tan θ =
4
, then find the value of sin θ + cos θ.
3
tan θ =
Sol.
⇒
If in ΔABC,
AB =
BC =
AC2 =
=
So,
AD
Hence,
3x
3
4x
= 4 and cos θ =
=
5x
5
5x
5
OWNL
D
O
E
sin θ + cos θ = 4 + 3E = 7 .
5 5 5
sin θ =
FR
Therefore,
4
3
4x, then
3x.
16x2 + 9x2
25x2 ⇒ AC = 5x
2. Find the value of
RELIABLE
sin 25
tan
23
.
+ ENTERPRISES
cos 65
cot 67
Sol.
sin 25°
tan 23°
sin 25° tan 23°
+
=
+
cos(90° − 25°) cot(90° − 23°)
cos 65° cot 67°
⇒
sin 25° tan 23°
= 1+1=2
+
sin 25° tan 23°
4. In a frequency distribution, if a = assumed mean = 55, Σfi = 100, h = 10 and
Σfi ui = – 30, then find the mean of the distribution.
Sol.
(− 30) × 10
Σfiui
× h = 55 +
100
Σfi
= 55 – 3 = 52.
Mean = a +
Contd...
Copyrighted Material