Real numbers class 10 questions with solutions

Q 1 :

If two positive integers p and q can be expressed as $p = a b^{2}$ and $q = a^{3} b;$ $a, b$ being prime numbers, then LCM (p, q) is

$a b$
$a^{2} b^{2}$
$a^{3} b^{2}$
$a^{3} b^{3}$

1

2

3

4

(3) $a^{3} b^{2}$

Q 2 :

Let a and b be two positive integers such that $a = p^{3} q^{4} and b = p^{2} q^{3}$ , where p and q are prime numbers. If $HCF (a, b) = p^{m} q^{n}$ and $LCM (a, b) = p^{r} q^{s},$ then $(m + n) (r + s) =$

15
30
35
72

1

2

3

4

(3)

$a = p^{3} q^{4} and b = p^{2} q^{3}$

$HCF (a, b) = p^{2} q^{3}$ ...(i)

and $LCM (a, b) = p^{3} q^{4}$ ...(ii)

But gives: $HCF (a, b) = p^{m} q^{n}$ and $LCM (a, b) = p^{r} q^{s}$

From eq. (i), $p^{m} q^{n} = p^{2} q^{3}$

So, $m = 2$ and $n = 3$

From eq. (ii), $p^{r} q^{s} = p^{3} q^{4}$

So, $r = 3$ and $s = 4$

$∴$ $(m + n) (r + s) = (2 + 3) (3 + 4) = 35 .$

Q 3 :

If two positive integers p and q can be expressed as $p = 18 a^{2} b^{4} and q = 20 a^{3} b^{2},$ where a and b are prime numbers, then LCM (p, q) is :

$2 a^{2} b^{2}$
$180 a^{2} b^{2}$
$12 a^{2} b^{2}$
$180 a^{3} b^{4}$

1

2

3

4

(4)

Given, $p = 18 a^{2} b^{4}$ and $q = 20 a^{3} b^{2}$

$LCM (p, q) = LCM (18 a^{2} b^{4}, 20 a^{3} b^{2}) = 180 a^{3} b^{4}$

Q 4 :

Two positive integers m and n are expressed as $m = p^{5} q^{2}$ and $n = p^{3} q^{4}$ , where p and q are prime numbers. The LCM of m and n is :

$p^{8} q^{6}$
$p^{3} q^{2}$
$p^{5} q^{4}$
$p^{5} q^{2} + p^{3} q^{4}$

1

2

3

4

(3) $p^{5} q^{4}$

Q 5 :

If the HCF(2520, 6600) = 40 and LCM(2520, 6600) = 252 × k, then the value of k is

1650
1600
165
1625

1

2

3

4

(1)

Given, HCF = 40 and LCM = 252 × k

We know that, LCM × HCF = Product of two number

$\Rightarrow 40 \times 252 \times k = 2520 \times 6600$

$\Rightarrow k = \frac{2520 \times 6600}{40 \times 252}$

$\Rightarrow k = 1650$

Q 6 :

If the prime factorisation of 2520 is $2^{3} \times 3^{a} \times 𝑏 \times 7$ , then the value of a + 2b is:

12
10
9
7

1

2

3

4

(1)

2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7
$= 2 ³ \times 3 ² \times 5 \times 7$
$C o m p a r i n g w i t h 2520 = 2 ³ \times 3 ᵃ \times b \times 7$
$W e g e t, a = 2, b = 5 ∴ a + 2 b = 2 + 2 \times 5 = 2 + 10 = 12$

Q 7 :

The LCM of the smallest prime number and the smallest odd composite number is:

10
6
9
18

1

2

3

4

(4)

We have, the smallest prime number = 2

and the smallest odd composite number = 9

∴ LCM of the smallest prime number and the smallest odd composite number

= LCM (2, 9) = 18

Q 8 :

If a = $2 ² \times 3 ˣ, b = 2 ² \times 3 \times 5, c = 2 ² \times 3 \times 7$ and LCM (a, b, c) = 3780, then x is equal to:

1
2
3
0

1

2

3

4

(3)

If LCM (a,b,c) = 3780
By prime factorisation of 3780

$3780 = 2^{2} \times 3^{3} \times 5 \times 7 L C M (a, b, c) = 2^{2} \times 3^{3} \times 5 \times 7 (i) a = 2^{2} \times 3^{x}; b = 2^{2} \times 3 \times 5; c = 2^{2} \times 3 \times 7 L C M (a, b, c) = 2^{2} \times 3^{x} \times 5 \times 7 (i i) C o m p a r i n g (i) a n d (i i), w e g e t 𝑥 = 3$

Q 9 :

The LCM of smallest 2-digit number and smallest composite number is:

12
4
20
40

1

2

3

4

(3)

Smallest 2-digit number = 10 and smallest composite number = 4
LCM of 10 and 4 = 20

Q 10 :

The total number of factors of a prime number is:

1
0
2
3

1

2

3

4

(3)

A prime number has only 2 factors, i.e. the number itself and 1.

Q 11 :

The sum of exponents of prime factors in the prime-factorisation of 196 is:

3
4
5
2

1

2

3

4

(2)

$W e h a v e, 196 = 2 ² \times 7 ² \Rightarrow S u m o f e x p o n e n t s = 2 + 2 = 4$

Q 12 :

Let a and b be two positive integers such that $a = p^{3} q^{4} a n d 𝑏 = 𝑝^{2} q^{3}$ where p and q are prime numbers. If HCF (a,b) = $p^{m} q^{n}$ and LCM (a,b) $= p^{r} q^{s}, t h e n$
(m+n) (r+s) =

15
30
35
72

1

2

3

4

(3)

Given, $a = p^{3} q^{4}, b = p^{2} q^{3} ∴ H C F (a, b) = p^{2} q^{3} a n d L C M (a, b) = p^{3} q^{4}$

Comparing with the HCF and LCM given in the question, we get

$H C F (a, b) = p^{m} q^{n} = p^{2} q^{3} \Rightarrow m = 2, n = 3 a n d L C M (a, b) = p^{r} q^{s} = p^{3} q^{4} \Rightarrow r = 3, s = 4 \Rightarrow (m + n) (r + s) = (2 + 3) (3 + 4) = 5 \times 7 = 35$

Q 13 :

The LCM of two prime numbers $p a n d q (p > q)$ is 221. Find the value of 3p−q.

4
28
38
48

1

2

3

4

(3)

p and q (p > q) are prime numbers, HCF (p,q) = 1

Given, LCM (p,q) = 221

$\Rightarrow L C M (p, q) \times H C F (p, q) = p \times q \Rightarrow 1 \times 221 = p \times q \Rightarrow p \times q = 221 S o, p = 17, q = 13 (∵ p > q) 3 p - q = 3 \times 17 - 13 = 51 - 13 = 38$

Q 14 :

If LCM(x, 18) = 36 and HCF(x, 18) = 2, then x is:

2
3
4
5

1

2

3

4

(3)

$G i v e n L C M (x, 18) = 36 a n d H C F (x, 18) = 2 ∴ L C M (x, 18) \times H C F (x, 18) = x \times 18 \Rightarrow 36 \times 2 = x \times 18 \Rightarrow x = (36 \times 2) \div 18 = 4$

Q 15 :

If the sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are:

2
3
4
5

1

2

3

4

(3)

HCF = 81, let the two numbers be 81x and 81y.
According to the question,

81 x +81 y = 1215
x + y = 1215

The gives 4 co-prime pairs (1, 14), (2, 13), (4, 11), (7, 8)

Q 16 :

If two positive integers p and q are written as $p = x^{2} y^{2} a n d 𝑞 = x y^{3}$ , where x and y are prime numbers, then HCF(p, q) is:

xy
$x y^{2}$
$x^{3} y^{3}$
$- x^{2} y^{2}$

1

2

3

4

(2)

We have,

$p = x^{2} y^{2} q = x y^{3} \Rightarrow H C F (p, q) = x y^{2}$

Q 17 :

If HCF and LCM of two numbers are respectively $(n - 1) a n d (n ² - 1) (n ² - 4)$ , then the product of the two numbers will be:

$(n ² - 1) (n ² - 4)$
$(n ² + 1) (n ² - 1) (n ² - 4)$
$(n ² - 4) (n + 1) (n - 1) ²$
$(n ² - 1) (n ² + 1) (n - 1)$

1

2

3

4

(3)

$W e h a v e, H C F = (n - 1) a n d L C M = (n ² - 1) (n ² - 4) W e k n o w t h a t p r o d u c t o f t w o n u m b e r s = L C M \times H C F S o p r o d u c t o f t w o n u m b e r s = (n - 1) (n ² - 1) (n ² - 4) = (n - 1) (n - 1) (n + 1) (n ² - 4) = (n + 1) (n - 1) ² (n ² - 4)$

Q 18 :

If the LCM of P and 18 is 36 and the HCF of P and 18 is 2, then P equals:

2
3
1
4

1

2

3

4

(4)

$W e k n o w t h a t L C M (a, b) \times H C F (a, b) = a \times b$

$\Rightarrow L C M (P, 18) \times H C F (P, 18) = P \times 18$

$\Rightarrow 36 \times 2 = P \times 18 \Rightarrow 72 = 18 P$

$\Rightarrow P = 72 \div 18 = 4$

Q 19 :

If HCF (26, 169) = 13, then LCM (26, 169) equals:

26
52
20
338

1

2

3

4

(4)

$H C F (26, 169) \times L C M (26, 169) = 26 \times 169 (∵ L C M (a, b) \times H C F (a, b) = a \times b)$

$\Rightarrow 13 \times L C M (26, 169) = 26 \times 169$

$\Rightarrow L C M (26, 169) = (26 \times 169) \div 13 = 338$

Q 20 :

The ratio between the LCM and HCF of 5, 15, 20 is:

9 : 1
4 : 3
11 : 1
12 : 1

1

2

3

4

(4)

$W e h a v e, 5 = 1 \times 5; 15 = 3 \times 5 a n d 20 = 2 \times 2 \times 5 ∴ L C M (5, 15, 20) = 3 \times 4 \times 5 = 60 a n d H C F (5, 15, 20) = 5 T h e r e f o r e, r a t i o b e t w e e n t h e L C M a n d H C F = 60 \div 5 = 12 : 1$

Q 21 :

If the product of two numbers a and b is 1152 and HCF of a and b is 12, then LCM of a and b is:

88
90
92
96

1

2

3

4

(4)

$W e h a v e, a \times b = 1152 a n d H C F (a, b) = 12 W e k n o w t h a t L C M (a, b) \times H C F (a, b) = a \times b \Rightarrow L C M (a, b) \times 12 = 1152 \Rightarrow L C M (a, b) = 1152 \div 12 = 96$

Q 22 :

What is the least number that is divisible by all the numbers from 1 to 10?

420
840
2520
3200

1

2

3

4

(3)

For any number to be divisible by all the numbers from 1 to 10, we need LCM of the numbers from 1 to 10. So, it must have maximum powers of each prime number that can be obtained in prime factorisation of numbers from 1 to 10.

$∴ 1 = 1 2 = 2 ¹ 3 = 3 ¹ 4 = 2 ² 5 = 5 ¹ 6 = 2 ¹ \times 3 ¹ 7 = 7 ¹ 8 = 2 ³ 9 = 3 ² 10 = 2 ¹ \times 5 ¹ S o, r e q u i r e d l e a s t n u m b e r = 2 ³ \times 3 ² \times 5 ¹ \times 7 ¹ = 8 \times 9 \times 5 \times 7 = 2520$

Topic Question Set

Q 1 :

If two positive integers p and q can be expressed as $p = a b^{2}$ and $q = a^{3} b;$ $a, b$ being prime numbers, then LCM (p, q) is

Q 2 :

Let a and b be two positive integers such that $a = p^{3} q^{4} and b = p^{2} q^{3}$ , where p and q are prime numbers. If $HCF (a, b) = p^{m} q^{n}$ and $LCM (a, b) = p^{r} q^{s},$ then $(m + n) (r + s) =$

Q 3 :

If two positive integers p and q can be expressed as $p = 18 a^{2} b^{4} and q = 20 a^{3} b^{2},$ where a and b are prime numbers, then LCM (p, q) is :

Q 4 :

Two positive integers m and n are expressed as $m = p^{5} q^{2}$ and $n = p^{3} q^{4}$ , where p and q are prime numbers. The LCM of m and n is :

Q 5 :

If the HCF(2520, 6600) = 40 and LCM(2520, 6600) = 252 × k, then the value of k is

Q 6 :

If the prime factorisation of 2520 is $2^{3} \times 3^{a} \times 𝑏 \times 7$ , then the value of a + 2b is:

Q 7 :

The LCM of the smallest prime number and the smallest odd composite number is:

Q 8 :

If a = $2 ² \times 3 ˣ, b = 2 ² \times 3 \times 5, c = 2 ² \times 3 \times 7$ and LCM (a, b, c) = 3780, then x is equal to:

Q 9 :

The LCM of smallest 2-digit number and smallest composite number is:

Q 10 :

The total number of factors of a prime number is:

Q 11 :

The sum of exponents of prime factors in the prime-factorisation of 196 is:

Q 12 :

Let a and b be two positive integers such that $a = p^{3} q^{4} a n d 𝑏 = 𝑝^{2} q^{3}$ where p and q are prime numbers. If HCF (a,b) = $p^{m} q^{n}$ and LCM (a,b) $= p^{r} q^{s}, t h e n$
(m+n) (r+s) =

Q 13 :

The LCM of two prime numbers $p a n d q (p > q)$ is 221. Find the value of 3p−q.

Q 14 :

If LCM(x, 18) = 36 and HCF(x, 18) = 2, then x is:

Q 15 :

If the sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are:

Q 16 :

If two positive integers p and q are written as $p = x^{2} y^{2} a n d 𝑞 = x y^{3}$ , where x and y are prime numbers, then HCF(p, q) is:

Q 17 :

If HCF and LCM of two numbers are respectively $(n - 1) a n d (n ² - 1) (n ² - 4)$ , then the product of the two numbers will be:

Q 18 :

If the LCM of P and 18 is 36 and the HCF of P and 18 is 2, then P equals:

Q 19 :

If HCF (26, 169) = 13, then LCM (26, 169) equals:

Q 20 :

The ratio between the LCM and HCF of 5, 15, 20 is:

Q 21 :

If the product of two numbers a and b is 1152 and HCF of a and b is 12, then LCM of a and b is:

Q 22 :

What is the least number that is divisible by all the numbers from 1 to 10?

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Topic Question Set

Q 1 : If two positive integers p and q can be expressed as p=ab2 and q=a3b; a,b being prime numbers, then LCM (p, q) is

Q 2 : Let a and b be two positive integers such that a=p3q4 and b=p2q3, where p and q are prime numbers. If HCF(a,b)=pmqn and LCM(a,b)=prqs, then (m+n)(r+s)=

Q 3 : If two positive integers p and q can be expressed as p=18 a2b4 and q=20 a3b2, where a and b are prime numbers, then LCM (p, q) is :

Q 4 : Two positive integers m and n are expressed as m=p5q2 and n=p3q4 , where p and q are prime numbers. The LCM of m and n is :

Q 5 : If the HCF(2520, 6600) = 40 and LCM(2520, 6600) = 252 × k, then the value of k is

Q 6 : If the prime factorisation of 2520 is 23 × 3a × 𝑏 × 7, then the value of a + 2b is:

Q 7 : The LCM of the smallest prime number and the smallest odd composite number is:

Q 8 : If a = 2² × 3ˣ, b = 2² × 3 × 5, c = 2² × 3 × 7 and LCM (a, b, c) = 3780, then x is equal to:

Q 9 : The LCM of smallest 2-digit number and smallest composite number is:

Q 10 : The total number of factors of a prime number is:

Q 11 : The sum of exponents of prime factors in the prime-factorisation of 196 is:

Q 12 : Let a and b be two positive integers such that a=p3 q4 and 𝑏 = 𝑝2 q3 where p and q are prime numbers. If HCF (a,b) = pm qn and LCM (a,b) =pr qs , then (m+n) (r+s) =

Q 13 : The LCM of two prime numbers p and q (p>q) is 221. Find the value of 3p−q.

Q 14 : If LCM(x, 18) = 36 and HCF(x, 18) = 2, then x is:

Q 15 : If the sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are:

Q 16 : If two positive integers p and q are written as p=x2 y2 and 𝑞 = xy3, where x and y are prime numbers, then HCF(p, q) is:

Q 17 : If HCF and LCM of two numbers are respectively (n – 1) and (n² – 1)(n² – 4), then the product of the two numbers will be:

Q 18 : If the LCM of P and 18 is 36 and the HCF of P and 18 is 2, then P equals:

Q 19 : If HCF (26, 169) = 13, then LCM (26, 169) equals:

Q 20 : The ratio between the LCM and HCF of 5, 15, 20 is:

Q 21 : If the product of two numbers a and b is 1152 and HCF of a and b is 12, then LCM of a and b is:

Q 22 : What is the least number that is divisible by all the numbers from 1 to 10?

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Q 1 :

If two positive integers p and q can be expressed as $p = a b^{2}$ and $q = a^{3} b;$ $a, b$ being prime numbers, then LCM (p, q) is

Q 2 :

Let a and b be two positive integers such that $a = p^{3} q^{4} and b = p^{2} q^{3}$ , where p and q are prime numbers. If $HCF (a, b) = p^{m} q^{n}$ and $LCM (a, b) = p^{r} q^{s},$ then $(m + n) (r + s) =$

Q 3 :

If two positive integers p and q can be expressed as $p = 18 a^{2} b^{4} and q = 20 a^{3} b^{2},$ where a and b are prime numbers, then LCM (p, q) is :

Q 4 :

Two positive integers m and n are expressed as $m = p^{5} q^{2}$ and $n = p^{3} q^{4}$ , where p and q are prime numbers. The LCM of m and n is :

Q 5 :

If the HCF(2520, 6600) = 40 and LCM(2520, 6600) = 252 × k, then the value of k is

Q 6 :

If the prime factorisation of 2520 is $2^{3} \times 3^{a} \times 𝑏 \times 7$ , then the value of a + 2b is:

Q 7 :

The LCM of the smallest prime number and the smallest odd composite number is:

Q 8 :

If a = $2 ² \times 3 ˣ, b = 2 ² \times 3 \times 5, c = 2 ² \times 3 \times 7$ and LCM (a, b, c) = 3780, then x is equal to:

Q 9 :

The LCM of smallest 2-digit number and smallest composite number is:

Q 10 :

The total number of factors of a prime number is:

Q 11 :

The sum of exponents of prime factors in the prime-factorisation of 196 is:

Q 12 :

Let a and b be two positive integers such that $a = p^{3} q^{4} a n d 𝑏 = 𝑝^{2} q^{3}$ where p and q are prime numbers. If HCF (a,b) = $p^{m} q^{n}$ and LCM (a,b) $= p^{r} q^{s}, t h e n$
(m+n) (r+s) =

Q 13 :

The LCM of two prime numbers $p a n d q (p > q)$ is 221. Find the value of 3p−q.

Q 14 :

If LCM(x, 18) = 36 and HCF(x, 18) = 2, then x is:

Q 15 :

If the sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are:

Q 16 :

If two positive integers p and q are written as $p = x^{2} y^{2} a n d 𝑞 = x y^{3}$ , where x and y are prime numbers, then HCF(p, q) is:

Q 17 :

If HCF and LCM of two numbers are respectively $(n - 1) a n d (n ² - 1) (n ² - 4)$ , then the product of the two numbers will be:

Q 18 :

If the LCM of P and 18 is 36 and the HCF of P and 18 is 2, then P equals:

Q 19 :

If HCF (26, 169) = 13, then LCM (26, 169) equals:

Q 20 :

The ratio between the LCM and HCF of 5, 15, 20 is:

Q 21 :

If the product of two numbers a and b is 1152 and HCF of a and b is 12, then LCM of a and b is:

Q 22 :

What is the least number that is divisible by all the numbers from 1 to 10?