Real numbers class 10 questions with solutions

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)

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

Topic Question Set

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:

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Topic Question Set

Q 11 : The sum of exponents of prime factors in the prime-factorisation of 196 is: .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Q 13 : The LCM of two prime numbers p and q (p>q) is 221. Find the value of 3p−q. .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Q 14 : If LCM(x, 18) = 36 and HCF(x, 18) = 2, then x is: .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Q 18 : If the LCM of P and 18 is 36 and the HCF of P and 18 is 2, then P equals: .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Q 19 : If HCF (26, 169) = 13, then LCM (26, 169) equals: .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Q 20 : The ratio between the LCM and HCF of 5, 15, 20 is: .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

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Q 11 :

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

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 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: