CBSE Class 10 Maths HOTs Real Numbers Set 07

Multiple Choice Questions  

Question. The HCF and the LCM of 12, 21, 15 respectively are
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 420, 3
Answer: (c) 3, 420
Sol. As \( 12 = 2^2 \times 3 \), \( 21 = 3 \times 7 \), \( 15 = 3 \times 5 \). HCF = 3, LCM = \( 2^2 \times 3 \times 7 \times 5 = 420 \).

Question. Three bulbs red, green and yellow flash at intervals of 80 seconds, 90 seconds and 110 seconds. All three flash together at 8:00 am. At what time will the three bulbs flash altogether again?
(a) 9:00 am
(b) 9:12 am
(c) 10:00 am
(d) 10:12 am
Answer: (d) 10:12 am
Sol. LCM of 80, 90 and 110 = 7920
\( \implies \) 7920 sec = 132 minutes
8:09 am + 132 minutes = 10:12 am

Question. If A = \( 2n + 13 \), B = \( n + 7 \), where \( n \) is a natural number, then HCF of A and B is:
(a) 2
(b) 1
(c) 3
(d) 4
Answer: (b) 1
Sol. Taking different values of \( n \) we find that A and B are coprime. \( \implies \) HCF = 1

Question. There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. The total number of sections thus formed are:
(a) 22
(b) 16
(c) 36
(d) 21
Answer: (b) 16
Sol. HCF of 576 and 448 = 64
Number of sections = \( \frac{576}{64} + \frac{448}{64} = 9 + 7 = 16 \)

Question. The HCF of 2472, 1284 and a third number N is 12. If their LCM is \( 2^3 \times 3^2 \times 5 \times 103 \times 107 \), then the number N is :
(a) \( 2^2 \times 3^2 \times 7 \)
(b) \( 2^2 \times 3^3 \times 103 \)
(c) \( 2^2 \times 3^2 \times 5 \)
(d) \( 2^4 \times 3^2 \times 11 \)
Answer: (c) \( 2^2 \times 3^2 \times 5 \)
Sol. \( 2472 = 2^3 \times 3 \times 103 \), \( 1284 = 2^2 \times 3 \times 107 \)
LCM = \( 2^3 \times 3^2 \times 5 \times 103 \times 107 \)
\( \implies \) N = \( 2^2 \times 3^2 \times 5 = 180 \)

Question. Two natural numbers whose difference is 66 and the least common multiple is 360, are:
(a) 120 and 54
(b) 90 and 24
(c) 180 and 114
(d) 130 and 64
Answer: (b) 90 and 24
Sol. Difference of 90 and 24 = 66
and LCM of 90 and 24 = 360
\( \implies \) Numbers are 90 and 24.

Question. The HCF and LCM of two numbers are 33 and 264 respectively. When the first number is completely divided by 2 the quotient is 33. The other number is
(a) 162
(b) 32
(c) 66
(d) 132
Answer: (d) 132
Sol. First number = \( 2 \times 33 = 66 \)
Other number = \( \frac{\text{HCF} \times \text{LCM}}{\text{1st number}} = \frac{33 \times 264}{66} = 132 \)

Question. The LCM of smallest prime and smallest odd composite natural number is
(a) 6
(b) 12
(c) 18
(d) 24
Answer: (c) 18
Sol. Smallest prime number = 2
Smallest composite odd number = 9
LCM of 2 and 9 = \( 2 \times 9 = 18 \)

Question. Decompose 32760 into prime factors, we get factors as
(a) \( 2^3 \times 3^3 \times 5 \times 7 \times 13 \)
(b) \( 2^3 \times 3^2 \times 5 \times 7 \)
(c) \( 2^3 \times 3^2 \times 5 \times 7 \times 13 \)
(d) \( 2^2 \times 3^3 \times 5 \times 7 \times 13 \)
Answer: (c) \( 2^3 \times 3^2 \times 5 \times 7 \times 13 \)
Sol. \( 32760 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 \times 13 = 2^3 \times 3^2 \times 5 \times 7 \times 13 \)

Question. The sum of exponents of prime factors in the prime factorisation of 250 is
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (b) 4
Sol. \( 250 = 2 \times 5^3 \)
\( \implies \) Sum of exponents = \( 1 + 3 = 4 \)

Question. The missing entries in the following factor tree are

(a) x = 21, y = 42
(b) x = 63, y = 21
(c) x = 21, y = 7
(d) x = 42, y = 21
Answer: (d) x = 42, y = 21

100% Scoring Tips

If question is “Find x, y, z from the diagram”
Then, find z first as “z = 5 × 7 = 35, then y = 3z = 3 × 35 = 105 and proceed”.
Note: But do not solve as “2x = 2y, y = 3z, etc,” it may take more time to solve.

Question. The HCF of smallest prime number and the smallest composite number is
(a) 1
(b) 2
(c) 4
(d) 8
Answer: (b) 2
Sol. The smallest prime number = 2
The smallest composite number = 4
HCF of 2 and 4 = 2

Question. If the prime factorisation of a natural number N is \( 2^4 \times 3^4 \times 5^3 \times 7 \), then number of consecutive zeroes in N are
(a) 5
(b) 4
(c) 3
(d) 2
Answer: (c) 3
Sol. Number of consecutive zeroes = zeroes in \( 2^3 \times 5^3 = \text{zeroes in } (10)^3 = 3 \)

Question. If product of two numbers is 3691 and their LCM is 3691, then their HCF =
(a) 2
(b) 3691
(c) 1
(d) 3
Answer: (c) 1
Sol. HCF = \( \frac{\text{Product of two numbers}}{\text{LCM}} = \frac{3691}{3691} = 1 \)

Question. If p and q are two co-prime numbers, then the HCF and LCM of p and q is
(a) HCF = p, LCM = pq
(b) HCF = 1, LCM = pq
(c) HCF = q, LCM = pq
(d) HCF = pq, LCM = pq
Answer: (b) HCF = 1, LCM = pq
Sol. Since p and q are co-prime numbers
Common factor of p and q = 1
\( \implies \) HCF of p and q = 1
Now, LCM(p, q) = \( \frac{p \times q}{\text{HCF}(p, q)} = \frac{pq}{1} = pq \)

Very Short Answer Type Questions 

Question. Two numbers are in the ratio 21 : 17. If their HCF is 5, find the numbers.
Answer: Let numbers are 21x and 17x.
Now, common factor of 21x and 17x = x
Also HCF = 5
\( \implies \) x = 5
Numbers are \( 21 \times 5 \) and \( 17 \times 5 \) i.e. 105 and 85.

Question. The HCF of two numbers is 29 and other two factors of their LCM are 16 and 19. Find the larger of the two numbers.
Answer: HCF of the two numbers is 29.
Numbers are \( 29 \times a \) and \( 29 \times b \) where a and b are co-prime.
Now other two factors of the LCM are 16 and 19.
LCM = \( 29 \times 16 \times 19 \)
\( \implies 29 \times 16 \times 19 = 29 \times a \times b \)
\( \implies a = 16 \) and \( b = 19 \)
So, larger of the two number is \( 29 \times 19 = 551 \).

Question. Three numbers are in the ratio 2 : 5 : 7. Their LCM is 490. Find the square root of the largest number.
Answer: Let numbers are 2x, 5x and 7x.
LCM of 2x, 5x and 7x = \( 2 \times 5 \times 7 \times x \)
Also LCM = 490
\( \implies 2 \times 5 \times 7 \times x = 490 \)
\( \implies x = 7 \)
So, numbers are \( 2 \times 7 \), \( 5 \times 7 \), \( 7 \times 7 = 14, 35 \) and 49
Largest number = 49
The square root of largest number = \( \sqrt{49} = 7 \)

Question. If least prime factor of a is 5 and least prime factor of b is 13, then what is the least prime factor of a + b?
Answer: Least prime factor of a = 5
\( \implies \) a is odd
Also, least prime factor of b = 11 (Note: correcting source OCR 13/11)
\( \implies \) b is also odd
Now a + b = sum of two odd numbers = even number
\( \implies \) Least prime factor of a + b is 2.

Question. Can we have any \( n \in N \), where \( 7^n \) ends with the digit zero?
Answer: For units digit to be 0, \( 7^n \) should have 2 and 5 as its prime factors, but \( 7^n \) does not contain 2 and 5 as its prime factors. Hence \( 7^n \) will not end with digit 0 for \( n \in N \).

100% Scoring Tips

If question is “Check whether \( 12^n \) can end with digit 0, for any \( n \in N \).”
Then, solve as “\( 12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n \)”
Here, we notice, prime factors are 2 and 3. Hence, \( 12^n \) does not end with 0. Because a number can end with digit zero, if it is divisible by both 2 and 5. So, check if \( 12^n \) has prime factors both 2 and 5.”
Note: But don’t follow as “For \( n = 1, 12^1 = 12 \), does not end with 0; \( n = 2, 12^2 = 144 \), does not end with 0...”. This is not a correct solution.

Question. Given that \( \sqrt{2} \) is irrational, prove that \( (5 + 3\sqrt{2}) \) is an irrational number. 
Answer: Given: \( \sqrt{2} \) is irrational.
To prove: \( 5 + 3\sqrt{2} \) is irrational.
Proof: Let us assume \( 5 + 3\sqrt{2} \) is Rational. So it is in form \( \frac{a}{b} \), \( b \neq 0 \), \( a, b \in \mathbb{Z} \), \( HCF(a, b) = 1 \).
\( 5 + 3\sqrt{2} = \frac{a}{b} \)
\( 3\sqrt{2} = \frac{a}{b} - 5 \)
\( 3\sqrt{2} = \frac{a - 5b}{b} \)
\( \sqrt{2} = \frac{a - 5b}{3b} \)
This shows that \( \sqrt{2} \) is rational (since a, b are integers).
But we know that \( \sqrt{2} \) is irrational.
This contradicts our assumption that \( 5 + 3\sqrt{2} \) is rational.
\( \implies 5 + 3\sqrt{2} \) is irrational, hence proved.

Question. The HCF and LCM of two numbers are 33 and 264 respectively. When the first number is completely divided by 2, the quotient is 33. Find the other number.
Answer: First number = \( 2 \times 33 = 66 \)
Other number = \( \frac{\text{HCF} \times \text{LCM}}{\text{1st number}} = \frac{33 \times 264}{66} = 132 \)

Short Answer Type Questions 

Question. The greatest number that will divide 76, 112, 172 and 184 so as to leave remainder 40 in each case is \( k^2 \times 3 \). Find the value of k.
Answer: Required number is HCF of (76 – 40), (112 – 40), (172 – 40) and (184 – 40) = HCF of 36, 72, 132 and 144
Now \( 36 = 2^2 \times 3^2 \),
\( 72 = 2^3 \times 3^2 \)
\( 132 = 2^2 \times 3 \times 11 \),
\( 144 = 2^4 \times 3^2 \)
HCF of 36, 72, 132 and 144 = \( 2^2 \times 3 \)
A.T.Q. HCF = \( k^2 \times 3 \)
\( \implies k^2 \times 3 = 2^2 \times 3 \)
\( \implies k = 2 \)

Question. LCM of two numbers is 10 times their HCF. Sum of HCF and LCM is 495. If one number is 90, then find the other number.
Answer: Let HCF = x
\( \implies \) LCM = 10x
Also HCF + LCM = 495
\( \implies x + 10x = 495 \)
\( \implies x = 45 \)
\( \implies \) HCF = 45, LCM = \( 10 \times 45 = 450 \)
Now HCF × LCM = Product of two numbers
\( \implies 45 \times 450 = 90 \times \text{Other number} \)
\( \implies \frac{45 \times 450}{90} = \text{Other number} \)
\( \implies \) Other number = 225

Question. Find the HCF and LCM of 180 and 288 by prime factorisation method.
Answer: The prime factorisation of 180 is:
\( 180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5 \)
The Prime factorisation of 288 is:
\( 288 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^5 \times 3^2 \)
The HCF of 180 and 288 is the product of common factors with their least exponent.
The HCF of 180 and 288 = \( 2^2 \times 3^2 = 36 \).
We know HCF (a, b) × LCM (a, b) = a × b
\( \implies 36 \times \text{LCM} (180 \text{ and } 288) = 180 \times 288 \)
\( \implies \text{LCM of } (180 \text{ and } 288) = \frac{180 \times 288}{36} = 1440 \)

Question. Find the largest number which divides 615 and 963 leaving remainder 6 in each case.
Answer: A.T.Q. Number divides 615 – 6 and 963 – 6 i.e. 609 and 957
Required number = HCF of 609 and 957.
\( 609 = 3 \times 7 \times 29 \)
\( 957 = 3 \times 29 \times 11 \)
HCF of 609 and 957 = \( 3 \times 29 = 87 \)
\( \implies \) Required number = 87

Question. Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers. 
Answer: Numbers: 404, 96. To find: HCF and LCM.
\( 404 = 2^2 \times 101 \)
\( 96 = 2^5 \times 3 \)
HCF = greatest common factor = \( 2^2 = 4 \).
LCM = all factors (least power) = \( 2^5 \times 3 \times 101 = 96 \times 101 = 9696 \).
Product of two numbers = \( 404 \times 96 = 38784 \)
Product of HCF & LCM = \( 4 \times 9696 = 38784 \).
Hence, HCF × LCM = product of two numbers.

Question. Find the LCM of 2.5, 0.5 and 0.175.
Answer: LCM of rational number = \( \frac{\text{LCM of numerators}}{\text{HCF of denominators}} \)
Numbers are \( \frac{25}{10}, \frac{5}{10}, \frac{175}{1000} \)
Now \( 25 = 5 \times 5; 5 = 5 \times 1; 175 = 5 \times 5 \times 7 \)
\( \implies \) LCM of (25, 5, 175) = \( 5 \times 5 \times 7 = 175 \)
Also \( 10 = 2 \times 5; 1000 = 2^3 \times 5^3 \)
\( \implies \) HCF of (10, 10, 1000) = 10
\( \implies \) LCM of (2.5, 0.5, 0.175) = \( \frac{175}{10} = 17.5 \).

Question. A forester wants to plant 66 apple trees, 88 banana trees and 110 mango trees in equal rows (in terms of number of trees). Also he wants to make distinct rows of trees (i.e., only one type of trees in one row). Find the number of minimum rows required.
Answer: \( 66 = 2 \times 3 \times 11; 88 = 2^3 \times 11; 110 = 2 \times 5 \times 11 \)
HCF of 66, 88 and 110 = 22
Number of trees in each row = 22
Number of rows = \( \frac{66}{22} + \frac{88}{22} + \frac{110}{22} = 3 + 4 + 5 = 12 \)

Long Answer Type Question 

Question. The HCF of 2472, 1284 and a third number N is 12. If their LCM is \( 2^3 \times 3^2 \times 5 \times 103 \times 107 \). Then find the number N.
Answer: \( 2472 = 2^3 \times 3 \times 103 \)
\( 1284 = 2^2 \times 3 \times 107 \)
LCM = \( 2^3 \times 3^2 \times 5 \times 103 \times 107 \)
\( \implies \) N = \( 2^2 \times 3^2 \times 5 = 180 \)

Question. Prove that \( 15 + 17\sqrt{3} \) be an irrational number.
Answer: Let \( \sqrt{3} = \frac{a}{b} \), where a and b are coprime integers, \( b \neq 0 \).
Squaring both sides, we get \( 3 = \frac{a^2}{b^2} \).
Multiplying with b on both sides, we get \( 3b = \frac{a^2}{b} \)
LHS = \( 3 \times b = \text{Integer} \)
RHS = \( \frac{a^2}{b} = \frac{\text{Integer}}{\text{Integer}} = \text{Rational number} \)
LHS \( \neq \) RHS. Our supposition is wrong. \( \implies \sqrt{3} \) is irrational.
Let \( 15 + 17\sqrt{3} \) be a rational number.
\( 15 + 17\sqrt{3} = \frac{a}{b} \), where a and b are coprime, \( b \neq 0 \)
\( \implies 17\sqrt{3} = \frac{a}{b} - 15 \)
\( \sqrt{3} = \frac{a - 15b}{17b} \)
\( \frac{a - 15b}{17b} \) is a rational number. But \( \sqrt{3} \) is irrational.
\( \implies \sqrt{3} \neq \frac{a - 15b}{17b} \)
Our supposition is wrong. \( \implies 15 + 17\sqrt{3} \) is an irrational number.

Practice Questions

Question. HCF of \( 5^2 \times 3^2 \) and \( 3^5 \times 5^3 \) is:
(a) \( 5^3 \times 3^5 \)
(b) \( 5 \times 3^3 \)
(c) \( 5^3 \times 3^2 \)
(d) \( 5^2 \times 3^2 \)
Answer: (d) \( 5^2 \times 3^2 \)

Question. 4 Bells toll together at 9.00 am. They toll after 7, 8, 11 and 12 seconds respectively. How many times will they toll together again in the next 3 hours ?
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (c) 5

Question. If n is a natural number, then \( 9^{2n} - 4^{2n} \) is always divisible by
(a) 11
(b) 4
(c) 5
(d) 9
Answer: (c) 5

Question. If n is any natural number, then \( 9^n - 5^n \) ends with
(a) 3
(b) 6
(c) 5
(d) 8
Answer: (b) 6

Question. If p is prime, then HCF and LCM of p and p + 1 would be
(a) HCF = p, LCM = p + 1
(b) HCF = p(p + 1), LCM = 1
(c) HCF = 1, LCM = p(p + 1)
(d) None of the options
Answer: (c) HCF = 1, LCM = p(p + 1)

Question. Two numbers are in the ratio of 15:11. If their H.C.F. is 13, then numbers will be
(a) 195 and 143
(b) 190 and 140
(c) 185 and 163
(d) 185 and 143
Answer: (a) 195 and 143

Question. The total number of factors of a prime number is
(a) 1
(b) 0
(c) 2
(d) 3
Answer: (c) 2

Question. Given that LCM (91, 26) = 182, then HCF (91, 26) is
(a) 11
(b) 26
(c) 13
(d) 91
Answer: (c) 13

Question. Can we have any \( n \in N \), where \( 4^n \) ends with the digit zero?
Answer: No, \( 4^n \) prime factors only contain 2. For it to end in 0, it needs both 2 and 5.

Question. The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one number is 280, then find the other number. 
Answer: Let HCF = x, LCM = 14x. \( x + 14x = 600 \implies 15x = 600 \implies x = 40 \).
HCF = 40, LCM = 560. \( \text{Other number} = \frac{40 \times 560}{280} = 80 \).

Question. Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively.
Answer: HCF of (2053 - 5) and (967 - 7) = HCF of 2048 and 960. Required number = 64.

Question. Show that \( 21^n \) can not end with the digits 0, 2, 4, 6 and 8 for any natural number n. 
Answer: \( 21^n = (3 \times 7)^n \). The units digit of \( 1^n \) is always 1. Thus it will always end with 1.

Question. If HCF of 144 and 180 is expressed in the form \( 13m - 3 \), find the value of m. 
Answer: HCF(144, 180) = 36. \( 13m - 3 = 36 \implies 13m = 39 \implies m = 3 \).

Question. Determine the values of p and q so that the prime factorisation of 2520 is expressible as \( 2^3 \times 3^p \times q \times 7 \).
Answer: \( 2520 = 2^3 \times 3^2 \times 5 \times 7 \). Comparing, \( p = 2 \) and \( q = 5 \).