Table of contents  
Previous Year Questions 2023  
Previous Year Questions 2022  
Previous Year Questions 2021  
Previous Year Questions 2020  
Previous Year Questions 2019 
Sol: Least composite number = 4
Least prime number = 2
∴ HCF = 2, LCM = 4
∴ Required ratio = 2/4
i.e. 1 : 2
Q2: Find the least number which when divided by 12, 16 and 24 leaves remainder 7 in each case. (2023)
Ans: 55
Given, least number which when divided by 12, 16 and 24 leaves remainder 7 in each case
∴ Least number = LCM( 12, 16, 24) + 7
= 48 + 7
= 55
Q3: Two numbers are in the ratio 2 : 3 and their LCM is 180. What is the HCF of these numbers? (2023)
Ans: 30
Let the two numbers be 2x and 3x
LCM of 2x and 3x = 6x, HCF(2x, 3x) = x
Now, 6x = 180
⇒ x = 180/6
x = 30
Q4: Prove that √3 is an irrational number. (2023)
Ans: Let us assume that √3 is a rational number.
Then √3 = a/b; where a and b ( ≠ 0) are coprime positive integers.
Squaring on both sides, we get
3 = a^{2}/b^{2} ⇒ a^{2} = 3b^{2 }
⇒ 3 divides a^{2}
⇒ 3 divides a _________(i)
= a = 3c, where c is an integer
Again, squaring on both sides, we get
a^{2} = 9c^{2}
⇒ 3b^{2} = 9c^{2} ⇒ b^{2} = 3c^{2} ⇒ 3 divides b^{2}
⇒ 3 divides b _________(ii)
From (i) and (ii), we get 3 divides both a and b.
⇒ a and b are not co prime integers.
This contradicts the fact that a and b are coprimes.
Hence, √3 is an irrational number.
Q5: Two positive numbers have their HCF as 12 and their product as 6336. The number of pairs possible for the numbers, is (2022)
(a) 2
(b) 3
(c) 4
(d) 1
Ans: (a)
Sol: Given, HCF = 12
Let two numbers be 12a and 12b
So. 12a x 12b = 6336 ⇒ ab = 44
We can write 44 as product of two numbers in these ways:
ab = 1 x 44 = 2 x 2 2 = 4x 11
Here, we will take a = 1 and b = 44 ; a = 4 and b = 11.
We do not take ab = 2 x22 because 2 and 22 are not copriine to each other.
For a = 1 and b = 44, 1^{st} no. = 12a = 12, 2^{nd} no. = 12b = 528
For a = 4 and b = 11, 1^{st} no. = 12a = 48, 2^{nd} no. = 12b = 132
Hence, we get two pairs of numbers, (12, 528) and (48, 132).
Q6: If 'n' is any natural number, then (12)^{n} cannot end with the digit (2022)
(a) 2
(b) 4
(c) 8
(d) 0
Ans: (d)
Sol: for n = 1, 2, 3, 4...
(12)^{n} cannot end with 0.
Q7: The number 385 can be expressed as the product of prime factors as (2022)
(a) 5 x 11 x 1 3
(b) 5 x 7 x 11
(c) 5 x 7 x 1 3
(d) 5 x 11 x 17
Ans: (b)
Sol: We have,
∴ Prime factorisation of 385 = 5 x 7 x 11
Q8: Explain why 2 x 3 x 5 + 5 and 5 x 7 x 11 + 7 x 5 are composite numbers. (2021)
View AnswerAns: We have, 2 x 3 x 5 + 5 and 5 x 7 x 11 + 7 x 5.
We can write these numbers as:
2 x 3 x 5+5 = 5(2 x 3 + 1)
=1 x 5 x 7
and 5 x 7 x 11 + 7 x 5 = 5 x 7(11 + 1)
= 5 x 7 x 12 = 1 x 5 x 7 x 12
Since, on simplifying. we find that both the numbers have more than two factors. So. these are composite numbers.
Q9: The HCF and the LCM of 12, 21 and 15 respectively, are (2020)
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 420, 3
Ans: (c)
Sol: We have, 12 = 2 x 2 x 3 = 2^{2}x 3
21= 3 x 7
15 = 3 x 5
∴ HCF (12, 21, 15) = 3
and LCM (12, 21 ,15 ) = 2^{2} x 3 x 5 x 7
= 420
Q10: The LCM of two numbers is 182 and their HCF is 13. If one of the numbers is 26. find the other. (2020)
Ans: 91
Let the other number be x
As, HCF (a, b) x LCM [a, b) = a x b
⇒ 13 x 182= 26x
⇒ x = 13 x 182 / 26
= 91
Hence, other number is 91.
Q11: If HCF (336, 54) = 6. find LCM (336, 54). (2019)
View AnswerAns: 3024
Since. HCF (a, b) x LCM (a, b) = a x b
∴ HCF (336, 54) x LCM {336, 54) = 336 x 54
⇒ 6 x LCM(336, 54) = 18144
⇒ LCM (336, 54) = 18144 / 6
= 3024
Q12: The HCF of two numbers a and b is 5 and their LCM is 200. Find the product of ab. (2019)
Ans: 1000
We know that HCF (a, b) x LCM (a, b)=a x b
⇒ 5 x 200 = ab
⇒ ab = 1000
Q13: 1f HCF of 65 and 117 is expressible in the form 65n117, then find the value of n. (2019)
Ans: 2
Since, HCF (65 ,117) = 13
Given HCF ( 65, 117 ) = 65n  117
13 = 65n  117
⇒ 65n = 13 +117
⇒ n = 2
Q14: Find the HCF of 612 and 1314 using prime factorisation. (2019)
Ans: 18
Prime factorisation of 612 and 1314 are
612 = 2 x 2 x 3 x 3 x 17
1314 = 2 x 3 x 3 x 73
∴ HCF (612, 1314) = 2 x 3 x 3
= 18
Q15: Prove that √5 is an irrational number. (2019)
Ans: Let us assume that √5 is a rational number.
Then √5 = a/b where a and b (≠ 0} are coprime integers,
if Squaring on both sides, we get
⇒ 5 divides a^{2}
⇒ 5 divides a (i)
⇒ a = 5c, where c is an integer
Again, squaring on both sides, we get
a^{2} = 25c^{2}
⇒ 5b^{2} = 25c^{2} ⇒ b^{2} = 5c^{2}
⇒ 5 divides b^{2} (ii)
⇒ 5 divides b
From (i) and {ii), we get 5 divides both a and b.
⇒ a and b are not coprime integers.
Hence, our supposition is wrong.
Thus, √5 is an irrational number.
Q16: Prove that √2 is an irrational number. (2019)
Ans: Let us assume √2 be a rational number.
Then, √2 = p/q where p, q (q ≠ 0) are integers and coprime. ;
On squaring both sides. we get
(i)
⇒ 2 divides p^{2}
⇒ 2 divides p (ii)
So, p = 2a, where a is some integer.
Again squaring on both sides, we get
p^{2} = 4a^{2}
⇒ 2q^{2} = 4a^{2} (using (i))
⇒ q^{2} = 2a^{2}
⇒ 2 divides q^{2}
⇒ 2 divides q (iii)
From (ii) and (iii), we get
2 divides both p and q.
∴ p and q are not coprime integers.
Hence, our assumption is wrong.
Thus √2 is an irrational number.
Q17: Prove that 2 + 5√3 is an irrational number given that √3 is an irrational number. (2019)
Ans: Suppose 2 + 5√3 is a rational number.
We can find two integers a, b (b ≠ 0) such that
2 + 5√3 = a/b, where a and b are co prime integers.
⇒ √3 is a rational number.
[ ∵ a, b are integers, so is a rational number]
But this contradicts the fact that √3 is an irrational number.
Hence, our assumption is wrong.
Thus, 2 + 5√3 is an irrational number.
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