Question 1

For integers a and b with HCF(a, b) = 1, what is HCF(a + b, a - b)?

Accepted Answer

Introduction: When a and b are coprime, the HCF of a + b and a - b depends on their parity. This is a classic result in number theory relating to sums and differences of coprime integers. Given Data / Assumptions: HCF(a, b) = 1 Concept / Approach: If one of a or b is even and the other is odd, then a + b and a - b are odd and hence coprime. If both are odd, then each of a + b and a - b is even, and their HCF includes a factor 2 but no higher power of 2 because a and b share no odd common factor. Step-by-Step Solution: Case 1: a and b of opposite parity → HCF(a + b, a - b) = 1 Case 2: a and b both odd → HCF(a + b, a - b) = 2 Verification / Alternative check: Examples: a = 2, b = 1 gives HCF(3, 1) = 1. a = 5, b = 3 gives HCF(8, 2) = 2. Why Other Options Are Wrong: Always 1 or always 2 is not correct across all parities; None of the above and always 0 are invalid here. Common Pitfalls: Ignoring parity, or assuming a fixed HCF independent of the parity of a and b. Final Answer: Either 1 or 2

Question 2

6 bells commence tolling together and toll of intervals are 2, 4, 6, 8, 10 and 12 s, respectively. In 1 h, how many times, do they toll together?

Accepted Answer

LCM of 2, 4, 6, 8, 10, 12 =(2 x 2 x 3 x 2 x 5) = 120 ∴ After every 2 min, they toll together. ∴ Number of times they toll together in one hour = (60/2) + 1 times = 31 times

Question 3

The HCF and LCM of two number are 21 and 4641, respectively. If one of the numbers lies between 200 and 300, then find the two numbers.

Accepted Answer

Let the numbers be 21a and 21b, where a and b are co-primes. Then, 21a x 21b = (21 x 4641) ⇒ ab = 221 Two co-primes with product 221 are 13 and 17. ∴ Required number = (21 x 13, 21 x 17) = (273, 357)

Question 4

If (x - 6) is the HCF of x 2 - 2x - 24 and x 2- kx - 6, then what is the value of k?

Accepted Answer

Given that, (x - 6) is the HCF of x2 - 2x - 24 and x2 - kx - 6 i.e., (x - 6) is a factor of both expressions. Let f (x1) = x12 - 2x1 - 24 and f (x2) = x22 - kx2 - 6 Now f(x1) = f (x2) at (x1 = x2 = 6) ⇒ (6)2 - 2(6) - 24 = (6)2 - k(6) - 6 [by condition] ⇒ 0 = 30 - 6k ⇒ 6 k = 30 ∴ k = 5

Question 5

If a and b are positive integers, then what is the value of (a / HCF (a,b) , b/HCF (a,b) )?

Accepted Answer

HCF (a /HCF (a, b) , b/HCF (a,b) it value is always equal to 1;IIIustration 1 Let the two positive integers be a = 24 and b = 36. ∴ HCF (24/HCF (24,36) , 36/HCF (24,36) ⇒ HCF (24/12 , 36/12) ⇒ HCF (2, 3) = 1IIIustration 2 Let the two positive integers be a = 13 and b = 17 ∴ HCF (13 / HCF (13,17) , 17/HCF (13,17) ⇒ HCF (13/1 , 17/1) = 1

Question 6

What is the HCF of 8 (N 5 - N 3 + N ) and 28 (N 6 + 1)?

Accepted Answer

Let p(N) = 8 (N5 - N3 + N) = 4 x 2 x N (N4 - N2 + 1) and q(N) = 28 (N6 + 1) = 7 x 4 [( N2)3 + (1)3] = 4 x 7 (N2 + 1) ( N4 - N2 + 1) ∴ HCF of p(N) and q (N) = 4 (N4 - N2 + 1)

Question 7

The HCF of (x 4 - y 4) and (x 6 - y 6) is

Accepted Answer

Let f (x) = (x4 - y4) = (x2 - y2) (x2 + y2) = (x - y) (x + y) (x2 + y2) and g(x) = (x6 - y6) = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y) (x2 - xy + y2) (x - y)(x2 + xy + y2) = (x - y) (x + y) (x2 - xy + y2) (x2 + xy + y2) ∴ HCF of [f(x), g(x)] = (x - y) (x + y) = x2 - y2

Question 8

The HCF of (x 3 - y 2 - 2x) and (x 3 + x 2) is

Accepted Answer

Let f(x) = x3 - y2 - 2x = x( x2 - x- 2) = x( x2 - 2x + x - 2) = x {x (x - 2) + 1 (x -2)} = x (x +1) (x - 2) and g(x) = x3 + x2 = x2 (x +1) = x. x (x +1) ∴ LCM of [ f(x), g(x)] = x (x +1) . x . (x - 2) = x2(x + 1)(x - 2) = x2(x2 - x - 2) = x4 - x3 - 2x2

Question 9

What is the HCF of a 2 b 4 + 2 a 2 b 2 and (ab) 7 - 4a 2 b 2?

Accepted Answer

a2 b4 + 2 a2 b2 = a2 b2 ( b2 +2).......................(i) and (ab)7 - 4 a2 b9 = a7 b7 - 4 a2 b9 = a2 b2(a5 b5 - 4b7)................(ii)from Eqs. (i) and (ii) HCF = a2 b2

Question 10

Find the largest number that divides 1305, 4665, and 6905 leaving the same remainder in each case. Also determine that common remainder.

Accepted Answer

Introduction / Context:Problems that ask for the largest number dividing several integers with the same remainder are classic Highest Common Factor (HCF) applications. The key insight is that if a number leaves the same remainder upon dividing multiple values, then it must exactly divide all pairwise differences of those values. Given Data / Assumptions: Numbers: 1305, 4665, 6905 Let the required divisor be D and the common remainder be r. Then 1305 = D*k1 + r, 4665 = D*k2 + r, 6905 = D*k3 + r for some integers k1, k2, k3. Concept / Approach:The same remainder implies D divides pairwise differences. Hence D = HCF of {4665 - 1305, 6905 - 4665, 6905 - 1305}. After D is found, compute r from any original number: r = number - D * floor(number / D). Step-by-Step Solution: Compute differences: 4665 - 1305 = 3360 Compute differences: 6905 - 4665 = 2240 Compute differences: 6905 - 1305 = 5600 Find D = HCF(3360, 2240, 5600) HCF(3360, 2240) = 1120; then HCF(1120, 5600) = 1120 ⇒ D = 1120 Common remainder r from 1305: r = 1305 - 1120 = 185 Verification / Alternative check:Check with 4665: 4665 / 1120 = 4 remainder 185 (since 1120*4 = 4480; 4665 - 4480 = 185). Check with 6905: 1120*6 = 6720; 6905 - 6720 = 185. Same remainder confirms correctness. Why Other Options Are Wrong: 1210, 158: 1210 does not divide all differences 3360, 2240, 5600. 1120, 158: 1120 is correct but r = 185, not 158. 1210, 185: 1210 is not the HCF of the differences. None of these: A matching correct pair exists, so this is invalid. Common Pitfalls:Using HCF of the original numbers instead of their differences, or computing the remainder with an incorrect dividend. Final Answer:Largest divisor = 1120 and common remainder = 185.

Question 11

Find the least common multiple (L.C.M.) of the two numbers 12 and 18.

Accepted Answer

Introduction: This is a straightforward L.C.M. question designed to test basic understanding of prime factorisation and the rule for finding the least common multiple of two numbers. Such questions are fundamental in aptitude tests and form the basis for more complex problems. Given Data / Assumptions: The two numbers are 12 and 18. We need to compute their L.C.M. Concept / Approach: To find the L.C.M. of two numbers using prime factorisation: Write each number as a product of prime powers. For each prime appearing in either number, take the highest exponent. Multiply these prime powers together to get the L.C.M. Step-by-Step Solution: Step 1: Factorise 12 and 18. 12 = 2^2 * 3. 18 = 2 * 3^2. Step 2: Collect all distinct primes: 2 and 3. Step 3: For L.C.M., choose the highest power of each prime. For 2: max exponent is 2 (from 12). For 3: max exponent is 2 (from 18). Step 4: L.C.M. = 2^2 * 3^2. Step 5: Compute: 2^2 = 4, 3^2 = 9. Step 6: L.C.M. = 4 * 9 = 36. Verification / Alternative check: Check that 36 is divisible by both 12 and 18: 36 ÷ 12 = 3 (integer). 36 ÷ 18 = 2 (integer). Also, there is no smaller positive number that is divisible by both 12 and 18, so 36 is indeed the least common multiple. Why Other Options Are Wrong: 42: Not divisible by 12, so cannot be the L.C.M. 12: Divisible by 12 but not by 18. 6: Divisible by neither 12 nor 18 as a multiple, it is a common factor, not L.C.M. 18: Divisible by 18 but not by 12. Common Pitfalls: A typical mistake is to confuse L.C.M. with H.C.F. and choose the smaller number. Another common error is to multiply the numbers directly (12 * 18 = 216) without reducing by their common factor, resulting in a non-minimal multiple. Final Answer: The least common multiple of 12 and 18 is 36.

Question 12

Find the least common multiple (LCM) of the three numbers 24, 36, and 40.

Accepted Answer

Introduction: This problem tests your understanding of how to find the least common multiple (LCM) of several numbers, which is a common topic in aptitude exams. You need to identify the smallest positive integer that is divisible by all the given numbers 24, 36, and 40. Given Data / Assumptions: Numbers: 24, 36, and 40 We assume all numbers are positive integers. We must find the LCM of these three numbers. Concept / Approach: To find the LCM using prime factorisation, we express each number as a product of prime factors. Then we take every prime that appears in any of the numbers, with the highest power that appears among them. Multiplying these highest powers gives the LCM. Step-by-Step Solution: Prime factorise each number: 24 = 2^3 × 3 36 = 2^2 × 3^2 40 = 2^3 × 5 Collect all prime factors: 2, 3, and 5 For 2: highest power among the three numbers is 2^3 For 3: highest power is 3^2 For 5: highest power is 5^1 Therefore, LCM = 2^3 × 3^2 × 5 Compute step by step: 2^3 = 8 3^2 = 9 8 × 9 = 72 72 × 5 = 360 So the LCM is 360 Verification / Alternative check: We can check divisibility: 360 ÷ 24 = 15, 360 ÷ 36 = 10, and 360 ÷ 40 = 9. All results are integers, so 360 is a common multiple. Also, there is no smaller positive integer that is divisible by 24, 36, and 40 together, because we already used the maximum prime powers. Why Other Options Are Wrong: 120 and 240 are multiples of some of the numbers but not of all three. For example, 120 is not divisible by 36, and 240 is not divisible by 36 either. 480 is a common multiple but not the least, so it cannot be the LCM. Common Pitfalls: Students sometimes add the numbers or multiply them directly instead of using prime factors or factorisation. Another frequent error is to take the minimum powers of primes (which gives the HCF) instead of the maximum powers required for LCM. Carefully separating HCF and LCM methods avoids such confusion. Final Answer: The least common multiple of 24, 36, and 40 is 360.

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