lcm and hcf Model Questions & Answers, Practice Test for ibps rrb clerk asst multipurpose prelims 2024

Answer: (c)

LCM × HCF = $1^{st} number × 2^{nd} number$

Let the $1^{st} number be 'A' and 2^{nd} number$ be 'B'.

L × H = A × B

L × 12 L = 93 × B

12 $L^2$ = 93 B --------------(i)

L + H = 403 --------------(ii)

L + 12L = 403

L = ${403}/{13}$ = 31

12 × 31 × 31 = 93B

B = ${12 × 31 × 31}/{93}$

B = 4 × 31

B = 124

So, option (a) is correct.

Answer: (a)

HCF of (22n + 7, 33n + 10) is always 1

Examples

For n = 1, HCF (29, 43)⇒ HCF = 1

For n = 2, HCF (51, 76)⇒ HCF = 1

For n = 3, HCF (73, 109)⇒ HCF = 1

since 22n and 33n are multiples of 11, therefore 22n + 7 and 33n + 10 are not the multiple of 11.

Hence, HCF of 22n + 7 and 33 + 10 will not be equal to 11 or n.

Answer: (c)

Let $f_1(x) = 8(x^5 – x^3 + x)$

= 4 × 2 × x ($x^4 – x^2$ + 1)

and $f_2(x) = 28(x^6 + 1) = 7 × 4[(x^2)^3 + (1)^3]$

= 4 × 7 × ($x^2 + 1) (x^4 – x^2$ + 1)

∴ HCF of $f_1(x) and f_2(x) = 4(x^4 – x^2$ + 1)

Answer: (c)

Given numbers are 392, 486 and 627.

For same remainder

486 – 392 = 94

627 – 486 = 141

627 – 392 = 235

HCF of (94, 141, 235) = 47

Answer: (d)

The least number which, when divided by 8, 12 and 16, leaves 3 as remainder = (LCM of 8, 12, 16) + 3 = 48 + 3 = 51

Other such numbers are 48 × 2 + 3 = 99, 48 × 3 + 3 = 147,

∴ the required number which is divisible by 7 is 147.