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

Answer: (d)

If (x + 1) is HCF then x = –1 will satisfy both the equation

A$(–1)^2$ + B(–1) + C = 0

⇒ A – B + C = 0 ----------(a)

B$(–1)^2$ + A(–1) + C = 0

⇒ B – A + C = 0 ----------(b)

Adding equ. (a) and (b)

A – B + C + B – A + C = 0

2C = 0

C = 0

So, option (d) is correct.

Answer: (d)

$p/q$ = 2.$\ov{52}$⇒$p/q = {252 - 2}/{99}⇒p/q= {250}/{99}$

∴ $p/q = {99}/{250}$ = 0.396

Answer: (c)

LCM of 2, 4, 6, 8 and 10 is 120s.

i.e., 2 min after tolling together.

Total in 20 min = ${\text"Total time"}/{\text"LCM t intervals"}$

In 20 min tolling = ${20 min}/{2 min}$ = 10 times.

Answer: (b)

Given function

$f_1 (x) = 4x^3 + 3x^2y – 9xy^2 + 2y^3$

= (x – y) ($4x^2 + 7xy – 2y^2$)

= (x – y) (x + 2y) (4x – y)

and another function

$f_2(x) = x^2 + xy – 2y^2$ = (x – y) (x + 2y)

HCF of $f_1 (x) and f_2$ (x) = (x – y) (x + 2y)

Answer: (d)

Let x be the remainder, then the numbers (55 – x), (127 – x) and (175 – x) are exactly divisible by the required number.

Now, we know that if two numbers are divisible by a certain number, then their difference is also divisible by the number.

Hence the numbers (127 – x) – (55 – x), (175 – x) – (127 – x) and (175 – x) – (55 – x) or, 72, 48 and 120 are divisible by the required number. HCF of 48, 72 and 120 = 24, therefore the required number = 24.