Lcm And Hcf Model Questions & Answers, Practice Test for ibps po prelims 2024 2025

Answer: (d)

Suppose the number is x

∴x ×$1/5$=81⇒x=81×5=405

∴${x × 68}/100$ = ${405 × 68}/100$

x = 275.4

Answer: (c)

Let $f_1(x) = x^3 + mx^2$ – x + 2m

and $f_2 (x) = x^2$ + mx – 2

Let m = 1

∴ $f_1(x) = x^3 + x^2$ – x + 2

and $f_2 (x) = x^2$ + x – 2 = (x + 2) (x – 1)

When x = 1,

f (1) = 1 + 1 – 1 + 2 ≠ 0

When x = – 2,

f (– 2) = $(–2)^3 + (–2)^2$ – (– 2) + 2 = 0

Required value of m is 1.

Answer: (a)

We have P = 0.8$\ov3$ and Q = 0.6$\ov2$

The distance between P and Q is

0.8$\ov3$ - 0.6$\ov2$ = 0.2$\ov1$

Expressing this distance in the form of rational number, we assume 0.2$\ov1$ = x

No. of digits with bar = 1

No. of digits without bar = 1

Therefore, the denominator would be 90.

No. of digits after the decimal = 2

Therefore, the numerator would be 21 – 2 = 19

Thus, x = ${19}/{90}$

Answer: (c)

HCF of two numbers is 98. It means that 98 is common in both the numbers. Therefore, the sum of these two numbers also be multiple of 98. So, 1372 is divided by 98.

Answer: (b)

Let $f_1 (x) = x^3 – x^2 – 2x = x (x^2 – x – 2)$

= x {$x^2$ – 2x + x – 2}

= x {x (x – 2) + 1 (x – 2)} = x (x + 1) (x – 2)

and $f_2 (x) = x^3 + x^2 = x^2$ (x + 1) = x . x (x + 1)

∴ LCM of [$f_1(x), f_2(x)$] = x (x + 1) . x (x – 2)

= $x^2 (x + 1) (x – 2) = x^2 (x^2$ – x – 2)

= $x^4 – x^3 – 2x^2$