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

Answer: (b)

Let the HCF of two number = x

The LCM of two numbers = 90x

According to question

LCM + HCF = 1456

90x + x = 1456

x = 16

HCF of two numbers = 16,

LCM of two number = 1440

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

⇒$2^{nd} Number = {1440 × 16}/{160}$ = 144

Answer: (d)

HCF of two natural numbers m and n = 24

m × n = 552

LCM of two natural numbers

= ${\text"Product of m and n"}/{\text"HCF of m and n"}$

= ${552}/{24}$ = 23

Therefore, no set of m and n is possible satisfying the given condition

Answer: (d)

Suppose number is x.

∴$x × 4/5 × 3/4 - x × 2/5 × 1/6 = 648$

${12x}/20 - {2x}/30 = 648⇒{36x-4x}/60$=648

⇒${32x}/60=648⇒x={648 × 60}/32$=81 × 15

x = 1215

Answer: (d)

∵ 1.$\ov{34}$ = ${134 - 1}/{99} = {133}/{99}$

and 4.1$\ov2$ = ${412 - 41}/{90} = {371}/{90}$

∴ 1.$\ov{34}$ + 4.$\ov{12}$ = ${133}/{99} + {371}/{90} = {1330 + 4081}/{990}$

= ${5411}/{990} = 5 {461}/{990}$

Answer: (c)

$x^4 – x^2 – 6 = x^4 – 3x^2 + 2x^2$ – 6

= $x^2 (x^2 – 3) + 2(x^2$ – 3)

= ($x^2 – 3) (x^2$ + 2)

and $x^4 – 4x^2 + 3 = x^4 – 3x^2 – x^2$ + 3

= $x^2 (x^2 – 3) – 1 (x^2$ – 3)

= ($x^2 – 3) (x^2$ – 1)

∴ HCF = ($x^2$ – 3)