simplification Model Questions & Answers, Practice Test for ibps rrb po officer scale 1 prelims 2024

Answer: (b)

1. Given, 3abc + $b^3 + c^3 – a^3$

= – $(a^3 – b^3 – c^3 – 3abc)$

= – $[a^3 + (–b)^3 + (–c)^3 – 3(a) (–b) (–c)]$

= – (a – b – c) ($a^2 + b^2+ c^2$ + ab – bc + ac)

Hence, (a – b – c) is a factor of 3abc + $b^3 + c^3 – a^3$

Therefore, Statement 1 is correct.

Given, 3bc + $b^3 + c^3$ – 1

= $b^3 + c^3 – (1)^3$ – 3bc (–1)

= (b + c – 1) [$b^2 + c^2 + 1^2$ – bc + c + b]

Hence, (b+c–1) is a factor of 3bc + $b^3 + c^3$ – 1.

Therefore, Statement 2 is also correct.

Answer: (c)

According to remainder theorem, when a polynomial say p (x) is divided by (x – a), the remainder is p(a). Therefore, when the given polynomial f (x) is divided by (ax – b), then the remainder is

f(ax - b = 0)⇒f(ax = b)⇒f$(x = b/a)$

Hence, the remainder is f$(b/a)$

Answer: (e)

? = ${558×45}/{18×4.5}$=310

Answer: (c)

f(x) = $x^3 + 4x^2$ + px + q

Now, (x –1) and (x + 2) divide

f(x). so, f(1) = f(–2) = 0

f(1) = $1^3 + 4 × 1^2$ + p × 1 + q = 0⇒p + q = –5

f(2) = –8 + 16 – 2p + q = 0 ⇒ q –2p = –8

on solving p = 1, q = –6

Answer: (e)

? = 999.99 + 99.99 + 9.99 = 1109.97