Simplification Aptitude Mcq Model Questions & Answers, Practice Test for ibps po prelims 2024 2025

Answer: (a)

$x^2 = y + z, y^2 = z + x, z^2$ = x + y

$x^2 + x = x + y + z; 1/{x + 1} = x/{x + y + z}$

$y^2 + y = x + y + z; 1/{y + 1} = y/{x + y + z}$

$z^2 + z = x + y + z; 1/{z + 1} = z/{x + y + z}$

⇒$x/{x + y + z} + y/{x + y + z} + z/{x + y + z} = {x + y + z}/{x + y + z}$ = 1

Answer: (a)

Answer: (c)

Given, x + y + z = 0

∴ (x + y) (y + z) (z + x) = (–z) (–x) (–y) = –xyz

Answer: (a)

$(x^2 + 1/{x^2}) = {17}/4$

⇒$x^2 + 1/{x^2} + 2 - 2 = {17}/4$

⇒$(x - 1/x)^2 + 2 = {17}/4$

⇒$(x - 1/x)^2 = {17}/4 - 2$

⇒$(x - 1/x)^2 = 9/4$

⇒$(x - 1/x) = 3/2$

On cubing both sides, we get

$(x - 1/x)^3 = (3/2)^3$

⇒$x^3 - 1/{x^3} - 3 × 1/x. x(x - 1/x) = {27}/8$

⇒$x^3 - 1/{x^3} = {27}/8 + 3 × (3/2)$

⇒$x^3 - 1/{x^3} = {27}/8 + 9/2$

∴ $x^3 - 1/{x^3} = {63}/8$

Answer: (a)

$x^2 + 2(1 + k) x + k^2$ = 0.

If it has equal roots, then D = 0

⇒$(2(1 + k))^2 - 4k^2$ = 0

⇒$4(1 + k^2 + 2k) - 4k^2$ = 0

⇒4 + $4k^2 + 8k - 4k^2$ = 0⇒4 + 8k = 0

⇒k = – $4/8$

∴ k = – $1/2$