simplification Model Questions & Answers, Practice Test for ssc steno grade c d 2024

Question :16

1256 × 3892 = ?

Answer: (b)

? = 1256 × 3892 = 4888352

Question :17

If $x^2 = y + z, y^2 = z + x, z^2 = x + y$, then what is the value of ? $1/{x + 1} + 1/{y + 1} + 1/{z + 1}$

Answer: (c)

Given, $x^2$ = y + z

⇒$x^2$ + x = x + y + z

⇒$x/{x + y + z} = 1/{x + 1}$

Similarly, $1/{y + 1} = y/{x + y + z}$

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

∴ $1/{x + 1} + 1/{y + 1} + 1/{z + 1}$

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

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

Question :18

What is the value of ${(2.3)^3 - 0.027}/{(2.3)^2 + 0.69 + 0.09}$ ?

Answer: (d)

${(2.3)^3 - 0.027}/{(2.3)^2 + 0.69 + 0.09} = {(2.3)^3 - (0.3)^3}/{(2.3)^2 + 0.69 + 0.09}$

= ${(2.3 - 0.3) [(2.3)^2 + (0.3)^2 + 2.3 × 0.3]}/{(2.3)^2 + 0.69 + 0.09}$

= ${2[(2.3)^2 + 0.09 + 0.69]}/{(2.3)^2 + 0.69 + 0.09}$ = 2

Question :19

For what value of k can the expression $x^3 + kx^2$ - 7x + 6 be resolved into three linear factors ?

Answer: (c)

For k= 0, $x^3$ – 7x + 6 = 0

$x^2$ (x – 1) + x(x – 1) –6(x – 1) = 0

(x – 1)($x^2$ + x – 6) = 0

(x – 1)(x + 3)(x – 2) = 0

∴ x = 1, 2, and –3

Hence, for k = 0, given expression can be resolved into three linear factors.

Question :20

What is ${{x^2 - 3x + 2}/{x^2 - 5x + 6}} ÷ {{x^2 - 5x + 4}/{x^2 - 7x + 12}}$ equal to?

Answer: (a)

${{x^2 - 3x + 2}/{x^2 - 5x + 6}} ÷ {{x^2 - 5x + 4}/{x^2 - 7x + 12}}$

= ${x^2 - 3x + 2}/{x^2 - 5x + 6} × {(x^2 - 7x + 12)}/{(x^2 - 5x + 4)}$

= ${(x - 1) (x - 2)}/{(x - 3) (x - 2)} × {(x - 4) (x - 3)}/{(x - 4) (x - 1)}$ = 1

ssc steno grade c d 2024 IMPORTANT QUESTION AND ANSWERS

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