simplification Model Questions & Answers, Practice Test for ssc chsl tier 1 2024

Answer: (e)

? = 21.25 + 47779.1944

= 4800.4444

Answer: (a)

$1/{1 × 4} + 1/{4 × 7} + 1/{7 × 10}............+ 1/{16 × 19}$

⇒$1/3[3/{1 × 4} + 3/{4 × 7} + 3/{7 × 10} + ........+ 3/{16 × 19}]$

⇒$1/3[1 - 1/4 + (1/4 - 1/7) + (1/7 - 1/{10}) +........+ (1/{16} - 1/{19})]$

⇒$1/3(1 - 1/{19})$

⇒$1/3 × {18}/{19} = 6/{19}$

So, option (b) is correct

Answer: (a)

We know that if a + b + c = 0 then $a^3 + b^3 + c^3$ = 3abc

Now, we have ${(x - y) (y - z) (z - x)}/{(x - y)^3 + (y - z)^3 + (z - x)^3}$

Since, here, (x - y) + (y - z) + (z - x) = 0 therefore,

$(x - y)^3 + (y - z)^3 + (z - x)^3$

= 3(x - y) (y - z) (z - x)

Thus,

${(x - y) (y - z) (z - x)}/{(x - y)^3 + (y - z)^3 + (z - x)^3}$

= ${(x - y) (y - z) (z - x)}/{3 (x - y) (y - z) (z - x)} = 1/3$

Answer: (c)

$({x^2 - 3x + 2}/{x^3 - 8}) ÷ ({x^2 - 9}/{x^2 + 7x + 12}) × ({x^3 + 2x^2 + 4x}/{x^2 + 3x - 4})$

= $({x^2 - 3x + 2}/{x^3 - 8} × {x^2 + 7x + 12}/{x^2 - 9}) × {x^3 + 2x^2 + 4x}/{x^2 + 3x - 4}$

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

= $x/{x - 3}$

Answer: (d)

$(96)^2 + (63)^2 = (?)^2 – (111)^2$ – 8350

or 9216 + 3969 = $(?)^2$ – 12321 – 8350

or 13185 = $?^2$ – 20671

or $?^2$ = 33856

or ? = $√33856$=184