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SSC CPO 2020 PYQ — Question 120

SSC CPO - 23 Nov 2020 - Shift 2 • Quantitative Aptitude • Q120

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The value of 0.325×0.325+0.175×0.175+25×0.004555×0.0065×3.25×3.25−7×0.175×0.025−0.51.5\dfrac{0.325 \times 0.325 + 0.175 \times 0.175 + 25 \times 0.00455}{5 \times 0.0065 \times 3.25\times 3.25 - 7 \times 0.175 \times 0.025 }-\dfrac{0.5}{1.5} is :

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Correct answer: A

Explanation

The correct answer is option 1 i.e. 3

 0.325×0.325+0.175×0.175+25×0.004555×0.0065×3.25×3.25−7×0.175×0.025−0.51.5\dfrac{0.325 \times 0.325 + 0.175 \times 0.175 + 25 \times 0.00455}{5 \times 0.0065 \times 3.25\times 3.25 - 7 \times 0.175 \times 0.025 }-\dfrac{0.5}{1.5} 

(a + b)2 = a2 + b2 + 2ab

= 25 × 0.00455 = 0.11375 = 2 × 0.325 × 0.175

= (0.325+0.175)20.0325×3.25×3.25−0.175×0.175−0.51.5\dfrac{(0.325+0.175)^2}{0.0325\times3.25\times3.25-0.175\times0.175}-\dfrac{0.5}{1.5}

= 2.50.3252−0.1752−0.51.5\dfrac{2.5}{0.325^2-0.175^2}-\dfrac{0.5}{1.5}

=2.5(0.325+0.175)(0.325−0.175)−0.51.5\dfrac{2.5}{(0.325+0.175)(0.325-0.175)}-\dfrac{0.5}{1.5}

= 0.50.15−0.51.5\dfrac{0.5}{0.15}-\dfrac{0.5}{1.5} = 3