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SSC CGL 2023 PYQ — Question 52

SSC CGL - 24 July 2023 - Shift 3 • Quantitative Aptitude • Q52

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Simplify the following.

0.01×0.01×0.01+0.003×0.003×0.0030.05×0.05−0.015×0.05+0.015×0.015\frac{0.01 \times 0.01 \times 0.01 + 0.003 \times 0.003 \times 0.003}{0.05 \times 0.05 - 0.015 \times 0.05 + 0.015 \times 0.015}

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Correct answer: D

Explanation

The correct answer is option 4 i.e. 1325×10−3\frac{13}{25} \times 10^{-3}.

0.01×0.01×0.01+0.003×0.003×0.0030.05×0.05−0.015×0.05+0.015×0.015\frac{0.01 \times 0.01 \times 0.01 + 0.003 \times 0.003 \times 0.003}{0.05 \times 0.05 - 0.015 \times 0.05 + 0.015 \times 0.015}

=0.013+0.003325(0.01×0.01−0.003×0.01+0.003×0.003)=\frac{0.01^3 + 0.003^3}{25(0.01 \times 0.01 - 0.003 \times 0.01 + 0.003 \times 0.003)}

=(0.01+0.003)(0.012−0.003×0.01+0.0032)25(0.012−0.003×0.01+0.0032)=\frac{(0.01 + 0.003)(0.01^2 - 0.003 \times 0.01 + 0.003^2)}{25(0.01^2 - 0.003 \times 0.01 + 0.003^2)}

=(0.01+0.003)25=\frac{(0.01 + 0.003)}{25}

=0.01325=\frac{0.013}{25}

=1325×10−3=\frac{13}{25} \times 10^{-3}

Therefore, the value of the given expression is 1325×10−3\frac{13}{25} \times 10^{-3}.