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SSC MTS 2019 PYQ — Question 70

SSC MTS Previous Year - 07 August 2019 - Shift 1 • Quantitative Aptitude • Q70

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What is the value of (0.7)2÷0.14+(0.6)2÷0.18+(0.5)2÷0.054(2.5 of 4−13×0.25×3)\dfrac{(0.7)^2 \div0.14+(0.6)^2\div0.18+(0.5)^2\div0.05}{4(2.5\ \text{of}\ 4 - 13\times0.25\times 3)}?

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Correct answer: D

Explanation

The correct answer is option 4 i.e. 21/2.

=(0.7)2÷0.14+(0.6)2÷0.18+(0.5)2÷0.054(2.5 of 4−13×0.25×3)=\dfrac{(0.7)^2 \div0.14+(0.6)^2\div0.18+(0.5)^2\div0.05}{4(2.5\ \text{of}\ 4 - 13\times0.25\times 3)}

=7/2+2+54(10−13×0.25×3)=\dfrac{7/2+2+5}{4(10 - 13\times0.25\times 3)}

=7/2+74(10−394)=\dfrac{7/2+7}{4(10 - \dfrac{39}{4})}

=21/24(40−394)=\dfrac{21/2}{4(\dfrac{40-39}{4})}=21/2=21/2