CTET 2 Math and Science Jan 2016 Paper(HM)

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Question : 48 of 150

Marks: +1, -0

यदि

0.001+1.01+1.001-(1.03 × 0.1)+(1.11 ÷ 0.1)+x

=1.4 ×(\;{1}/{10})^{-1}

है, तो

x

का मान है

0.919
0.785
0.758
0.991

Solution:

0.001+1.01+1.001-(1.03 × 0.1)+(1.11 ÷ 0.1)+x

\;\; =1.4 ×(\;{1}/{10})^{-1}

⇒ \;\; 2.012-.103+11.1+x=1.4 × \;{10}/{1}

⇒ \;\; 13.112-.103+x=14

⇒ \;\; 13.009+x=14

⇒ \;\; x=14-13.009

=0.991

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