\[{{(0.49)}^{4}}\times {{(0.343)}^{4}}\div {{(0.2401)}^{4}}={{(70\div 100)}^{?+3}}\] [IBPS (SO) 2012] |
A) 3
B) 1
C) 4
D) 7
E) None of these
Correct Answer: B
Solution :
\[{{(70\div 100)}^{?+3}}={{(0.49)}^{4}}\times {{(0.343)}^{4}}\div {{(0.2401)}^{4}}\] |
\[\Rightarrow \]\[{{(0.7)}^{?+3}}={{\left( \frac{0.49\times 0.343}{0.2401} \right)}^{4}}\] |
\[\Rightarrow \]\[{{(0.7)}^{?+3}}={{(0.7)}^{4}}\] |
On comparing the exponents both sides, we get |
\[?\,+3=4\] |
\[\therefore \] \[?=4-3=1\] |
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