A) 8
B) 6
C) 4
D) 2
Correct Answer: B
Solution :
| [b] Let n erasers be available for a rupee. Reduced price \[=\left( \frac{75}{100}\times 1 \right)=\frac{3}{4}\] \[\text{Rs}\text{.}\,\,\frac{3}{4}\] fetch n erasers Rs. 1 will fetch \[\left( n\times \frac{4}{3} \right)\] erasers. \[\therefore \] \[\frac{4n}{3}=n+2\] \[\Rightarrow \] \[4n=3n+6\] \[\Rightarrow \] \[n=6\] |
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