A) Rs. 1600
B) Rs. 2000
C) Rs. 1800
D) Rs. 2100
Correct Answer: B
Solution :
[b] Here, SI =540 and t = 3 yr |
\[\therefore \] \[\frac{P\times 3\times r}{100}=540\]\[\Rightarrow \]\[r=\frac{540\times 100}{3P}=\frac{18000}{P}\] |
Now, CI = 376.20 and t = 2 yr |
\[\Rightarrow \]376.20 |
\[=P\,\,\left[ {{\left( 1+\frac{18000}{100P} \right)}^{2}}-1 \right]=P\left[ {{\left( 1+\frac{180}{P} \right)}^{2}}-1 \right]\] \[=P\,\,\left[ \frac{{{P}^{2}}+32400+360P-{{P}^{2}}}{{{P}^{2}}} \right]\] |
\[376.20=\frac{32400+360P}{P}\] |
\[\Rightarrow \] \[\,376.20P=32400+360P\] |
\[\Rightarrow \] \[16.20P=32400\]\[\Rightarrow \]\[P=Rs.\,\,2000\] |
You need to login to perform this action.
You will be redirected in
3 sec