| A principal amounts to Rs.944 in 3 yr and to Rs.1040 in 5 yr, each sum being invested at the same rate of simple interest. The principal was |
A) Rs. 800
B) Rs. 992
C) Rs. 750
D) Rs. 900
Correct Answer: A
Solution :
| Let the principal be Rs. x. |
| Rate of interest = R% |
| Case I \[P=\text{Rs}.\,\,x,\]\[T=3\,\,yr\] |
| \[R= R\]% \[SI=\text{Rs}.\,\,(944-x)\] |
| We know that, \[SI=\frac{P\times R\times T}{100}\] |
| \[\Rightarrow \] \[(944-x)=\frac{x\times R\times 3}{100}\] |
| \[\Rightarrow \] \[\frac{100\,\,(944-x)}{3x}=R\] (i) |
| Case II \[P=\text{Rs}.\,\,x,\]\[T=5\,\,yr,\]\[R= R\]% |
| \[SI=\text{Rs}.\,\,(1040-x)\] |
| \[\therefore \] \[SI=\frac{P\times R\times T}{100}\] |
| \[\Rightarrow \] \[(1040-x)=\frac{x\times R\times 5}{100}\] |
| \[\Rightarrow \] \[R=\frac{(1040-x)\times 100}{5x}\] ... (ii) |
| From Eqs. (i) and (ii), we get |
| \[\frac{(1040-x)\times 100}{5x}=\frac{100\,\,(944-x)}{3x}\] |
| \[\Rightarrow \]\[3\,\,(1040-x)=5\,\,(944-x)\] |
| \[\Rightarrow \]\[3120-3x=4720-5x\] |
| \[\Rightarrow \] \[2x=4720-3120\] |
| \[\Rightarrow \] \[2x=1600\]\[\Rightarrow \]\[x=Rs.\,\,800\] |
| Alternate Method |
| Let the principal be Rs. x. |
| Rate of interest = R% |
| Case I |
| Here, \[P=x,\]\[T=2\,\,yr\]and \[SI=(880-x)\] |
| \[SI=\frac{P\times R\times T}{100}\] |
| \[\Rightarrow \] \[(880-x)=\frac{x\times R\times 2}{100}\] |
| \[\Rightarrow \] \[R=\frac{100\,\,(880-x)}{2x}\] (i) |
| Case II |
| Similarly, \[R=\frac{100\,\,(920-x)}{3x}\] |
| From Eqs. (i) and (ii), we get |
| \[\frac{100\,\,(880-x)}{2x}=\frac{100\,\,(920-x)}{3x}\] |
| \[\Rightarrow \]\[2640-3x=1840-2x\] |
| \[\Rightarrow \] \[x=800\] |
You need to login to perform this action.
You will be redirected in
3 sec
