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 |
\[\frac{100\,\,(880-x)}{2x}=\frac{100\,\,(920-x)}{3x}\] |
\[\Rightarrow \]\[2640-3x=1840-2x\] |
\[\Rightarrow \] \[x=800\] |
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