Two persons P and Q borrowed Rs.40,000/- and Rs.60,000/- respectively from R at different rates of simple interest. The interest payable by P at the end of the first four years and that payable by Q at the end of the first three years is the same. If the total interest payable by P and Q for one year is Rs.8,400/- then at what rate did Q borrow the money from R?

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Solution

Let $R_{1}$ & $R_{2}$ be the different rates
$\Rightarrow \frac{40, 000 \times 4 \times R_{2}}{100} = \frac{60, 000 \times 3 \times R_{2}}{100}$
$\Rightarrow 16 R_{1} = 18 R_{2}$
$\Rightarrow R_{1} = \frac{9}{8} R_{2}$
1 year interest = $\frac{40000 \times 1 \times R_{1}}{100} + \frac{60, 000 \times 1 \times R_{2}}{100}$
$= 8400$
$\Rightarrow 400 R_{1} + 600 R_{2} = 8400$
$\Rightarrow 4 R_{1} + 6 R_{2} = 84$ $(\div 2)$
$\Rightarrow 2 R_{1} + 3 R_{2} = 42$
$\Rightarrow 2 (\frac{9}{8} R_{2}) + 3 R_{2} = 42$ since $R_{1} = \frac{9}{8} R_{2}$
$\Rightarrow 9 R_{2} + 12 R_{2} = 168$
$\Rightarrow 21 R_{2} = 168$
$R_{2} = \frac{168}{21} = 8$

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Let R1 & R2 be the different rates⇒40,000×4×R2100=60,000×3×R2100⇒16R1=18R2⇒R1=98R21 year interest = 40000×1×R1100+60,000×1×R2100=8400⇒400R1+600R2=8400⇒4R1+6R2=84 (÷2)⇒2R1+3R2=42⇒2(98R2)+3R2=42 since R1=98R2⇒9R2+12R2=168⇒21R2=168R2=16821=8