Pratap borrowed a sum of money from Arun at simple interest, at the rate of $12 % p . a .$ for the first three years, $16 % p . a .$ for the next five years and $20 % p . a .$ for a period beyond eight years. If at the end of $11$ years, the total interest is $R s . 6080$ more than the sum, what was the sum borrowed?

R s . 3000

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R s . 8000

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R s . 7000

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R s . 4000

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Solution

The correct option is B $R s . 8000$ .
Let the sum borrowed be $R s . x$ . Then,
$\frac{x \times 12 \times 3}{100} + \frac{x \times 16 \times 5}{100} + \frac{x \times 20 \times (11 - 8)}{100} = x + 6080$
$\Rightarrow \frac{36 x}{100} + \frac{80 x}{100} + \frac{60 x}{100} = x + 6080 \Rightarrow \frac{176 x}{100} = x + 6080$
$\Rightarrow 176 x = 100 x + 608000 \Rightarrow 76 x = 608000 \Rightarrow x = \frac{608000}{76} = R s . 8000$ .

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Pratap borrowed a sum of money from Arun at simple interest, at the rate of 12%p.a. for the first three years, 16%p.a. for the next five years and 20%p.a. for a period beyond eight years. If at the end of 11 years, the total interest is Rs.6080 more than the sum, what was the sum borrowed?

The correct option is B Rs.8000.Let the sum borrowed be Rs.x. Then,x×12×3100+x×16×5100+x×20×(11−8)100=x+6080⇒36x100+80x100+60x100=x+6080⇒176x100=x+6080⇒176x=100x+608000⇒76x=608000⇒x=60800076=Rs.8000.