Question

astha borrowed some money at the rate of 7% for the first two years, at the rate of 6% for the next 4 years, and at the rate of 9% for last years. if she pays a totals interest of rs 14110 at the end of 11 years, how much did she borrowed. ​

Answer 1

Answer:

Astha borrowed approx. $\text{Rs}30021$.

Step-by-step explanation:

Let the sum Astha borrowed be $x$.

Then, using $S.I=\frac{p*r*t}{100}$, where $p,r,t$ are principle, rate  & time respectively.

$\begin{aligned}&\left(\frac{x \times 7 \times 2}{100}\right)+\left(\frac{x \times 6 \times 4}{100}\right)+\left(\frac{x \times 9 \times 1}{100}\right)=14110 \\&\left(\frac{14 x}{100}+\frac{24 x}{100}+\frac{9 x}{100}\right)=14110 \\\frac{47 x}{100}=14110 \\& x=\frac{14110\times 100}{47}=30021 .27\end{aligned}$

Hence, sum borrowed$= Rs 30021.27$ or approximately Rs. $30021$.

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Answer 2

Answer:

Total amount of money Astha borrowed= Rs 17000

Step-by-step explanation:

In the above question:

Given:

1. R(rate)= 7%

   T(time)= 2 years

2. R(rate)= 6%

    T(time)= 4 years

3.    R(rate)= 9%

    T(time)= 5 years

To find:

Total amount borrowed?

Solution:

Total interest= Rs 14110

We know, the formula for Simple interest as:

S.I= P×R×T/100

P= principle amount(here borrowed amount)

R= Rate of interest

T= Time

Let the total amount borrowed be 'x'

Then, total amount borrowed at the end of 11years will be;

S.I= P1×R1×T1/100+P2×R2×T2/100+P3×R3×T3/100

14110= x×7×2/100+x×6×4/100+x×9×5/100

14110= 14x/100+24x/100+45x/100(common denominator)(x is also common)

14110= 14x+24x+45x/100          

14110= 83x/100

  83x= 14110×100

      x= 1411000/83

      x= 17000

The amount of money she borrowed was Rs17000

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