Question

A person borrowed 8,000 at a certain rate of interest for 2 years and than 10,000 at 1% lower then the first in all paid 2500 as interest in 3 years. find the two rates at which he borrowed the amount.

Answer 1

Answer:

He borrowed Rs. 8000 on 10% interest and Rs. 10000 on 9% interest

Step-by-step explanation:

Let the rate at which he borrowed initially be r%

Interest in two years on Rs. 8000

I=\frac{8000*2*r}{100} =160rI=1008000∗2∗r=160r

After two years he borrows Rs. 10,000 on r-1 % interest

Therefore,

Interest paid in one year on Rs. 10000

I'=\frac{10000*1*(r-1)}{100} =100(r-1)I′=10010000∗1∗(r−1)=100(r−1)

Total interest paid = Rs. 2500

or I + I' = 2500

\implies 160r+100(r-1)=2500⟹160r+100(r−1)=2500

\implies 160r+100r-100=2500⟹160r+100r−100=2500

\implies 260r=2600⟹260r=2600

\implies r=2600/260=10⟹r=2600/260=10

Thus

He borrowed Rs. 8000 on 10% interest

He borrowd Rs. 10000 on 9% interest