NCERT Class 7 Mathematics

(Ron) #1

170 MATHEMATICS


SOLUTIONThe sum borrowed = Rs 5,000, Rate of interest = 15% per year.
This means if Rs 100 is borrowed, she has to pay Rs 15 as interest for one year. If she has
borrowed Rs 5,000, then the interest she has to pay for one year
= Rs^155000
100

× = Rs 750
So, at the end of the year she has to give an amount of Rs 5,000 + Rs 750 = Rs 5,750.
We can write a general relation to find interest for one year. Take P as the principal or
sum and R % as Rate per cent per annum.
Now on every Rs 100 borrowed, the interest paid is Rs R

Therefore, on Rs P borrowed, the interest paid for one year would be
100

RP×
=
PR×
100
.

8 .6.1 Interest for Multiple Years
If the amount is borrowed for more than one year the interest is calculated for the period
the money is kept for. For example, if Anita returns the money at the end of two years and
the rate of interest is the same then she would have to pay twice the interest i.e., Rs 750 for
the first year and Rs 750 for the second. This way of calculating interest where principal is
not changed is known as simple interest. As the number of years increase the interest
also increases. For Rs 100 borrowed for 3 years at 18%, the interest to be paid at the end
of 3 years is 18 + 18 + 18 = 3 × 18 = Rs 54.
We can find the general form for simple interest for more than one year.
We know that on a principal of Rs P at R% rate of interest per year, the interest paid
for one year is. Therefore, interest I paid for T years would be

And amount you have to pay at the end of T years is A = P + I


  1. Rs 10,000 is invested at 5% interest rate p.a. Find the interest at the end of one
    year.

  2. Rs 3,500 is given at 7% p.a. rate of interest. Find the interest which will be received
    at the end of two years.

  3. Rs 6,050 is borrowed at 6.5% rate of interest p.a.. Find the interest and the amount
    to be paid at the end of 3 years.

  4. Rs 7,000 is borrowed at 3.5% rate of interest p.a. borrowed for 2 years. Find the
    amount to be paid at the end of the second year.
    Just as in the case of prices related to items, if you are given any two of the three
    quantities in the relation
    100


I=PT R××, you could find the remaining quantity.

TRY THESE

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