Page 199 - Start Up Mathematics_8 (Non CCE)

\ Amount at the end of the first quarter = ` (6,000 + 240) = ` 6,240

Principal for the second quarter = ` 6,240, Rate of interest = 4%

,
¥¥
Ê 6 240 41ˆ
Interest for the second quarter = ` Á Ë 100 ˜ = ` 249.60
¯
\ Amount at the end of the second quarter = ` (6,240 + 249.60) = ` 6,489.60
Principal for the third quarter = ` 6,489.60, Rate of interest = 4%

,
.
¥¥
Ê 6 489 60 41ˆ
Interest for the third quarter = ` Á 100 ˜ = ` 259.584
Ë
¯
\ Amount at the end of the third quarter = ` (6,489.60 + 259.584) = ` 6,749.184
Principal for the fourth quarter = ` 6,749.184, Rate of interest = 4%
.
,
¥¥
Ê 6 749 184 41ˆ
Interest for the fourth quarter = ` Á Ë 100 ˜ = ` 269.97
¯
\ Amount at the end of the fourth quarter = ` (6,749.18 + 269.97) = ` 7,019.15
\ Compound interest = ` (7,019.15 – 6,000) = ` 1,019.15

EXERCISE 13.1

1. Rahul invested ` 7,000 in shares for 2 years at the rate of 12% per annum compounded annually.
What is the compound interest that Rahul gets after 2 years?
1
2. Find the compound interest on ` 3,000 at the rate of 6% per annum for 1 years compounded half-
yearly. 2
1
3. Veena borrowed ` 20,000 from a money lender at the rate of 4% per annum for 1 years compounded
2
1
half-yearly. Find the amount she has to pay at the end of 1 years.
2
4. Find the compound interest on a sum of ` 12,000 at the rate of 8% per annum for 1 year compounded
quarterly.
5. Ruchi borrowed ` 18,000 from her friend at the rate of 12% per annum at simple interest. She lent it
further to another friend Aarti at the same rate but compounded annually. Find the profit that Ruchi
earns after 2 years.
6. Kavya deposited ` 8,500 in a bank which pays her 12% interest per annum compounded quarterly.
What is the amount which she receives after 9 months?

Computation of Compound Interest Using Formulae
Computation of compound interest and the amount when interest is compounded annually

If, P = Principal R% = Rate of interest per year

Ê R ˆ n
n = Number of years A = P 1+ 100¯ ˜
Á
Ë

Ô
Ê R ˆ n Ï Ê Ô R ˆ n ¸
Then, C.I. = A – P = P 1+ 100¯ ˜ – P = P Ì Á Ë 1+ 100¯ ˜ - 1 ˝
Á
Ë
˛ Ô
Ó Ô
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