Page 207 - Start Up Mathematics_8 (Non CCE)

7. Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3
years at the rate of 5% per annum gives ` 1,200 as the simple interest.
1
8. A sum of ` 31,250 amounts to ` 35,152 in 1 years. Find the rate per annum, interest being
compounded semi-annually. 2
1
9. Hariti invested a certain sum of money for 1 years at the rate of 15% per annum compounded
1 2
six-monthly. At the end of 1 years she receives a compound interest of ` 1,891.50. Find the sum
invested by Hariti. 2
10. Find the rate at which a certain sum of money will almost double itself in 2 years, if the interest is
compounded annually.
11. The difference between the compound interest and the simple interest on a sum of ` 15,000 for
2 years is ` 96. What is the rate of interest?
12. Pallavi deposited ` 46,875 in a bank. In how many years will this sum yield a compound interest of
` 5,853 at the rate of 8% per annum compound semi-annually?
13. Sanju took a loan from the bank to buy a motorcycle at the rate of 15% per annum compounded
1
half-yearly. He paid a compound interest of ` 15,507 after 1 years. Find his loan amount.
2
14. In how much time will a sum of money double if invested at the rate of 8% simple interest per
annum?

Appreciation

In our daily life, many things and assets like property value, population of city, etc. grow or appreciate over
a period of time. The relative increase in value is called appreciation and the appreciation per unit of time is
called the rate of appreciation.
(a) If P is the value of the asset at the beginning of a certain period and R% is the rate of appreciation/

Ê R ˆ n
growth per annum, then asset value after n year = P 1+ 100¯ ˜ .
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Ë
(b) If P is the value of the asset at the beginning of a certain year and R %, R %, ..., R % are the rates of
2
n
1
appreciation/growth in the 1st, 2nd, ...... nth year respectively, then the asset value after n years
Ê R ˆ Ê R ˆ Ê R ˆ
2
n
1
= P 1+ 100¯ Ë 1+ 100¯ ... 1+ 100¯ ˜ .
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Á
˜ Á
˜
Ë
Ë
In some cases, the asset value reduces during certain years or rate of appreciation/growth is negative. In
such problems, the rate of appreciation/growth (R) is taken as negative, i.e., (–R) in the formula. An example
for such a case will be appreciation in population over the years. But the rate of appreciation can turn
negative over certain years due to, say, an epidemic.
Example 17: The population of a town is 50,000. If the annual birth rate is 5% and the annual death rate
is 3%, find the population after two years.
Solution: Annual birth rate = 5%, Annual death rate = 3%
\ Annual growth = (5 – 3)% = 2%
Initial population (P) = 50,000, Rate of growth (R) = 2% per annum, Time (n) = 2 years

Ê R ˆ n Ê 2 ˆ 2
\ Population after 2 years = P 1+ 100¯ ˜ = 50,000 1+ 100¯ ˜
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