Page 127 - Maths Skills - 8
P. 127
Comparing Quantities 125
Exercise 8.6
1. Find the amount and the compound interest on ` 18000 for 2 years at 8% per annum, compounded
half-yearly.
2. Find the amount and the compound interest on ` 10000 for 3 years at 12% per annum, compounded half-
yearly.
3. Find the compound interest on ` 31250 for 9 months, at 16% per annum compounded quarterly.
4. What will be the compound interest on ` 15000 for 1 year at 4% per annum compounded quarterly?
GROWTH (APPRECIATION) AND DEPRECIATION
Growth means increase counted as increase in size, amount or degree of anything over a period of time. For
example, the growth of height and weight of a child, the growth of a tree, the growth of population density etc.
The decrease in value of an article over a period of time is called depreciation. For example, the value of a
machine decreases over a period of time due to its wear and tear.
Following formulae are used in growth and depreciation problems:
Rule 1: When growth takes place at a constant rate
Let P = Initial value, R = Rate of growth and n = Time of growth in years.
R n
Then, for constant growth, the final value of growth (A) after n years is given by A= P 1
100
R n
Net growth = A – P = P 1+ − 1
100
Rule 2: When rate of growth is different for every year
Let, P = initial value and R , R , R , ... be the rates of growth for the first year, second year, third year and
1
3
2
so on.
Then, the final value of growth (A) is given by: A= P 1+ R 1+ R 1+ R ... 1+ R .
n
2
1
3
100 100 100 100
Rule 3: When depreciation takes place at a constant rate
When depreciation takes place at a constant rate, then we use (– R) in the place of R in the growth
formula. n
In case of depreciation, final value (A) is given by A= P 1 R
100
Rule 4: When depreciation rate is different for every year
When depreciation rate is R for the first year and R for the second year, then final value (A) is given by
1 2
R
R
A= P 1− 100 1− 100 , , where P = initial value.
2
1