Quantitative Aptitude :: Percentage

Concept of Percentage:

By a certain percent, we mean that many hundredths.Thus, X percent means x hundredths, written as X%.

To express X% as a fraction: We have, X% = \(\frac{X}{100}\).

Thus, 20% = \(\frac{20}{100}\) = \(\frac{1}{5}\).

To express \(\frac{a}{b}\) as a percent: We have, \(\frac{a}{b}\) = \(\frac{a}{b} \times 100\)%.

Thus, \(\frac{1}{4}\) = \(\left(\frac{1}{4} \times 100 \right) \)% = 25%.

Percentage Increase/Decrease:

If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is: \(\left[\frac{R}{(100+R)} \times 100\right]\)%

If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:

\(\left[\frac{R}{(100-R)} ]\times 100\right]\)%

Results on Population:

Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:

1. Population after n years = \(P \times \left(1 + \frac{R}{100}\right)^{n}\)

2. Population n years ago = \(\frac{P}{\left(1 + \frac{R}{100}\right)^n}\)

Results on Depreciation:

Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:

1. Value of the machine after n years = \(P \times \left(1 - \frac{R}{100}\right)^{n}\)

2. Value of the machine n years ago = \(\frac{P}{\left(1 - \frac{R}{100}\right)^n}\)

3. If A is R% more than B, then B is less than A by

\(\left[\frac{R}{(100 + R)} \times 100\right]\)%.

4. If A is R% less than B, then B is more than A by

\(\left[\frac{R}{(100 - R)} \times 100\right]\)%.