Aptitude Questions and Answers

- 1. Percentage
- 2. Why Percentage?
- 3. Percentages - Fractions Conversions
- 4. Converting Decimals
- 5. Important Points to Note

The word percent can be understood as follows:

Per cent $\Rightarrow$ for every 100.

So, when percentage is calculated for any value, it means that you calculate the value for every 100 of the reference value. When you see the word "percent" or the symbol %, remember it means $\dfrac{1}{100}$.

$20 \text{ percent} = 20\% = 20 \times \left(\dfrac{1}{100}\right) = \dfrac{1}{5}$

Percentage is a concept evolved so that there can be a uniform platform for comparison of various things. (Since each value is taken to a common platform of 100)

To compare three different students depending on the marks they scored we cannot directly compare their marks until we know the maximum marks for which they took the test. But by calculating percentages they can directly be compared with one another.

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 , $x\% = \dfrac{x}{100}.$

Thus,$ 20\% = \dfrac{20}{100}=\dfrac{1}{5};$

$48\% =\dfrac{48}{100} =\dfrac{12}{25}$ etc.

To express $\dfrac{a}{b}$ as a percent: We have,$ \dfrac{a}{b} =\left(\dfrac{a}{b}\right) \times 100\%$

Thus, $\dfrac{1}{4} =\left[\dfrac{1}{4} \times 100\right] = 25\%;$

$0.6 =\dfrac{6}{10} =\dfrac{3}{5} =\left[\dfrac{3}{5} \times 100\right]\% =60\%.$

If the price of a commodity increases by $R\%$, then the reduction in consumption so as not to increase the expenditure is:

$=\left[\left(\dfrac{R}{(100+R)}\right) \times 100\right]\%$

If the price of the commodity decreases by $R\%$, then to maintain the same expenditure by increasing the consumption is:

$=\left[\left(\dfrac{R}{(100-R)}\right) \times 100\right]\%$

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

1. Population after $n$ years $= P \left[1+\left(\dfrac{R}{100}\right)\right]^n$

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

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

1. Value of the machine after $n$ years $= P\left[1-\left(\dfrac{R}{100}\right)\right]^n$

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

5. If A is $R\%$ more than B, then B is less than A by

$=\left[\left(\dfrac{R}{(100+R)}\right) \times 100\right]\%$

If A is $R\%$ less than B , then B is more than A by

$=\left[\left(\dfrac{R}{(100-R)}\right) \times 100\right]\%$

For faster calculations we can convert the percentages or decimal equivalents into their respective fraction notations. The following is a table showing the conversions of percentages and decimals into fractions:

Table below can be scrolled horizontally

Percentage | Decimal | Fraction |
---|---|---|

10% |
0.1 |
1/10 |

12.5% |
0.125 |
1/8 |

16.66% |
0.1666 |
1/6 |

20% |
0.2 |
1/5 |

25% |
0.25 |
1/4 |

30% |
0.3 |
3/10 |

33.33% |
0.3333 |
1/3 |

40% |
0.4 |
2/5 |

50% |
0.5 |
1/2 |

60% |
0.6 |
3/5 |

62.5% |
0.625 |
5/8 |

66.66% |
0.6666 |
2/3 |

70% |
0.7 |
7/10 |

75% |
0.75 |
3/4 |

80% |
0.8 |
4/5 |

83.33% |
0.8333 |
5/6 |

90% |
0.9 |
9/10 |

100% |
1.0 |
1 |

We can go for converting decimals more than 1 from the knowledge of the above cited conversions as follows:

We know that $12.5\% = 0.125 = \dfrac{1}{8}$

Then, $1.125 = \dfrac{[8(1)+1]}{8} = \dfrac{9}{8}$ (i.e., the denominator will add to numerator once, denominator remaining the same.

Also, $2.125 = \dfrac{[8(2)+1]}{8} = \dfrac{17}{8}$ (here the denominator is added to numerator twice)

$3.125 = \dfrac{[8(3)+1]}{8} = \dfrac{25}{8}$ and so on.

Thus we can derive the fractions for decimals more than 1 by using those less than 1.

We will see how use of fractions will reduce the time for calculations:

What is 62.5% of 320?

Value = $\left(\dfrac{5}{8}\right) \times 320 $ (since $62.5\% = \dfrac{5}{8}$) $= 200$.

When any value increases by

10%, it becomes 1.1 times of itself. (since 100+10 = 110% = 1.1)

20%, it becomes 1.2 times of itself.

36%, it becomes 1.36 times of itself.

4%, it becomes 1.04 times of itself.

Thus we can see the effects on the values due to various percentage increases.

When any value decreases by

10%, it becomes 0.9 times of itself. (Since 100-10 = 90% = 0.9)

20%, it becomes 0.8 times of itself

36%, it becomes 0.64 times of itself

4%, it becomes 0.96 times of itself.

Thus we can see the effects on a value due to various percentage decreases.

1. When a value is multiplied by a decimal more than 1 it will be increased and when multiplied by less than 1 it will be decreased.

2. The percentage increase or decrease depends on the decimal multiplied.

When the actual value is $x$, find the value when it is 30% decreased.

30% decrease => 0.7 $x$.

A value after an increase of 20% became 600. What is the value?

$1.2x = 600$ (since 20% increase)

=> $x = 500$.

If 600 is decrease by 20%, what is the new value?

new value $= 0.8 \times 600 = 480$. (Since 20% decrease)

Thus depending on the decimal we can decide the % change and vice versa.

When a value is increased by 20%, by what percent should it be reduced to get the actual value?

(It is equivalent to 1.2 reduced to 1 and we can use % decrease formula)

$\% \text{ decrease} = \left(\dfrac{1.2 - 1}{1.2}\right) \times 100 = 16.66\%$

When a value is subjected multiple changes, the overall effect of all the changes can be obtained by multiplying all the individual factors of the changes.

The population of a town increased by 10%, 20% and then decreased by 30%. The

new population is what % of the original?

The overall effect $= 1.1 \times 1.2 \times 0.7$ (Since 10%, 20% increase and 30% decrease)

$= 0.924 = 92.4\%$.

Two successive discounts of 10% and 20% are equal to a single discount of ___

Discount is same as decrease of price.

So, decrease $= 0.9 \times 0.8 = 0.72 \Rightarrow 28\%$ decrease (Since only 72% is remaining)