# Percentage Questions

**FACTS AND FORMULAE FOR PERCENTAGE QUESTIONS**

**I.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% = x/100.

Thus, 20% = 20/100 = 1/5;

48% = 48/100 = 12/25, etc.

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

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

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

If the price of the commodity decreases by R%,then the increase in consumption so as to decrease the expenditure is $\left[\frac{R}{\left(100-R\right)}\times 100\right]\%$

**III. Results on Population : **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 = $P{\left(1+\frac{R}{100}\right)}^{n}$

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

**IV. Results on Depreciation :** 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 = $P{\left(1-\frac{R}{100}\right)}^{n}$

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

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

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

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

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

A) 10 kg | B) 15 kg |

C) 20 kg | D) 25 kg |

Explanation:

Let the original quantity be x kg. Vanaspati ghee in x kg = (40x / 100 )kg = (2x / 5) kg.

Now, (2x/5)/(x + 10) = 20/100

=> 2x / (5x + 50) = 1/5

=> 5x = 50

=> x = 10.

A) 1096.30 | B) 1226.70 |

C) 1124.20 | D) 1186.70 |

Explanation:

52.5% of 800 + 30.5% of 2800 = ? + 87.30

420 + 854 - 87.30 = ?

1274 - 87.30 = ?

? = **1186.70**

A) 1.4 | B) 1.5 |

C) 1.6 | D) 1.7 |

Explanation:

Quantity of pure acid = 20% of 8 litres = ((20/100)*8) litres = 1.6 litres.

A) 4000 | B) 8000 |

C) 12000 | D) 16000 |

Explanation:

Let the number of candidates appeared from each state be x.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

A) 65380 | B) 68921 |

C) 70987 | D) 72345 |

Explanation:

Population afer 3 years = 64000 * (1 + 5/(2 *100) )^3 = (64000 * 41/40 * 41/40 * 41/40 ) = 68921..

A) increase by 2% | B) decrease by 3% |

C) decrease by 4% | D) increase by 5% |

Explanation:

Here, x = 20 and y = - 20

Therefore, the net % change in value

=**$x+y+\frac{xy}{100}$ %**

= $\mathbf{20}\mathbf{}\mathbf{+}\mathbf{}\left(\mathbf{-}\mathbf{20}\right)\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{20}\mathbf{}\mathbf{\times}\mathbf{}\mathbf{-}\mathbf{20}}{\mathbf{100}}$% or - 4%

Since the sign is negative, there is a decrease in value by 4%.

A) 75 | B) 85 |

C) 95 | D) 105 |

Explanation:

Let the number be x. Then, x-(2/5)x = 510

=> 3x/5 =510

=>x =[510 * ( 5/3)] =850

10 % 0f 850 = 85.

A) 52% | B) 64% |

C) 84% | D) 77% |

Explanation:

Let the number be 'x'

Then, according to the given data,

$\frac{{\displaystyle \frac{\mathbf{5}}{\mathbf{2}}\mathbf{x}\mathbf{-}\frac{\mathbf{2}}{\mathbf{5}}\mathbf{x}}}{{\displaystyle \frac{\mathbf{5}}{\mathbf{2}}\mathbf{x}}}\mathit{x}\mathbf{100}$

= $\frac{21}{25}x100$

= 84%