# 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) 32.5 % | B) 43 % |

C) 46.6 % | D) 53.2 % |

Explanation:

Number of males = 60% of 1000 = 600. Number of females = (1000 - 600) = 400.

Number of literates = 25% of 1000 = 250.

Number of literate males = 20% of 600 = 120.

Number of literate females = (250 - 120) = 130.

Required pecentage = (130/400 * 100 ) % = 32.5 %.

A) 7500 | B) 6200 |

C) 5600 | D) 4700 |

Explanation:

Let the candidates be A & B

35%-----------A

65%-----------B

Difference of votes (65-35)% = 30%

=> 30% ------- 2250

Then for 100% -------- ? => 7500

A) 65.5 % | B) 64.5 % |

C) 63.5 % | D) 62.5 % |

Explanation:

x = 80 % of y

=> x =(80/100 )y

=> y/x = 5/4

Required percentage = [(y/2x)*100] % = (5/8*100) % =62.5%

A) 155679 | B) 167890 |

C) 179890 | D) 177366 |

Explanation:

Present population = 160000 * (1 + 3/100)(1 + 5/200)(1 + 5/100)= 177366.

A) 20 % | B) 40 % |

C) 60 % | D) 80 % |

Explanation:

Let the total original sale be Rs. 100. Then, original number of visitors = 100.

New number of visitors = 120/0.75 = 160.

Increase % = 60 %.

A) 324138 | B) 349680 |

C) 509940 | D) None of these |

Explanation:

Total Population = 728400

Migrants = 35 % of 728400 = 254940

local population = (728400 - 254940) = 473460.

Rural migrants = 20% of 254940 = 50988

Urban migrants = (254940 - 50988) = 203952

Female population = 48% of 473460 + 30% of 50988 + 40% of 203952 = 324138

A) (16 + 2/3) % | B) (15 + 2/3)% |

C) (14 +1/3)% | D) (13 + 4/5)% |

Explanation:

Let the orginal salary = Rs. 100. New's salary = Rs.120.

Decrease on 120 = 20. Decrease on 100 = [(20/120)*100]% = (16 + 2/3) %.