# 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) 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) 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) (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) %.

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) 21.6 % | B) 15.3 % |

C) 14.6 % | D) 12.5 % |

Explanation:

Total money = Rs.[600*(25/100)+1200*(50/100)]= Rs. 750.

25 paise coins removed = Rs. (600*12/100) = 72.

50 paise coins removed = Rs. (1200*24/100)= 288.

Money removed =Rs.(72*25/100+288*50/100) = Rs.162.

Required percentage = (162/750*100)% = 21.6 %.

A) 777 | B) 666 |

C) 606 | D) 789 |

Explanation:

56% of 870 = 56x870/100 = 487.20

82% of 180 = 82x180/100 = 147.60

32% of 90 = 32x90/100

487.20 + 147.6 - 28.8 = ** ?**

**?** = 634.8 - 28.8

**?** = 606.

A) 60 | B) 70 |

C) 80 | D) 90 |

Explanation:

We have : x + (x% of 150 )= 150

=> x+(x/100)*150] = 150

=> $\frac{5}{2}x=150$

=> x = (150*2)/5 = 60

A) 15 | B) 25 |

C) 35 | D) 45 |

Explanation:

Let the large number be x.

Then x - 20 = 20x/100

=> x - x/5 = 20 => x = 25.