# 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) 2500 | B) 2700 |

C) 2900 | D) 3100 |

Explanation:

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = 7500*(80/100)

1st candidate got 55% of the total valid votes.

Hence the 2nd candidate should have got 45% of the total valid votes

=> Valid votes that 2nd candidate got = total valid votes x (45/100)

7500*(80/100)*(45/100) = 2700

A) 4% of a | B) 6% of a |

C) 8% of a | D) 10% of a |

Explanation:

20% of a = b

=> (20/100) * a = b

b% of 20 =(b/100) x 20 = [(20a/100) / 100] x 20= 4a/100 = 4% of a.

A) 54 % | B) 64 % |

C) 74 % | D) 84 % |

Explanation:

Let the number be x.

Then, ideally he should have multiplied by x by 5/3. Hence Correct result was x * (5/3)= 5x/3.

By mistake he multiplied x by 3/5 . Hence the result with error = 3x/5

Then, error = (5x/3 - 3x/5) = 16x/15

Error % = (error/True vaue) * 100 = [(16/15) * x/(5/3) * x] * 100 = 64 %

A) 20 | B) 30 |

C) 40 | D) 50 |

Explanation:

The fruit content in both the fresh fruit and dry fruit is the same.

Given, fresh fruit has 68% water.so remaining 32% is fruit content. weight of fresh fruits is 100kg

Dry fruit has 20% water.so remaining 80% is fruit content.let weight if dry fruit be y kg.

Fruit % in freshfruit = Fruit% in dryfruit

Therefore, (32/100) x 100 = (80/100 ) x y

we get, y = 40 kg.

A) 10000 | B) 12000 |

C) 14000 | D) 16000 |

Explanation:

Purchase price = $Rs.\left[\frac{8748}{{{}^{\left(1-{\displaystyle \frac{10}{100}}\right)}}^{3}}\right]$ = Rs. [8748 * (10/9) * (10/9 )* (10/9)] = Rs.12000

A) 6 % | B) 8 % |

C) 10 % | D) 12 % |

Explanation:

Let the total income be x.

Then, income left = (100 - 80)% of [100 - (35 + 25)] % of x = 20% of 40% of x = [(20/100) * (40/100) * 100] % of x = 8 % of x.

A) 10 | B) 20 |

C) 30 | D) 40 |

Explanation:

Let the original price be Rs. 100.

New final price = 120 % of (75 % of Rs. 100) = Rs. [(120/100) * (75/100) * 100] = Rs. 90.

Decrease = 10%