# 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) 8 | B) 9 |

C) 10 | D) 11 |

Explanation:

Actual price = Rs. (25 + 2.50) = Rs. 27.50.

Saving = 2.50

Percentage Saving = [(2.50/27.50)*100]

= 2500/275

= $9\frac{1}{11}$%

$\approx 9$%

A) 1000 | B) 2000 |

C) 3000 | D) 4000 |

Explanation:

Let the original number be x.

Final number obtained = 110% of (90% of x) =(110/100 * 90/100 * x) = (99/100)x.

x-(99/100)x=10

=> x =1000

A) 337.50 | B) 217.50 |

C) 376.21 | D) 120 |

Explanation:

Given expression = (45 x 750/100) - (25 x 480/100) = (337.50 - 120) = 217.50

A) .25% | B) .5% |

C) .75% | D) 1% |

Explanation:

Required percentage = (18/7200 * 100)% = 1/4% = 0.25%

A) 66.66% | B) 76.66% |

C) 96.66% | D) 86.66% |

Explanation:

Excess of B's height over A's = [(40/(100 - 40)] x 100% = 66.66%

A) 10 | B) 20 |

C) 30 | D) 35 |

Explanation:

From the given data,

$\frac{\mathbf{40}}{\mathbf{100}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{75}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{80}}{\mathbf{100}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{25}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{K}}{\mathbf{100}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{250}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}30+20=\frac{5\mathrm{K}}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}50\mathrm{x}2=5\mathrm{K}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{K}=\frac{100}{5}=20$

A) 60% | B) 40% |

C) 50% | D) 30% |

Explanation:

Lets say shyam sells at 100,

Since Ram sells 25% cheaper than Shyam,

Therefor Ram sells at less than 25% of100 or 75 rs.

Ram sells 25% dearer than Bram or 125% of Ram =100% of Bram or

125% of x (say x price of bram )=75rs.

or 100% of x =60rs.

hence price of bram is 60rs.

now Bram's good is cheaper than Shyam's as (100-60)x100/100% or 40%.

Hence Brams Good is 40% cheaper than that of Shyam's good.

A) 15 | B) 20 |

C) 25 | D) 30 |

Explanation:

Let original consumption = 100 kg and new consumption = x kg,

So, 100 * 6 = x * 7.50 => x = 80kg

Reduction in consumption = 20%.