# Percentage

Q1. 50% of ? = 250

(a) 400

(b) 500

(c) 800

(d) 1000

**Answer:** (b) 500

Let, 50% of x = 250

⇒ (50/100) X x = 250

⇒ x = (250 X 100)/50 = 500

Q2. Rs. 395 are divided among A, B and C in such a manner that B gets 25% more than A and 20% more than C. The share of A is

(a) Rs. 80

(b) Rs. 95

(c) Rs. 120

(d) Rs. 135

**Answer:** (c) Rs. 120

Let A gets Rs. x

Then B gets 125% of x = 125/100 X x = Rs. 5x/4

If B gets Rs. 5x/4, then C gets = Rs. ( | 100 | X | 5x | ) = Rs. | 25x |

120 | 4 | 24 |

∴ x + | 5x | + | 25x | = 395 |

4 | 24 |

⇒ 24x + 30x + 25x = 395 X 24

⇒ 79x = 395 X 24

⇒ x = 120 = Rs. 120

Q3. The price of a certain item is increased by 10%. If a consumer wants to keep his expenditure on the item same as before, how much percent must he reduce his consumption of that item

(a) 3(1/11)%

(b) 5(1/11)%

(c) 9(1/11)%

(d) 12(1/11)%

**Answer:** (c) 9(1/11)%

Reduction % in consumption = { | r | X 100% } |

100 + r |

= ( | 10 | X 100 )% |

100 + 10 |

= | 10 | X 100 % |

110 |

= | 100 | % |

11 |

= 9 | 1 | % |

11 |

Q4. Three candidates in all election received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the wining candidate get?

(a) 45%

(b) 51%

(c) 53%

(d) 57%

**Answer:** (d) 57%

Total votes = 1136 + 7636 + 11628 = 20400

Required Percentage = (11628/20400) X 100 % = 57%

Q5. If the numerator of a fraction is increased by 150% and the denominator is increased by 300% the resultant fraction is 5/16. What is the original fraction

(a) ½

(b) ¼

(c) ¾

(d) 1/5

**Answer:** (a) ½

Let the original fraction be x/y, then

x + x X 150% | = | 5 |

y + y X 300% | 16 |

⇒ | x + (150x/100) | = | 5 |

y + (300y/100) | 16 |

⇒ | 250x | = | 5 |

400y | 16 |

⇒ | x | = | 5 | X | 400 | = ½ |

y | 16 | 250 |

Q6. The value of a plant depreciates by 10% annually. If the present value of the plant is Rs. 100000, then what will be its value after 3 years

(a) Rs. 43800

(b) Rs. 55900

(c) Rs. 68100

(d) Rs. 72900

**Answer:** (d) Rs. 72900

Value after 3 years = Rs. [100000 X ( 1 - | 10 | )^{3}] |

100 |

= Rs. [100000 X | 90 | X | 90 | X | 90 | ] |

100 | 100 | 100 |

= Rs. 72900

Q7. In an election between two candidates, 75% of the voters cast their votes, out of which 4% of the voters were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. The total number of voters enrolled in that election was

(a) 16832

(b) 17150

(c) 18460

(d) 18865

**Answer:** (b) 17150

Let the enrolled vote be x

Votes cast = (75/100) X x = 3x/4

Valid votes = (3x/4) X (96/100) = 18x/25

A/Q | 18x | X 75% = 9261 |

25 |

⇒ | 18x | X | 75 | = 9261 |

25 | 100 |

⇒ x = (9261 X 100)/(18X3) = 17150

Q8. Fresh grapes contains 80% water while dry grapes contain 10% water. If the weight of dry grapes is 250 kg, what is the total weight when it was fresh ?

(a) 980 Kg

(b) 1020 Kg

(c) 1125 Kg

(d) 1235 Kg

**Answer:** (c) 1125 Kg

Let the weight of the fresh grapes be x kg.

Quantity of water in it = (80/100) X x = 4x/5 Kg

Quantity of pulp in it = (x - 4x/5) Kg = x/5 Kg

Quantity of water in 250 kg dry grapes = (10/100) X 250 Kg = 25 Kg

Quantity of pulp in it = 250 - 25 = 225 kg

∴ x/5 = 225

⇒ x = 1125

Weight of fresh grapes = 1125 Kg

Q9. If the numerator of a fraction is increased by 120% and the denominator is increased by 130%, the resultant fraction is 4/15. What is the original fraction ?

(a) 33/165

(b) 41/165

(c) 46/165

(d) 51/165

**Answer:** (c) 46/165

Let, the original function be x/y

A/Q | 220x | = | 4 |

230y | 15 |

⇒ | x | = | 4 | X | 230 |

y | 15 | 220 |

= 46/165

Q10. Fresh fruit contains 60% water and dry fruit contains 20% water. How much dry fruit can be obtained from 200Kg of fresh fruit ?

(a) 90 Kg

(b) 100 Kg

(c) 110 Kg

(d) 120 Kg

**Answer:** (b) 100 Kg

Quantity of water in 200 kg of fresh fruits = 60/100 X 200 Kg = 120 Kg

Quantity of pulp in it = (200 - 120) Kg = 80 Kg

Let the dry fruit be x Kg

Water in it = (20/100) X x = x/5 Kg

Quantity of pulp in it = (x - x/5) = 4x/5 Kg

Now 4x/5 = 80

⇒ x = (80X5)/4 = 100 Kg