# Percentage

Q11. A reduction of 20% in the price of rice enables a person to buy 3.5 Kg more rice for Rs. 385. The original price of the rice is

(a) Rs. 25.50 /Kg

(b) Rs. 26 /Kg

(c) Rs. 27.50 /Kg

(d) Rs. 28.50 /Kg

**Answer:** (c) Rs. 27.50 /Kg

Let, the original price of rice be Rs. x per Kg

New price of 1 Kg rice = Rs.(x - (20/100) X x) = Rs. (x - x/5) = Rs. 4x/5

A/Q | 5 X 385 | - | 385 | = 3.5 |

4x | x |

⇒ | 1925 - 1540 | = 3.5 |

4x |

⇒ | 385 | = 3.5 |

4x |

⇒ x = 385/(4X3.5) = 27.5 = Rs. 27.50/Kg

Q12. The quantity of water that should be added to reduce 10ml lotion containing 60% alcohol to a lotion containing 40% alcohol is

(a) 5 ml

(b) 8 ml

(c) 11 ml

(d) 13 ml

**Answer:** (a) 5 ml

Alcohol in 10ml lotion = (60/100) X 10 ml = 6 ml

Water in it = 10ml - 6ml = 4 ml

Let x ml of water be added to it

Now | 6 | = 40% |

10 + x |

⇒ | 6 | = | 40 | = | 2 |

10 + x | 100 | 5 |

⇒ 20 + 2x = 30

⇒ 2x = 10

⇒ x = 5 ml

Q13. In two successive years, 200 and 150 students of a school appeared at the final examination. Respectively 70% and 60% of them passed. The average rate of pass is

(a) 63.25%

(b) 65.12%

(c) 65.71%

(d) 66.10%

**Answer:** (c) 65.71%

Total candidates = 200 + 150 = 350

Total passed = 200 X | 70 | + 150 X | 60 |

100 | 100 |

= 140 + 90 = 230

Pass% = (230/350) X 100 = 65.71%

Q14. In two successive years, 100 and 80 students of a school appeared at the final examination. Respectively 75% and 60% of them passed. Teh average rate of fail is

(a) 28.75%

(b) 31.66%

(c) 32.54%

(d) 37.45%

**Answer:** (b) 31.66%

Total candidate = 100 + 80 = 180

Total passed = 100 X | 75 | + 80 X | 60 |

100 | 100 |

= 75 + 48 = 123

Total failed = 180 - 123 = 57

Fail% = (57/180) X 100 = 31.66

Q15. The price of petrol is increased by 20%. By how much percent a car owner should reduce his consumption of petrol so that the expenditure on the petrol would not be increased ?

(a) 13(2/3)%

(b) 14(1/5)%

(c) 16(2/7)%

(d) 16(2/3)%

**Answer:** (d) 16(2/3)%

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

100 + r |

= | 20 | X 100 % = | 200 | % = 16 | 8 | % = 16 | 2 | % |

120 | 12 | 12 | 3 |

Q16. After reducing the salary of a man by 50% and subsequently it is increased by 40%. The loss is

(a) 10%

(b) 20%

(c) 25%

(d) 30%

**Answer:** (d) 30%

Let the salary of the man be Rs. 100

After reducing salary = 100 - 100 X(50/100) = Rs. 50

After increasing salary = Rs. 50 + 50 X (40/100) = Rs. 70

∴ Loss% = 100% - 70% = 30%

Q17. 10% of the electorate did not cast their votes in an election between two candidates. 10% of the votes polled were found invalid. The successful candidate got 54% of the valid votes and won by a majority of 1620 votes. The number of voters enrolled on the voters list was

(a) 15000

(b) 18000

(c) 22000

(d) 25000

**Answer:** (d) 25000

Let the number of voters enrolled be x

Votes cast = (90/100) X x = 9x/10

Valid votes = 90% of | 9x | = | 90 | X | 9x | = | 81x |

10 | 100 | 10 | 100 |

Votes polled by successful candidate = | 54 | X | 81x | = | 4374x |

100 | 100 | 10000 |

Votes polled by defeated candidate = | 46 | X | 81x | = | 3726x |

100 | 100 | 10000 |

4374x - 3726x = 16200000

⇒ x = 25000

Q18. The sum of two numbers is 2490. If 6.5% of one number is equal to 8.5% of the other, then the numbers are

(a) 1025

(b) 1158

(c) 1411

(d) 1735

**Answer:** (c) 1411

Let, the numbers be x and 2490 - x

Now 6.5% of x = 8.5% of (2490 - x)

⇒ 6.5x/100 = 8.5/100 (2490 - x)

⇒ 65x/100 = 85/100(2490 - x)

⇒ 65x = 85 X 2490 - 85x

⇒ 150x = 85 X 2490

⇒ x = 1411

Q19. In measuring the sides of a rectangle errors of 5% and 3% in excess are made. The error percent in the calculated area is

(a) 5.25%

(b) 8.15%

(c) 9.12%

(d) 11.23%

**Answer:** (b) 8.15%

Let, length = x units

breadth = y units

∴ Actual area = xy sq. units

Length shown = (105/100) X x =21x/20 units

Breadth shown = (103/100) X y = 103y/100 units

Calculated area = (21x/100) X (103y/100) sq. unit = 2163xy/2000 sq. unit

Error = 2163xy/2000 - xy = 163xy/2000

Error % = | 163xy | X | 1 | X 100% = | 163 | % = 8.15% |

2000 | xy | 20 |

Q20. Of the 1000 inhabitants of a town, 60% are females of whom 20% are literate. If of all the inhabitants 25% are literate, then what percent of the males of the town are literate

(a) 32.5%

(b) 33.4%

(c) 35.9%

(d) 42.5%

**Answer:** (a) 32.5%

Females = (60/100) X 1000 = 600

Males = 1000 - 600 = 400

Literate Females = (20/100) X 600 = 120

Total literates = (25/100) X 1000 = 250

∴ Male literates = 250 - 120 = 130

∴ Required % = (130/400) X 100% = 32.5%

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