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## Arithmetic: Percentage Test -5

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Question 1 |

A person bought two tables for Rs. 2200. He sells one at 5% loss and the other at 6% profit and thus on the whole he neither gains nor loses. Find the cost price of each table.

Rs 1500, Rs 700 | |

Rs 2000, Rs 200 | |

Rs 1200, Rs 1000 | |

Rs 1100, Rs 1100 |

Question 1 Explanation:

Let the cost of table = t

Then the cost price of other will be = (2200 –t)

=> t x (95/100) + (2200 – t) x 106/100 =2200

=> 95t + 233200 – 106 t =220000

=> 11t = 13200

=> t = 1200 Rs and 2200 – t =1000 Rs

Then the cost price of other will be = (2200 –t)

=> t x (95/100) + (2200 – t) x 106/100 =2200

=> 95t + 233200 – 106 t =220000

=> 11t = 13200

=> t = 1200 Rs and 2200 – t =1000 Rs

Question 2 |

A tree was planted three years ago. The rate of its growth is 30% per annum. If at present, the height of the tree is 670 cm, what was it when the tree was planted?

305 cm | |

500 cm | |

405 cm | |

625 cm |

Question 2 Explanation:

Let the height of the tree be (h)

=> h x {1+(30/100)

=> h = 670 x {( 10 x 10 x10 )/(13 x 13 x 13)} = 305 approx.

=> h x {1+(30/100)

^{3}} =670=> h = 670 x {( 10 x 10 x10 )/(13 x 13 x 13)} = 305 approx.

Question 3 |

When the price of sugar was increased by 32%, a family reduced its consumption in such a way that the expenditure on sugar was only 10% more than before. If 30 kg per month were consumed before, find the new monthly consumption.

42 kg | |

35 kg | |

25 kg | |

16 kg |

Question 3 Explanation:

Let the price of the sugar be Rs(p) / kg

=> Initial expenditure = 30p

=> new expenditure = 33p

=> new monthly consumption = 33p/1.32p = 25kg

=> Initial expenditure = 30p

=> new expenditure = 33p

=> new monthly consumption = 33p/1.32p = 25kg

Question 4 |

A man's income is increased by Rs. 1200 and at the same time, the rate of tax to be paid is reduced from 12% to 10%. He now pays the same amount of tax as before. What is his increased income, if 20% of his income is exempted from tax in both cases?

Rs 6300 | |

Rs 7200 | |

Rs 4500 | |

Rs 6500 |

Question 4 Explanation:

Let the increased income be P

=> (P – 1200) x 80/100 x 12/100 = P x 80/100 x10/100

=> 12P – 14400 =10p

=> P = Rs. 7200

=> (P – 1200) x 80/100 x 12/100 = P x 80/100 x10/100

=> 12P – 14400 =10p

=> P = Rs. 7200

Question 5 |

Mr. Bharat Toni buys a generator for Rs. 100000 and rents it. He puts 12.5% of each month's rent aside for upkeep and repairs, pays Rs. 325 per year as taxes and realizes 5.5% annually on his investment. Find the monthly rent.

Rs 634.76 | |

Rs 654.76 | |

Rs 554.76 | |

Rs 456.32 |

Question 5 Explanation:

Let Monthly rent be = (r)

=> 12r – 12r x (12.5/100) -325 = 100000 x (5.5/100)

=> 12r – 1.5r – 325 = 5500

=> r = (5500 +325)/ 10.5 = Rs 554.76

=> 12r – 12r x (12.5/100) -325 = 100000 x (5.5/100)

=> 12r – 1.5r – 325 = 5500

=> r = (5500 +325)/ 10.5 = Rs 554.76

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### Table of Contents

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### How to study this lesson?

It is actually simple to study this lesson.

Go through the given concepts and their corresponding exercises.

Then, solve the percentage practice tests and perfect the skills you have learnt