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## Arithmetic: Percentage Test -2

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Question 1 |

In an examination, 40% marks are required to pass. A obtains 10% less than the number of marks required to pass. B obtains 11

^{1}/_{9}% less than A and C obtained 41^{3}/_{17}% less than the number of marks obtained by A and B together. What marks did C get? 50 | |

40 | |

35 | |

45 |

Question 1 Explanation:

Let the total marks be 100

Minimum marks required to pass the exam =40

Marks obtained by A = 40 – (40 x 10/100) = 36 marks

Marks obtained by B= 36 – (100/9 x 36/100) = 32 marks

Marks obtained by C ==(36 +32) x 700/(17 x100)=40 marks

Minimum marks required to pass the exam =40

Marks obtained by A = 40 – (40 x 10/100) = 36 marks

Marks obtained by B= 36 – (100/9 x 36/100) = 32 marks

Marks obtained by C ==(36 +32) x 700/(17 x100)=40 marks

Question 2 |

An article is listed at Rs 65. A customer bought this article for Rs 56.16 with two successive discounts, out of which one is 10%. The other discount of this discount scheme that was allowed by the shopkeeper is:

4% | |

3% | |

6% | |

2.5% |

Question 2 Explanation:

let the discount be D

Therefore 65 x (90)/100 x (100 –D)/100 = 56.16

100 – D = (56.16 x 100 x100) / (65 x 9)

100 – D = 96

D = 4%

Therefore 65 x (90)/100 x (100 –D)/100 = 56.16

100 – D = (56.16 x 100 x100) / (65 x 9)

100 – D = 96

D = 4%

Question 3 |

Mohan spends 40% of his salary on food items, 50% of the remaining on transport, 30% of the remaining on clothes , after spending on food and transport; and saves the balance. If he saves Rs 630 every month, what is his monthly salary?

Rs 1500 | |

Rs 3000 | |

Rs 5000 | |

Rs 6500 |

Question 3 Explanation:

4s/10 + (10s-4s)/10 x 50/100 +(6s-3s)/10 x 30/100 + saving = s

4s/10 + 6s/10 x 50/100 +3s/10 x 30/100 + saving = s

=> 4s/10 + 3s/10 + 9s/100 + 630 =s

=> 630 = (100s - 70s – 9s)/100

=> 630 = 21s/100

=> s = Rs 3000

4s/10 + 6s/10 x 50/100 +3s/10 x 30/100 + saving = s

=> 4s/10 + 3s/10 + 9s/100 + 630 =s

=> 630 = (100s - 70s – 9s)/100

=> 630 = 21s/100

=> s = Rs 3000

Question 4 |

The wheat sold by a grocer contained 10% low quality wheat. What quantity of good quantity wheat should be added to 150 kg of wheat so that the percentage of low quality wheat becomes 5%?

85 kg | |

50 kg | |

135 kg | |

150 kg |

Question 4 Explanation:

First, let’s calculate the low quality wheat: 10% of 150 = 15

This 15kg should now be 5% of the total amount of wheat.

Therefore the total amount of wheat becomes 300 kg

Originally, the wheat which is of good quality was 150 -15 =135kg

To become 5% low quality of wheat, we add the 150kg of good quality wheat.

This 15kg should now be 5% of the total amount of wheat.

Therefore the total amount of wheat becomes 300 kg

Originally, the wheat which is of good quality was 150 -15 =135kg

To become 5% low quality of wheat, we add the 150kg of good quality wheat.

Question 5 |

The present population of a village is 5500. If the number of males increases by 11 % and the number of females increases by 20%, then the population will become 6330. What is the present population of females in the village?

3000 | |

3500 | |

2500 | |

2000 |

Question 5 Explanation:

Let the present population of male = m

Let the present population of female = f

So m + f= 5500................................................1

111/100m +120/100f =6330

111m +120f =633000....................................2

On solving 1 and 2

f = 2500

Let the present population of female = f

So m + f= 5500................................................1

111/100m +120/100f =6330

111m +120f =633000....................................2

On solving 1 and 2

f = 2500

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### How to study this lesson?

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