- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.

## Arithmetic: Profit and Loss Test -2

Please wait while the activity loads.

If this activity does not load, try refreshing your browser. Also, this page requires javascript. Please visit using a browser with javascript enabled.

If this activity does not load, try refreshing your browser. Also, this page requires javascript. Please visit using a browser with javascript enabled.

Congratulations - you have completed

*Arithmetic: Profit and Loss Test -2*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

Postcards costing 30 paise each and inland letters costing Rs 1.50 each were purchased for Rs. 66. Total number of postcards and inland letters purchased was 60. If the number of postcards and inland letters is interchanged, then the cost is:

Rs 42 | |

Rs 30 | |

Rs 21 | |

Rs 60 |

Question 1 Explanation:

Let us assume the number of postcards = n

Let the assume the number of letters = m

Therefore,

0.30 x n + 1.50 x m = 66………………………i

n +m=60 ...............................................ii

On solving equation i and ii

We get n = 20 , and m= 40

Therefore, the required cost price

= 0.30 x40 +1.5 x 20

= 12 + 30 =Rs 42

Let the assume the number of letters = m

Therefore,

0.30 x n + 1.50 x m = 66………………………i

n +m=60 ...............................................ii

On solving equation i and ii

We get n = 20 , and m= 40

Therefore, the required cost price

= 0.30 x40 +1.5 x 20

= 12 + 30 =Rs 42

Question 2 |

Maninder bought a certain number of pen drives for for 1800. Keeping one to himself, he sold the rest at a profit of Rs 6 each. In total, he earned a profit of Rs. 114. The number of pen drives he bought are:

20 | |

28 | |

32 | |

30 |

Question 2 Explanation:

Let Maninder buy (A) number of pen drives

So

(A-1) x 6=114

A-1 = 19

A=20

So

(A-1) x 6=114

A-1 = 19

A=20

Question 3 |

A garment company declared 15% discount for wholesale buyers. Bharat bought garments from the company for Rs. 25000 after getting a discount. He fixed up the selling price of garments in such a way that he earned a profit of 8% on original company price. What is the approximate total selling price?

Rs 28000 | |

Rs 29000 | |

Rs 32000 | |

Rs 29500 |

Question 3 Explanation:

Given cost price of garments = Rs25000
Company price = 25000/85 x 100
Profit = 8%
Therefore Bharat sold the Garments
= 25000/85 x 100 x 108/100
= Rs 31764.71 = Rs. 32000

Question 4 |

A shopkeeper sold an article at a profit of 17.5%. If he had bought it at 8% less and sold it at 30% profit, he would have earned Rs. 11.55 more as profit. Cost price of the article is:

Rs 550 | |

Rs 675 | |

Rs 750 | |

Rs 1475 |

Question 4 Explanation:

Let the cost price of the article be Rs (P)

So according to question

Selling price = P x (117.5)/100

If he had bought it at 8% less and sold it at 30% profit, he would have earned Rs. 11.55 more as profit.

Therefore P x (117.5)/100 + 11.55

= P x (92/100) x130/100

P = (11.55)/(0.99 x 130 -117.5) x 100

P= Rs 550

So according to question

Selling price = P x (117.5)/100

If he had bought it at 8% less and sold it at 30% profit, he would have earned Rs. 11.55 more as profit.

Therefore P x (117.5)/100 + 11.55

= P x (92/100) x130/100

P = (11.55)/(0.99 x 130 -117.5) x 100

P= Rs 550

Question 5 |

A manufacturer sold a machine to a wholesale dealer at a profit of 10%. The wholesale dealer sold it to a retailer at a profit of 20%. While transporting some defect occurred in the machine and hence the retailer sold it at a loss of 5%. The customer paid Rs672. Find the cost of the machine for the manufacturer.

Rs 672 | |

Rs 500 | |

Rs 572 | |

None |

Question 5 Explanation:

This is a very simple problem.

Let cost price of the machine = p

p x (110/100) x (120/100) x (95/100) = 672

p=535.88

Let cost price of the machine = p

p x (110/100) x (120/100) x (95/100) = 672

p=535.88

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.

List |