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## Arithmetic : Level 3 Test -2

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Question 1 |

A person who has a certain amount with him goes to market. He can buy 50 oranges or 40 mangoes. He retains 10% of the amount for taxi fares and buys 20 mangoes and of the balance he purchases oranges. Number of oranges he can purchase is

36 | |

40. | |

15 | |

20 |

Question 1 Explanation:

Let us assume that he has Rs. 100.

Fare = 10% = Rs. 10

Therefore money left is Rs. 90

In Rs. 100 he can buy 50 oranges or 40 mangoes.

The price of an orange is Rs. 2 and that of a mango is Rs. 2.50.

Now if he buys 20 mangoes, he would spend Rs. 50

and will be left with Rs. 40, in which he can buy 20 oranges.

Fare = 10% = Rs. 10

Therefore money left is Rs. 90

In Rs. 100 he can buy 50 oranges or 40 mangoes.

The price of an orange is Rs. 2 and that of a mango is Rs. 2.50.

Now if he buys 20 mangoes, he would spend Rs. 50

and will be left with Rs. 40, in which he can buy 20 oranges.

Question 2 |

The price of a Maruti car rises by 30% while the sales of the car come down by 20%. What is the percentage change in the total revenue?

â€“4% | |

â€“2% | |

+4% | |

+2% |

Question 2 Explanation:

Lets suppose that the price was Rs 100

And the sale was also 100

Total revenue = 10000

Price raised by 30% i.e new price = Rs. 130

New sale = 80

This can simply be solved by multiplying the two

multiplication factors to get the effective New revenue = 10400

Which is 400 more than the old one so there is increase in the revenue by 4%

And the sale was also 100

Total revenue = 10000

Price raised by 30% i.e new price = Rs. 130

New sale = 80

This can simply be solved by multiplying the two

multiplication factors to get the effective New revenue = 10400

Which is 400 more than the old one so there is increase in the revenue by 4%

Question 3 |

A man earns r% on the first Rs. 2,000 and s% on the rest of his income. If he earns Rs. 700 from income of Rs. 4,000 and Rs. 900 from Rs. 5,000 of Income , find r%.

20% | |

15% | |

25% | |

None of these |

Question 3 Explanation:

The two equations can be written

2000 (r/100) + 2000(s/100) = 700

and 2000(r/1000) + 3000(s/100) = 900

The equations can be simplified to

r + s = 35 and 2r + 3s = 90. Solving these two equations

Simultaneously, we get r = 15%.

2000 (r/100) + 2000(s/100) = 700

and 2000(r/1000) + 3000(s/100) = 900

The equations can be simplified to

r + s = 35 and 2r + 3s = 90. Solving these two equations

Simultaneously, we get r = 15%.

Question 4 |

A man buys spirit at Rs. 60 per liter, adds water to it and then sells it at Rs. 75 per liter. What is the ratio of spirit to water if his profit in the deal is 37.5%?

9 : 1 | |

10 : 1 | |

11 : 1 | |

None of these |

Question 4 Explanation:

Since his SP of (spirit + water) = Rs.75/l

Profit of 37.5%,

CP of (spirit + water) = 75/1.375 = Rs.54.54.

Assume that the cost price of water = 0

So if we alligate,

We come to know that

Ratio will be 10 : 1

Spirit (60) -Â 54.54 = 5.545

Water(0) â€“ 54.54 = 54.54

So the ratio of spirit to water is 10 : 1

Profit of 37.5%,

CP of (spirit + water) = 75/1.375 = Rs.54.54.

Assume that the cost price of water = 0

So if we alligate,

We come to know that

Ratio will be 10 : 1

Spirit (60) -Â 54.54 = 5.545

Water(0) â€“ 54.54 = 54.54

So the ratio of spirit to water is 10 : 1

Question 5 |

There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the second container. Then

A > B | |

A < B | |

A = B | |

Cannot be determined |

Question 5 Explanation:

Let the capacity of each cup be 100 ml.

The ratio of mixture remained in the vessel

after taking out mixture from first and by pouring it to the second is 5 : 3.

Hence, proportion of alcohol in B = 3/8

Now after the second process of taking out 300 ml of mixture from the second container

The quantity of water will be (300 x 5/18 ) = 187.5 ml of water

The quantity of alcohol will be (300 x 3/18 ) = 112.5 ml of alcohol.

Now by pouring this mixture in the second vessel the quantity would be

(200 + 112.5) = 312.5 ml of alcohol and 187.5 ml of water.

Hence, ratio of alcohol to water in this container = 312.5: 187.5 = 5 : 3

Hence, proportion of water = A = 3/8 Hence, we find that A = B

The ratio of mixture remained in the vessel

after taking out mixture from first and by pouring it to the second is 5 : 3.

Hence, proportion of alcohol in B = 3/8

Now after the second process of taking out 300 ml of mixture from the second container

The quantity of water will be (300 x 5/18 ) = 187.5 ml of water

The quantity of alcohol will be (300 x 3/18 ) = 112.5 ml of alcohol.

Now by pouring this mixture in the second vessel the quantity would be

(200 + 112.5) = 312.5 ml of alcohol and 187.5 ml of water.

Hence, ratio of alcohol to water in this container = 312.5: 187.5 = 5 : 3

Hence, proportion of water = A = 3/8 Hence, we find that A = B

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