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## Arithmetic : Level 2 Test -5

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Question 1 |

The sum of money is to be divided amongst A, B and C in the respective ratio of 3 : 4 : 5 and another sum of money is to be divided between E and F equally. If F got Rs. 1050 less than

*A,*how much amount did B*receive?*Rs.750 | |

Rs. 2000. | |

Rs. 1400 | |

Cannot be determined |

Question 1 Explanation:

Since the sum of money, which is to be divided,is not known, so we cannot find the answers.

Question 2 |

What should be subtracted from 15, 28, 20 and 38 so that the remaining numbers are proportion?

6 | |

4 | |

2 | |

None of these |

Question 2 Explanation:

Let N is the number which is to be subtracted from each of the number

Therefore the equations becomes

(15 – N)/(28 – N) = (20 – N)/(38 – N)

By solving

N = 2

Hence option (c)

Therefore the equations becomes

(15 – N)/(28 – N) = (20 – N)/(38 – N)

By solving

N = 2

Hence option (c)

Question 3 |

One test tube contains some acid and another test tube contains an equal quantity of water. To prepare a solution, 20 g of the acid is poured into the second test tube. Then, two-thirds of the so-formed solution is poured from the second tube into the first. If the fluid in the first test tube is four times that in the second, what quantity of water was taken initially?

80 g | |

60 g | |

40 g | |

None of these |

Question 3 Explanation:

Water taken = w gm

First test tube = w gm

Second test tube w gm

After first process it becomes

First tube water left is (w – 20)

Second test tube water left is (w +20)

After second process

First tube water left is {(w – 20) + 2/3(x + 20)}

Second test tube water left is 1/3(w +20)

(w – 20) + 2/3(x + 20) = 4 x 1/3(w +20)

Solving for w it becomes

W = 100 gm

First test tube = w gm

Second test tube w gm

After first process it becomes

First tube water left is (w – 20)

Second test tube water left is (w +20)

After second process

First tube water left is {(w – 20) + 2/3(x + 20)}

Second test tube water left is 1/3(w +20)

(w – 20) + 2/3(x + 20) = 4 x 1/3(w +20)

Solving for w it becomes

W = 100 gm

Question 4 |

Sita and Gita enter into a partnership, Sita contributes Rs. 5000 while Gita contributes Rs. 4000. After 1 month, Gita withdraws 1/4 part of her contribution and after 3 months from the starting, Sita puts Rs

*.*2000 more. When Gita withdraws her money Rita also joins them with Rs. 7000. If at the end of 1 yr, there is profit of Rs. 1218, what will be the share of Rita in the profit? Rs. 844.37 | |

Rs. 488.47 | |

Rs. 588.47 | |

None of these |

Question 4 Explanation:

Share of Sita = 5000 x 3 + 7000 x 9 = 78000

Share of Gita = 4000 x 1 + 3000 x 11 = 37000

Share of Rita = 7000 x 11 = 77000

Therefore share in profit of Rita is = {77/78+37 +77} x 1218 =Rs. 488.47

Hence option b is the right answer

Share of Gita = 4000 x 1 + 3000 x 11 = 37000

Share of Rita = 7000 x 11 = 77000

Therefore share in profit of Rita is = {77/78+37 +77} x 1218 =Rs. 488.47

Hence option b is the right answer

Question 5 |

In a mixture of 60 L, the ratio of milk and water is 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is

30 L | |

20 L | |

40 L | |

60 L |

Question 5 Explanation:

Milk in the mixture is = 2/3 x 60 = 40

Quantity of water will be = 20 l

Let w be the quantity of water is to be added so,

40 /(20 + w) = ½

w = 60 L

Quantity of water will be = 20 l

Let w be the quantity of water is to be added so,

40 /(20 + w) = ½

w = 60 L

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