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## Arithmetic: Alligation and Mixture Test-3

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Question 1 |

A container has 30 litres of water. If 3 litres of water is replaced by 3 litres of spirit and this operation is repeated twice, what will be the quantity of water in the new mixture?

24 litres | |

23 litres | |

24.3 litres | |

23.3 litres |

Question 1 Explanation:

After 1

This amount would be replaced by sprit

=27-2.7=24.3

^{st}operation: Water=27litres,Â Â Sprit=3litres Now water in 3 litres= 27/30Ã—3=2.7,This amount would be replaced by sprit

=27-2.7=24.3

Question 2 |

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

150 kgs | |

135 kgs | |

50 kgs | |

85 kgs |

Question 2 Explanation:

Bad quality wheat in 150 kgs=15kg

Let the good quantity wheat added be x kgs

Then We want,{15/(150+x)}Ã—100=5

X=150

Let the good quantity wheat added be x kgs

Then We want,{15/(150+x)}Ã—100=5

X=150

Question 3 |

A man purchased 35 kg of rice at the rate of Rs.9.50 per kg and 30 kg at the rate of Rs.10.50 per kg. He missed the two. Approximately, at what price (in Rupees) per kg should he sell the mixture to make 35 per cent profit in the transaction?

12 | |

12.50 | |

13 | |

13.50 |

Question 3 Explanation:

Total cost of the bought rice=35Ã—9.50+30Ã—10.5=647.50

Total amount he should earn for a profit of

35%=1.35Ã—647.50=874.125

So the SP per kg is 874.125/65=13.45â‰…13.5

Total amount he should earn for a profit of

35%=1.35Ã—647.50=874.125

So the SP per kg is 874.125/65=13.45â‰…13.5

Question 4 |

When one litre of water is added to a mixture of acid and water, the new mixture contains 20% acid. When one litre of acid is added to the new mixture, then the resulting mixture contains $ \displaystyle 33\frac{1}{3}%$ acid. The percentage of acid in the original mixture was

20% | |

22% | |

24% | |

23% |

Question 4 Explanation:

$ \displaystyle \begin{array}{l}Let\text{ }the\text{ }quantity\text{ }of\text{ }water\text{ }in\text{ }\\mixture\text{ }be\text{ }x\text{ }and\text{ }acid\text{ }be\text{ }y\\so,\frac{y}{x+y+1}=\frac{20}{100}\\4y=x+1...............a\\Also\,\,\frac{y+1}{x+y+1}=\frac{100}{300}\\2y+2=x...............b\\By\,\,a\,\,and\,\,b\,\,we\,\,get\\y=1.5\\x=5\\Acid\,in\,original\,mixture\,=\frac{1.5}{6.5}\times 100=23\end{array}$

Question 5 |

A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7: 2 and 7: 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be:

5: 7 | |

5: 9 | |

7: 5 | |

9: 5 |

Question 5 Explanation:

Let the amount added of both be 1 kg

Amount of gold in A=7/9

Amount of copper in A=2/9

Amount of gold in B=7/18

Amount of copper in B=11/18

SO ratio of gold and copper is

$ \frac{\frac{7}{9}+\frac{7}{18}}{\frac{2}{9}+\frac{11}{18}}=\frac{7}{5}$

Amount of gold in A=7/9

Amount of copper in A=2/9

Amount of gold in B=7/18

Amount of copper in B=11/18

SO ratio of gold and copper is

$ \frac{\frac{7}{9}+\frac{7}{18}}{\frac{2}{9}+\frac{11}{18}}=\frac{7}{5}$

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