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## Arithmetic: Alligation and Mixture Test-4

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*Arithmetic: Alligation and Mixture Test-4*.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 |

In a mixture of milk and water the proportion of water by weight was 75%. If in the 60 gms mixture 15 gms water was added, what would be the percentage of water in the new mixture?

75% | |

88% | |

90% | |

None |

Question 1 Explanation:

Water in 60 gms=75/100×60=45

Milk in 60 gms=60-45=15

Water in new mixture=45+15=60

So percentage of water=60/75×100=80%

Milk in 60 gms=60-45=15

Water in new mixture=45+15=60

So percentage of water=60/75×100=80%

Question 2 |

In 1 kg mixture of sand and iron, 20% is iron. How much sand should be added, so that the proportion of iron becomes 5%?

3 kg | |

4 kg | |

5 kg | |

6 kg |

Question 2 Explanation:

Iron present in 1 kg mixture= 200gms

Let the quantity of sand added be x kgs

Then We want, $ \frac{0.2}{1+x}\times 100=5$ x=3

Let the quantity of sand added be x kgs

Then We want, $ \frac{0.2}{1+x}\times 100=5$ x=3

Question 3 |

A grocer purchased 2 kg. of rice at the rate of Rs.15 per kg and 3 kg of rice at the rate of Rs.13 per kg. At what price per kg should he sell the mixture to earn$ \displaystyle 33\frac{1}{3}%$ profit on the cost price?

Rs.28.00 | |

Rs.20.00 | |

Rs.18.40 | |

Rs.17.40 |

Question 3 Explanation:

Total cost of the bought rice=15×2+13×3=69

Total amount he should earn for a profit of $ \frac{100}{3}%=\frac{4}{3}\times 69=92$

So the SP per kg is 92/4=18.40

Total amount he should earn for a profit of $ \frac{100}{3}%=\frac{4}{3}\times 69=92$

So the SP per kg is 92/4=18.40

Question 4 |

To x litres of an x% solution of acid, y litres of water is added to get (x –10)% solution of acid. If x > 20, then value of y is

$ \displaystyle \frac{{{x}^{2}}}{100}$ | |

$ \displaystyle \frac{10x}{x-10}$ | |

$ \displaystyle \frac{10x}{x+10}$ | |

$ \displaystyle \frac{10{{x}^{2}}}{x-10}$ |

Question 4 Explanation:

$ \begin{array}{l}Acid\text{ }present\text{ }in\text{ }x\text{ }litres=\frac{{{x}^{2}}}{100}\\After\text{ }adding\text{ }y\text{ }litres\text{ }of\text{ }water\\Acid\text{ }percentage=\frac{\frac{{{x}^{2}}}{100}}{x+y}\times 100=x-10\\{{x}^{2}}={{x}^{2}}-10x+xy-10y\\Y=\frac{10x}{x-10}\\Y=\frac{10x}{x-10}\end{array}$

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