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## Arithmetic: Alligation and Mixture Test-2

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Question 1 |

Given that 24 carat gold is pure gold, 18 carat gold is 3/4 gold and 20 carat gold is 5/6 gold, the ratio of the pure gold in 18 carat gold to the pure gold in 20 carat gold is

A | 5 : 8 |

B | 9 : 10 |

C | 15 : 24 |

D | 8 : 5 |

Question 1 Explanation:

18 carat gold = ¾ pure gold = ¾ x 24 = 18

20 carrot = 5/6 pure gold = 5/6 x 24 = 20

Therefore the required ratio = 18 :20 = 9:10

20 carrot = 5/6 pure gold = 5/6 x 24 = 20

Therefore the required ratio = 18 :20 = 9:10

Question 2 |

A sink contains exactly 12 L of water. If water is drained from the sink until it holds exactly 6 L of water less than the quantity drained away, how many liters of water were drained away?

A | 2 L |

B | 6 L |

C | 3 L |

D | 9 L |

Question 2 Explanation:

Sink Capacity = 12 L

Effectively this means:

Remaining water (R)+ Drained water (D) = 12 L….(i)

Remaining Water (R) = D – 6

Using (i) and (ii)

D – 6 +D =12

D = 9

Effectively this means:

Remaining water (R)+ Drained water (D) = 12 L….(i)

Remaining Water (R) = D – 6

Using (i) and (ii)

D – 6 +D =12

D = 9

Question 3 |

Several liters of acid were drawn off a 54 L vessel full of acid and an equal amount of water -added. Again, the same volume of the mixture was drawn off and replaced by water. As a result the vessel contained 24 L of pure acid. How much of the acid was drawn off initially?

A | 12 L |

B | 16 L |

C | 18 L |

D | 24 L |

Question 3 Explanation:

Let the container contain a volume of v L (54 L in the given case)

And also let us suppose that the liquid which is drained out = u L

So the final quantity will be = v(1-u/v)

Where n is the number of time of operation is carried out

So from the given information we can say that

24 = 54(1-u/54)

(1-u/54)

u/54 = 1/3

u =18L

And also let us suppose that the liquid which is drained out = u L

So the final quantity will be = v(1-u/v)

^{n}Where n is the number of time of operation is carried out

So from the given information we can say that

24 = 54(1-u/54)

^{2}(1-u/54)

^{2}= 24/54 = 4/9 (1-u/54) = 2/3u/54 = 1/3

u =18L

Question 4 |

A tin of oil was four- fifths full. When six bottles of oil were taken out and four bottles of oil were poured into it, it was three-fourths full. How many bottles of oil were contained by the tin?

A | 10 |

B | 20 |

C | 30 |

D | 40 |

Question 4 Explanation:

Let us assume the total capacity of the tin = P bottles

In this case we assume that a single bottle represents one unit of tin.

So according to the question

(4/5P) – 6 +4=3/4P

(4/5P) – 3/4P = 2

P=40

In this case we assume that a single bottle represents one unit of tin.

So according to the question

(4/5P) – 6 +4=3/4P

(4/5P) – 3/4P = 2

P=40

Question 5 |

A vessel contains 5 liters of a mixture of milk and water, the quantity of milk was 36% of the total mixture. A few liters of mixture was taken out and replaced by the same amount of water. This process is repeated 2 times and the quantity of milk becomes 16% only. How many liters of mixture was taken out every time?

A | 16 |

B | 20 |

C | 1.6 |

D | 1.98 |

Question 5 Explanation:

Total amount of the mixture = 5L

Number of operations = 2

Amount of ingredient left/actual amount of milk = {1 – x/amount of mixture}

16%/36% = (1 – x/5)

4/6 = 1-x/5

X = approximately 1.66L

Number of operations = 2

Amount of ingredient left/actual amount of milk = {1 – x/amount of mixture}

^{n}16%/36% = (1 – x/5)

^{2}4/6 = 1-x/5

X = approximately 1.66L

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