- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.

## Arithmetic: Alligation and Mixture Test-1

Congratulations - you have completed

*Arithmetic: Alligation and Mixture Test-1*.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 |

One-third dettol is taken out from the bottle of dettol and then an equal amount of water is poured into the bottle to fill it. This operation is done four times. Find the final ratio of dettol and water in the bottle?

13 : 55 | |

20 : 74 | |

16 : 65 | |

10 : 48 |

Question 1 Explanation:

After the first operation, the amount of Dettol in the bottle = 2/3

After the second operation, the amount of Dettol in the bottle = 2/3 x 2/3 = 4/9

After the third operation, the amount of Dettol in the bottle = 4/9 x 2/3 = 8/27

After the fourth operation, the amount of Dettol in the bottle = 8/27 x 2/3 = 16/81

The amount of water after the forth operation = 1 – 16/81 = 65/81

Therefore the required ratio = 16/81 : 65/81 = 16 : 65

Therefore the right answer is option (c)

After the second operation, the amount of Dettol in the bottle = 2/3 x 2/3 = 4/9

After the third operation, the amount of Dettol in the bottle = 4/9 x 2/3 = 8/27

After the fourth operation, the amount of Dettol in the bottle = 8/27 x 2/3 = 16/81

The amount of water after the forth operation = 1 – 16/81 = 65/81

Therefore the required ratio = 16/81 : 65/81 = 16 : 65

Therefore the right answer is option (c)

Question 2 |

How many kgs of sugar costing Rs. 5.75 per kg should be mixed with 75 kg of cheaper variety of sugar costing Rs. 4.50 per kg so that the mixture is worth Rs. 5.50 per kg?

350 kg | |

300 kg | |

250 kg | |

325 kg |

Question 2 Explanation:

Here we use the mixture rule/Allegation method

We have two types of sugar, one is of Rs. 5.75 and other is of Rs. 4.50

The final price mixture price will be in the ratio:

(5.50 – 4.50) : (5.75 – 5.50) = 1:0.25

That means 4:1

Thus, the required quantity of sugar = 75 x 4 = 300 kg

(read allegation concepts for greater clarity)

We have two types of sugar, one is of Rs. 5.75 and other is of Rs. 4.50

The final price mixture price will be in the ratio:

(5.50 – 4.50) : (5.75 – 5.50) = 1:0.25

That means 4:1

Thus, the required quantity of sugar = 75 x 4 = 300 kg

(read allegation concepts for greater clarity)

Question 3 |

5 L of water is added to be a certain quantity of pure milk costing Rs. 3 /L. If by selling the mixture is same price as before, a profit of 20% is made, then what is the amount of pure milk in the mixture?

20 L | |

30 L | |

25 L | |

35 L |

Question 3 Explanation:

Let the quantity of milk be x liters

Then, before adding water, x liters was sold at 3 per liters,

thus the total price was = 3x

Adding 5 liters of water takes the total volume to (x + 5) liters.

Therefore, at 3/liter, the total value of the mixture becomes 3(x+5)

Now it's given that after adding the water, there is a profit of 20%, therefore

20% increase in value BEFORE adding water = Value AFTER adding water

1.2(3x) = 3(x+5)

Solve for x

x = 25

Alternate solution

Let us assume the quantity of the milk = x L

Now 5 L of water is added

Therefore, the cost of the milk will be = 3(x+5)

Now profit = 20%

1.2 (3x ) = 3(x+5)

3.6x = 3x +15

0.6x= 15

x = 25

Then, before adding water, x liters was sold at 3 per liters,

thus the total price was = 3x

Adding 5 liters of water takes the total volume to (x + 5) liters.

Therefore, at 3/liter, the total value of the mixture becomes 3(x+5)

Now it's given that after adding the water, there is a profit of 20%, therefore

20% increase in value BEFORE adding water = Value AFTER adding water

1.2(3x) = 3(x+5)

Solve for x

x = 25

Alternate solution

Let us assume the quantity of the milk = x L

Now 5 L of water is added

Therefore, the cost of the milk will be = 3(x+5)

Now profit = 20%

1.2 (3x ) = 3(x+5)

3.6x = 3x +15

0.6x= 15

x = 25

Question 4 |

A chemist has 10 L of a solution that is 10% nitric acid by volume. He wants to dilute the solution to 4% strength by adding water. How many liters of water must be added?

15 | |

20 | |

18 | |

25 |

Question 4 Explanation:

Total solution = 10L

Nitric acid = 10%

Therefore quantity of the nitric acid = 10 x 10/100 = 1L

Therefore the quantity of water =10 – 1 = 9

Now let us assume the quantity of the water which is to be added = x L

Therefore (10 +x) x 4/100 = 1

x= 15 L

Nitric acid = 10%

Therefore quantity of the nitric acid = 10 x 10/100 = 1L

Therefore the quantity of water =10 – 1 = 9

Now let us assume the quantity of the water which is to be added = x L

Therefore (10 +x) x 4/100 = 1

x= 15 L

Question 5 |

Three containers A, B and C have mixtures of milk and water in the ratio 1 : 5, 3 : 5 and 5 : 7 respectively. If the capacities of the containers are in the ratio 5 : 4 : 5, then find the ratio of the milk to the water if the mixtures of all the three containers are mixed together.

51 : 115 | |

52 : 115 | |

53 : 115 | |

54 : 115 |

Question 5 Explanation:

Ratio of milk: water

Let us assume that the containers have 120L, 96L and 120L of the mixture (5:4:5 ratio)

Container 1, Container 2, Container 3 combined milk quantity will be

= (1/6 x 120 + 3/8 x 96 + 5/12 x 120) = 106

Container 1, Container 2, Container 3 combined water quantity is

= (5/6 x 120 + 5/8 x 96 + 7/12 x 120) = 230

Therefore the ratio will become : 53 : 115

Let us assume that the containers have 120L, 96L and 120L of the mixture (5:4:5 ratio)

Container 1, Container 2, Container 3 combined milk quantity will be

= (1/6 x 120 + 3/8 x 96 + 5/12 x 120) = 106

Container 1, Container 2, Container 3 combined water quantity is

= (5/6 x 120 + 5/8 x 96 + 7/12 x 120) = 230

Therefore the ratio will become : 53 : 115

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.

List |