# QUESTIONS ON ALLEGATION AND MIXTURE (PART-II)

## QUESTIONS ON ALLEGATION AND MIXTURE (PART-II)

**VERY IMPORTANT:** Remember that the weighted average/combined average/mean value is always written in the middle and the averages are written at the top. We write the smaller average/value in the left and greater one in the right. The mean value is subtracted from the value written in the right, and the value written in the left is subtracted from the mean value, figure thus got is considered positive (negative signs are ignored).

But in questions having multiple ratios as are given in queries 1, 2 and 3, we write the value of the thing that is first in the asked ratio in the left and the other value in the right.

#### QUERY 11

**A shopkeeper sells milk with 5% water. In this 2 liters of milk how much milk should be mixed so that water become 4%?**

A) 0.4 liter

B) 0.5 liter

C) 1 liter

D) 1.5 liter

**MAHA GUPTA
**Change has only to be made in milk, let’s convert the percentage according to milk.

Milk initially = 100 – 5 = 95%

Milk after change = 100 – 4 = 96%

**MIXTURE MILK ADDED**

95% | 100% | |||

96% | ||||

100 – 96 = 4 | 96 – 95 = 1 |

So, MIXTURE : MILK ADDED = 4 : 1

Mixture = 4 ratios = 2 litre = 2000 ml

Hence, 1 ratio = ^{2000}⁄_{4} = 500 ml

Therefore milk to be added = 1 ratio = 500 ml = 0.5 litre (option ‘B’)

ANOTHER METHOD

Total mixture (pure milk and water) = 2 liters = 2,000 ml

But 5% of it is water, so water in it = 5% of 2,000 = 100 ml

Therefore the milk initially = 1,900 ml

Now the water should be 4% of the mixture after milk has been added to it; obviously the amount of water will remain the same i.e. 100 ml. Now if the water is 4%, milk will be 96%

If 4% is equal 100 ml

96% = (^{100}⁄_{4})*96 = 2,400 ml

So the milk now to be added = 2,400 – 1,900 = 500 ml = 1/2 liter = 0.5 liter (option ‘B’)

#### QUERY 12

**How much pure alcohol has to be added to 400 ml of a solution containing 15% of alcohol to change the concentration of alcohol in the mixture to 32% ?**

A) 60 ml

B) 100 ml

C) 128 ml

D) 68 ml

**MAHA GUPTA**

**MIXTURE ALCOHOL ADDED**

15% | 100% | |||

32% | ||||

100 – 32 = 68 | 32 – 15 = 17 |

So, MIXTURE : MILK ADDED = 68 : 17 = 4 : 1

Mixture = 4 ratios = 400 ml

Hence, 1 ratio = ^{400}⁄_{4} = 100 ml

Therefore milk to be added = 1 ratio = 100 ml (option ‘B’)

ANOTHER METHOD

Total mixture (alcohol and other things) = 400 ml

Therefore alcohol initially = 15% of 400 = 60 ml

And other things in the mixture = 400 – 60 = 340 ml

Obviously the amount of other things will remain the same i.e. 340 ml as the change has only to be made in alcohol. Now alcohol is to be 32% of the mixture, means 340 is 100 – 32 = 68% of the mixture.

If 68% is equal 340 ml

32% = (^{340}⁄_{68})*32 = 160 ml

So alcohol now to be added = 160 – 60 = 100 ml (option ‘B’)

#### QUERY 13

**600 grams of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution**

A) 160 grams

B) 120 grams

C) 130 grams

D) 140 grams

**MAHA GUPTA**

**MIXTURE SUGAR ADDED**

40% | 100% | |||

50% | ||||

100 – 50 = 50 | 50 – 40 = 10 |

So, MIXTURE : MILK ADDED = 50 : 10 = 5 : 1

Mixture = 5 ratios = 600 ml

Hence, 1 ratio = ^{600}⁄_{5} = 120 ml

Therefore sugar to be added = 1 ratio = 120 ml (option ‘B’)

ANOTHER METHOD**
**Case-I

Quantity of sugar 40% of 600 = 240 gm

Quantity of the other things = 60% of 600 = 360 gm

Case-II

Only the quantity of sugar is to be changed; quantity of other thing/s is to remain the same. As the sugar now is to be 50% of the solution, so quantity of the other thing/s is also 50%, means both equal, means sugar also needs to be 360 gm in the new solution.

So additional sugar required = 360 – 240 = 120 gm (option ‘B’)

#### QUERY 14

**The ratio of water and spirit in the mixture is 1 : 3. If he volume of the solution is increased by 25% by adding spirit only; what is the resultant ratio of water and spirit?**

A) 2 : 3

B) 4 : 1

C) 1 : 4

D) 2 : 5

**MAHA GUPTA**

According to the ratio the solution is = 1+3 = 4 units

After the increase of the solution = 4 + 25% of 4 i.e. 5 units; means 1 more unit than earlier

But only spirit is added; therefore it is 1 + (3+1) = 1 + 4

So the ration now = 1 : 4 (option ‘C’)

#### QUERY 15

**A sample of 50 litres of glycerine is found to be adulterated to the extent of 20%. How much glycerine should be added to it so as to bring down the percentage of impurity to 5% ?**

A) 155 litres

B) 150 litres

C) 150.4 litres

D) 149 litres

**MAHA GUPTA**

In a sample of adulterated mixture glycerine is not to be added as the mixture will be more adulterated then. So glycerine added = 0.

**GLYCERINE ADDED SAMPLE **

0% | 20% | |||

5% | ||||

20 – 5 = 15 | 5 – 0 = 5 |

So, GLYCERINE ADDED : SAMPLE = 15 : 5 = 3 : 1

SAMPLE = 1 ratio = 50 litres

Therefore, glycerine added = 3*50 = 150 (option ‘B’)

ANOTHER METHOD

We can also do this sum considering quantity of glycerine in the mixture. See how

Quantity of glycerine in the mixture initially = 100 -20 = 80%

Percentage of glycerine in the mixture when the impurity brought down to 5% = 100 – 5 = 95%

**SAMPLE GLYCERINE ADDED**

80% | 100% | |||

95% | ||||

100 – 5 = 5 | 95 – 80 = 15 |

So, SAMPLE : GLYCERINE ADDED = 5 : 15 = 1 : 3

SAMPLE = 1 ratio = 50 litres

Therefore, glycerine added = 3*50 = 150 (option ‘B’)

#### QUERY 16

**How many litres of water must be evaporated from 20 litres of 3% sugar solution to get a 5% solution?**

A) 6

B) 8

C) 10

D) 12

**MAHA GUPTA**

3% sugar solution of 20 litres means the quantity of sugar in it is 3%.

So quantity of sugar in 20 litres solution = ^{3}⁄_{100} × 20 = ^{3}⁄_{5} litre

But now we want this solution to be 5%, obviously, quantity of sugar will remain the same and only water will be evaporated.

Now, let the 5% sugar solution = x litres

Hence, 5% of x = ^{3}⁄_{5}

=> x/20 = ^{3}⁄_{5}

=> x = 12

So, quantity of the new solution = 12 litres

Therefore, water to be evaporated = 20 – 12 = 8 litres (option ‘B’)

#### QUERY 17

**Pure ghee cost Rs 100 per kg. After adulterating it with vegetable oil costing Rs. 50 per kg, a shopkeeper sells the mixture at Rs 96 per kg thereby making a profit of 20%. In what ratio does he mix the two?**

A) 2 : 3

B) 3 : 2

C) 1 : 2

D) 2 : 1

**MAHA GUPTA
**Cost Price per kg of the mixture = 96 ×

^{100}⁄

_{120}= Rs 80

**PURE GHEE VEGETABLE OIL**

100 | 50 | |||

80 | ||||

50 – 80 = 30 | 80 – 100 = 20 |

Hence, ratio of the two quantities = 30 : 20 = 3 : 2 (option ‘B’)

ANOTHER METHOD

Let he mixes pure ghee and the vegetable oil in the ratio = x : y

Therefore the CP of above = 100x + 50y

And the SP of above= 96(x + y)

Hence, profit = 96(x + y) – (100x + 50y)

But profit also = (^{20}⁄_{100})(100x + 50y)

So, (^{20}⁄_{100})(100x + 50y) = 96(x + y) – (100x + 50y)

=> 20x + 10y = -4x + 46y

=> 24x = 36y

=> ^{x}⁄_{y} = ^{36}⁄_{24}

=> x : y = 3 : 2 (option ‘B’)

#### QUERY 18

**Pure milk costs Rs 10 per litre. A milkman adds some water to 20 litres of pure milk and sells the mixture at Rs 8 per litre and hence making no profit no loss. How many litres of water does he add?**

A) 2 litres

B) 5 litres

C) 8 litres

D) 10 litres

**MAHA GUPTA**

As water is free, cost of water = Re 0 per litre

As there is no profit no loss, selling price of the mixture will be the cost price of the mixture. Hence cost price of the mixture = Rs 8 per litre

We know water is free, so cost price of water = Re 0 per litre

**COST OF MILK COST OF WATER**

10 | 0 | |||

8 | ||||

0 – 8 = 8 | 8_{ }– 10 = 2 |

We see, MILK : WATER = 8_{ }: 2_{ }= 4 : 1

MILK = 4 ratio = 20 litre

Hence, WATER = 1 ratio = ^{20}⁄_{4 }= 5 litre (option ‘B’)

ANOTHER METHOD

As water is free, cost of water = Re 0 per litre

Let he mixes milk and water in the ratio = x : y

Therefore the CP of above = 10x + 0y

And the SP of above= 8(x + y)

Hence, profit = 8(x + y) – (10x + 0y)

But profit also = 0

So, 0 = 8(x + y) – (10x + 0y)

=> 0 = -2x + 8y

=> 2x = 8y

=> ^{x}⁄_{y} = ^{8}⁄_{2}

=> x : y = 4 : 1

MILK = 4 ratio = 20 litre

Hence, WATER = 1 ratio = ^{20}⁄_{4 }= 5 litre (option ‘B’)

#### QUERY 19

**A liquid P is 10/7 as heavy as water and water is 7/5 times as heavy as another liquid Q. The amount of liquid P that must be added to 7 litres of liquid Q so that the mixture may weigh as much as an equal volume of water will be?**

A) 7 litres

B) 31/6 litres

C) 5 litres

D) 14/3 litres

**Vaibhav Vats**

If P is 10/7 times as heavy as water and water is 7/5 as heavy as Q; then 7 litres of Q is equal in weight to 5 litres of W (water).

Therefore if 1 litre of P is added to 7 litres of Q; then 10/7 litres of W can be added to 5 litres of W to keep the same weight in both. In other words we can say that if ‘x’ litres of P is added to 7 litres of Q; then (10/7)x litres of W can be added to 5 litres of W to keep the same weight in both.

Hence

x + 7 = 5 + (10/7)x

=> 7x + 49 = 35 +10x

=> 14 = 3x

=> x = 14/3 (option ‘D’)

#### QUERY 20

**One liquid contains 22.5% of water, another contains 27% water. A glass if filled with 5 parts of the first liquid and 7 parts of the second. How much per cent of it is water?**

A) 50.25

B) 25.125

C) 25

D) 25.12

**MAHA GUPTA**

The required percentage of water in the new mixture should be:

(Quantity of water in the new mixture/Quantity of the new mixture)*100

= [(5 parts of 22.5% of water + 7 parts of 27% of water)/(5 parts + 7 parts) of the LIQUID]*100

= 301.5/12 = 25.125 (option ‘B’)