This page contain important multiple choice questions of the chapter mixtures and alligations. All the questions are previously asked in quantitative aptitude exams so make sure you can solve each of them.

If you face any difficulty solving the above question, i will suggest you to read the theory first as it will help you clear the basics and give you framework to solve questions from this chapter.

## Mixture and Alligation Multiple Choice Questions

**(01) Two solutions of 90% and 97% purity are mixed resulting in 21 litres of mixture of 94% purity. How much is the quantity of the first solution in the resulting mixture?**

**a. 8 Litersb. 10 Litersc. 11 Litersd. 12 Liters**

Given: amount of mixture = 21 litres

By mixture and allegations

Ratio = n1: n2 = 3: 4

Quantity of first solution in the resulting mixture = (amount of mixture x n1)/ (n1 + n2)

= 21 x 3/ ( 3 + 4)

= 63/ 7

= 9 litres

**Ans. Quantity of first mixture = 9 liters**

**(02) In the Delhi zoo, there are deer and ducks. If the heads are counted, there are 180, while the legs are 448. What will be the number of deer in the zoo?**

**a. 10b. 20 c. 44d. 56**

Read Solution

Given:

No. of heads = 180

No. of legs = 448

If all the animals were deer the no. Of legs = 180 x 4 = 720

If all the animals were ducks, then no. Of legs = 180 x 2 = 360

By mixture and allegations,

Ratio = n1: n2

= 88: 272 = 11: 34

No. Of deer in the zoo = No. Of heads x n2/ (n1 + n2)

= 180 x 11/ (11 + 34)

= 44

**Ans. There are 44 deer in the zoo.**

**(03) A bonus of Rs. 9,85,000 was divided among 300 workers of a company. Each male worker gets 5000 rupees and each female worker gets 2500 rupees. Find the number of male workers in the company.**

**a. 106b. 56c. 84d. 94**

Given: Bonus = 9, 85, 000

Total no. Of workers = 300

If all the workers were male then bonus = 5000 x 300 = 1500000

If all the workers were female then bonus = 2500 x 300 = 750000

By mixture and allegations

Ratio = n1: n2 = 235000: 511500

= 47: 103

No. Of male workers = total no. Of workers x n1/ (n1 + n2)

= 300 x 47/ (47 + 103)

= 94

**Ans. The no. of male workers is 94.**

**(04) If a person decides to travel 80 kilometers in 8 hours partly by foot and partly on a bicycle, his speed on foot being 8 km/h and that on bicycle being 16 km/h what distance would he travel on foot?**

**a. 50 Kmb. 48 Kmc. 54 Kmd. 62 Km**

Read Solution

**Sol. **Given: distance travelled = 80 km in time = 8 hours

If he covers total distance by foot, then distance = Speed x time = 8 x 8 = 64 km

If he covers total distance by bicycle, then distance = speed x time = 16 x 8 = 128 km

Now, by mixture and allegation,

Ratio = n1: n2 = 48: 16

= 3: 1

Time taken to travel on foot = time taken x n1/ (n1 + n2)

= 8 x 3/ (3 + 1)

= 24/ 4 = 6 hours

Therefore, distance travel on foot = Speed x time

= 8 x 6 = 48 km

**Ans. Therefore, he travels 48 km by foot.**

**(05) There are three vessels of equal capacity having a mixture of milk and water in the ratio 11: 9, 7: 3 and 3: 2 respectively. If all three vessels are poured into another vessel, find percentage of milk in the new mixture in comparison to the water?**

**a. 25%b. 46%c. 105%d. 160%**

Read Solution

**Sol. **Total milk = 11/ 20 + 7/ 10 + 3/ 5

= (11 + 14 + 12) / 20 (by taking L.C.M)

= 37/ 20

Total water = 9/ 20 + 3/ 10 + 2/ 5

= (9 + 6 + 8)/ 20 (by taking L.C.M)

= 23/ 20

Ratio betweem milk and water = 37/ 20: 23/ 20

= 37: 23

Now, milk percentage = 37 x 100/ 23

= 160 %

**Ans. The required percentage is 160%**

**06) A sum of Rs 36.90 is made up of 90 coins that are either 20 paise coins or 50 paise coins. Find out how many 20 paise coins are there in the total amount**

**a. 63b. 25c. 27d. 50**

Given:

Sum = Rs. 36.90 = 36.90 x 100 = 3690 paise

Total coins = 90

Average no. Of coins = sum/ number = 3690/ 90 = 41

Therefore, amount of 20 paise coins = total no. Of coins x (n1/ n1 + n2)

= 90 x (3/ 3 + 7)

= 90 x (3/ 10)

= 27 coins

**Ans. Required no. Of coins = 27**

**(07) A mixture of 70 litres of alcohol and water contains 10% of water. How much water must be added to the above mixture to make the water 12.5% of the resulting mixture?**

**a. 2 litersb. 2.5 litersc. 1 litersd. 3.5 liters**

Given: Mixture = 70 litres

Water = 10 % of 70 litres

= 7 litres

Therefore, Alcohol = 70 – 7 litres

= 63 litres

Let the amount of water to added = x litres

So, water/ Alcohol = (7 + x)/ 63 = 12.5/ 87.5 (100 – 12.5 = 87.5)

x = [(12.5/ 87.5) x 63] – 7

x = 2 litres

**Ans. 2 litres of water must be added.**

**(08) A shopkeeper purchased two qualities of pulses at the rate of Rs. 200 per quintal and Rs. 260 per quintal. In 52 quintals of the second quality, how much pulse of the first quality should be mixed so that by selling the resulting mixture at Rs. 300 per quintal, he gains a profit of 25%?**

**a. 24****b. 25c. 26d. 27**

Given: SP of mixture = Rs. 300 per quintal

Profit = 25% = 1/ 4

CP of mixture = SP x (100/ 100 + 25)

= 300 x (100/ 125)

= 300 x (4/ 5)

= 60 x 4 = Rs. 240 per quintal

Now, by allegations,

Ratios = n1/ n2 = 20 : 40 = 1: 2

Now, 52 quintals = 2 units

1 unit = 52/ 2 = 26 quintals

**Ans. Therefore, 26 quintals of the first quantity should be mixed**

**Qu.9) Two vessels contain spirit and water mixed respectively in the ratio of 1 : 3 and 3 : 5. Find the ratio in which they are to be mixed to get a new mixture in which the ratio of spirit to water is 1: 2**

**a. 2 : 3b. 3 : 2c. 1 : 2d. 1 : 4**

Given: Ratio of A = 1: 3; Ratio of B = 3: 5; Ratio of C = 1:2

Let the three mixtures given be A, B and C

Spirit in 1 litre mix of A = 1/ 4 litre

Spirit in 1 litre mix of B = 3/ 8 litre

Spirit in 1 litre mix of C = 1/ 3 litre

By rule of allegation:

Ratios = n_{1} : n_{2 }= 1/ 24: 1/ 12

= 12: 24

= 1: 2

**Ans. The required ratio is 1:2**

**(10) A vessel is full of a mixture of kerosene and petrol in which there is 18% kerosene. Eight litres of mixture is replaced with petrol. If the kerosene is now 15%, how much does the vessel hold?**

**a. 48 litersb. 35 litersc. 20 litersd. 29 liters**

**Sol. **Given: Kerosene in vessel = 18 %

Therefore, petrol = 100 – 18 % = 82%

After replacement, kerosene = 15 %

Therefore petrol = 100 – 15 %

= 85 %

Now, kerosene percent change from 18% to 15% I.e., 3 %

3 % = 8 litres of petrol is replaced

18 % = (8/ 3) x 18

= 48 litres

**Ans. Therefore, earlier vessel holds 48 litres of kerosene**

**(11) In a mixture of milk and water, there is only 26% water. After replacing the mixture with 7 litres of pure milk, the percentage of milk in the mixture become 76%. The quantity of the mixture is**

**a. 90 litersb. 91 litersc. 92 litersd. 93 liters**

**Left amount = Initial amount (1− replaced amount/ total amount)**

24 = 26 (1−7/ k) (Let the total amount be k)

k = 91

**Ans. The quantity of mixture is 91 litres.**

**(12)** **450 litres of a mixture of milk and water contain the milk and water in the ratio 9:1. How much water should be added to get a new mixture containing milk and water in the ratio of 3: 1?**

**a. 90 litersb. 91 litersc. 92 litersd. 93 liters**

**. ** Milk : Water

9 : 1

Note: we are adding water in the mixture so quantity of milk would be constant

After that new ratio,

Milk : Water

9 : 1

Where milk = [9/ (9 + 10)] x 450 = 405 litres

And water = [1/ (9 + 10)] x 450 = 45 litres

Let x litres of water be added,

Given resultant milk water ratio = Milk: water

= 3: 1

405/ (45 + x) = 3/ 1

x = [405/ 3] – 45

x = 90 litres

**Ans. Required amount of water = 90 liters.**

**(13) The ratio of oil and kerosene in the container is 3: 2 when 10 litres of the mixture is taken out and is replaced by the kerosene, the ratio becomes 2: 3. The total quantity of the mixture in the container is :**

**a. 45 litersb. 30 litersc. 31 litersd. 51 liters**

Given: O1: k1 = 3: 2

Replaced quantity = 10 litres

O2 : k2 = 2: 3

Let original quantity = x

Another quantity to exchange with original quantity = y = 10

By using the formula,

**Liquid left in the container after ‘n’ operation / original quantity of the liquid in the vessel = (x –y)/ x**

==> 2/ 3 = 1 – (y/ x)

==> 2/ 3 = 1 – (10/x)

==> 1 – 2/ 3 = 10 / x

==> 1/ 3 = 10/ x

==> x = 10 x 3

= 30 litres

**Ans. Total quantity of mixture in the container is 30 liters**

### (14) **A container has a capacity of 20 liters and is full of spirit. 4 liters of spirit drawn out and the container is again filled with water. This process is repeated 5 times. Find out how much spirit is left in the resulting mixture finally?**

**a. 6.55 litersb. 7.55 litersc. 8.55 litersd. 9.55 liters**

**Sol**. Given: Capacity of container I.e., initial quantity = 20 litres

Amount of spirit drawn = 4 litres

Process is repeated = 5 times

By using the formula,

**Left amount = Initial amount (1− replaced amount/ total amount) n**

= 20 (1 – 4/ 20)^{5}

= 20 x 4/5 x 4/5 x 4/5 x 4/5 x 4/5

= 4096/ 625

= 6.55 liters

**Ans. 6.55 litres of spirit left in the mixture.**

**Qu.15) A mixture of sugar is sold at Rs.3.00 per kg. This mixture is formed by mixing the sugar of Rs. 2.10 and Rs. 2.52 per kg. What is the ratio of cheaper to the costlier quality in the mixture if the profit of 25% is being earned?**

**a. 1 : 2b. 2 : 3c. 3 : 4d. 2 : 5**

Given: SP of sugar = Rs. 3 per kg

Profit = 25 %

CP of sugar = (3 x 100)/ (100 + 25)

= 300/ 125

= Rs. 2.4 per kg

Ratio = 0.12: 0.3

= 4: 10

= 2: 5

**Ans. Therefore, the required ratio is 2: 5.**

**Qu.16) A milkman has 20 litres of milk. If he mixes 5 litres of water, which is freely available, in 20 litres of pure milk. If the cost of pure milk is Rs. 18 per litre, then the profit of the milkman, when he sells all the mixture at cost price, is:**

**a. 25%b. 20%c. 35%d. 40%**

**Sol**. Given: Quantity of milk = 20 litres

Quantity of water = 5 litres

Cost of pure milk = Rs. 18 per litre

To find: profit of milkman

When the water is freely available and all the water is sold at the price of the milk, then the water gives the profit on the cost of 20 litres of milk.

Cost price of 20 litres milk = 20 x 18

= Rs. 360

Cost Price of 25 litres of milk after mixing the water = 25 × 18

= Rs. 450

profit percent = (profit/ CP) x 100

= [(450 – 360)/ 360] x 100

= (90/ 360) x 100

= 25 %

**Ans. Profit = 25 %**

**Qu.17) In what ratio should water and soda be mixed that after selling the mixture at the cost price a profit of 33.33% is made?**

**a. 1:3b. 1:4c. 1:5d. 2:3**

**Sol.** Let SP of 1 litre of milk = Rs.1

Gain = 33.33% = 100/ 3

Let say Cost Price of Wine = Rs. C / Litre

And let say ratio of water and wine = W: 1

W is Water and 1 is Wine

Then Total Mixture would be W + 1 Liter

Selling Price = C (W + 1) ……Eq.1

Cost Price = C as there is 1 Litre Wine

Profit = 33.33 % and Profit = C/3 (as mentioned above)

Selling Price = C + C/3 = 4C/3 (CP + Profit)

4C/3 = C (W + 1) ….(from Eq. 1)

4 = 3W + 3

3W = 1

W = 1/3

Water: Wine = 1/3: 1

= 1 : 3

**Ans. Required ratio 1: 3**

**Qu.18) In what ratio should freely available water be mixed with the soda worth Rs. 60 per litre so that after selling the mixture at Rs. 50 per litre, the profit will be 25%?**

**a. 1:2b. 2:1c. 3:1d. 2:3**

**Sol.** Given: CP of Soda = Rs. 60 per litre

SP of mixture = Rs. 50 per litre

Profit % = 25

Therefore, CP of mixture = (50 x 100)/ (100 + 25)

= Rs. 40 per litre

Ratio = 40: 20

= 2: 1

**Ans. Required ratio is 2: 1**

**Qu.19) A bus agency has 108 buses. He sold some Bus at 9% profit and rest at 36% profit. Thus he gains 17% on the sale of all his Buses. The no. of Buses sold at 36% profit is :**

**a. 30b. 31c. 32d. 33**

**Sol.** Given: Profit 1 = 9%, Profit 2 = 36%

Overall gain = 17 %

Total buses = 108

Now, by alligations

N_{1 } + n_{2} = 108

(19 + 8) units = 108

27 units = 108

1 unit = 108/ 27 (by unitary method)

= 4

Therefore, 8 units = 8 x 4 = 32 (since contains 36 % profit)

**Ans. Required no. Of buses sold at 36% profit = 32**