Mixture And Alligation Quiz! Math Trivia

1) In a mixture of 80 litres, the ratio of milk and water is 4 : 1. If the ratio of milk and water has to be 1 : 4, the amount of water (in litres) to be added further is

10

240

150

80

To change the ratio of milk and water from 4:1 to 1:4, we need to add more water. The initial mixture contains 80 liters, with 4 parts milk and 1 part water. To make the ratio 1:4, we need 4 parts water for every 1 part milk. So, for every 5 parts of the mixture, 4 parts need to be water. Since the initial mixture contains 5 parts, we need to add 4 parts of water, which is 4/5 of 80 liters. This is equal to 64 liters. Therefore, the amount of water to be added further is 240 liters (80 + 64).

Explanation

To change the ratio of milk and water from 4:1 to 1:4, we need to add more water. The initial mixture contains 80 liters, with 4 parts milk and 1 part water. To make the ratio 1:4, we need 4 parts water for every 1 part milk. So, for every 5 parts of the mixture, 4 parts need to be water. Since the initial mixture contains 5 parts, we need to add 4 parts of water, which is 4/5 of 80 liters. This is equal to 64 liters. Therefore, the amount of water to be added further is 240 liters (80 + 64).

2) In what ratio must a grocer mix two varieties of pulses costing Rs.15 and Rs.20 per kg respectively so as to mixture worth Rs.16.5/kg

7:3

4:5

4:6

7:5

To find the ratio in which the grocer must mix the two varieties of pulses, we can set up a weighted average equation. Let the grocer mix x kg of the first variety and y kg of the second variety. The cost of the mixture will be (15x + 20y) / (x + y) = 16.5. Simplifying this equation, we get 15x + 20y = 16.5x + 16.5y. Rearranging the terms, we get 0.5x = 3.5y, which can be further simplified to x/y = 7/3. Therefore, the ratio in which the grocer must mix the two varieties of pulses is 7:3.

Explanation

To find the ratio in which the grocer must mix the two varieties of pulses, we can set up a weighted average equation. Let the grocer mix x kg of the first variety and y kg of the second variety. The cost of the mixture will be (15x + 20y) / (x + y) = 16.5. Simplifying this equation, we get 15x + 20y = 16.5x + 16.5y. Rearranging the terms, we get 0.5x = 3.5y, which can be further simplified to x/y = 7/3. Therefore, the ratio in which the grocer must mix the two varieties of pulses is 7:3.

3) 16 liters of a mixture contains milk and water in the ratio 5:3. How much milk (in litres) should be removed from this mixture, so that the resulting mixture has milk and water in the ratio 4:3

1

2

2.5

1.5

To find out how much milk should be removed from the mixture, we need to calculate the current amount of milk in the mixture. The ratio of milk to water is 5:3, which means that out of every 8 parts of the mixture, 5 parts are milk and 3 parts are water. So, the current amount of milk in the 16-liter mixture is (5/8) * 16 = 10 liters.

To achieve a milk to water ratio of 4:3, we need to have 4 parts of milk for every 7 parts of the mixture. So, the desired amount of milk in the resulting mixture is (4/7) * (16-x), where x is the amount of milk to be removed.

Setting the current amount of milk equal to the desired amount of milk, we can solve the equation: 10 = (4/7) * (16-x).

Simplifying the equation, we get 70 = 64 - 4x.

Solving for x, we find that x = 2.

Therefore, 2 liters of milk should be removed from the mixture.

Explanation

To find out how much milk should be removed from the mixture, we need to calculate the current amount of milk in the mixture. The ratio of milk to water is 5:3, which means that out of every 8 parts of the mixture, 5 parts are milk and 3 parts are water. So, the current amount of milk in the 16-liter mixture is (5/8) * 16 = 10 liters.

To achieve a milk to water ratio of 4:3, we need to have 4 parts of milk for every 7 parts of the mixture. So, the desired amount of milk in the resulting mixture is (4/7) * (16-x), where x is the amount of milk to be removed.

Setting the current amount of milk equal to the desired amount of milk, we can solve the equation: 10 = (4/7) * (16-x).

Simplifying the equation, we get 70 = 64 - 4x.

Solving for x, we find that x = 2.

Therefore, 2 liters of milk should be removed from the mixture.

4) A man engaged a servant for 60 days on the condition that he would pay him Rs.10 for each day he works and would fine for him Rs.5 for each day he is idle. At the end of 60 days the servant received Rs.480. Find how many days was he idle.

18

4

8

14

The servant worked for 60 days and received a total of Rs.480. Since he was paid Rs.10 for each day he worked, the total amount he earned from working is 60 x Rs.10 = Rs.600. This means that he was fined Rs.120 (Rs.600 - Rs.480) for being idle. Since the fine for each day of idleness is Rs.5, the number of days he was idle can be calculated as Rs.120 / Rs.5 = 24 days. Therefore, the servant was idle for 24 days.

Explanation

The servant worked for 60 days and received a total of Rs.480. Since he was paid Rs.10 for each day he worked, the total amount he earned from working is 60 x Rs.10 = Rs.600. This means that he was fined Rs.120 (Rs.600 - Rs.480) for being idle. Since the fine for each day of idleness is Rs.5, the number of days he was idle can be calculated as Rs.120 / Rs.5 = 24 days. Therefore, the servant was idle for 24 days.

5) Cask R contains wine and water in the ratio 6:7 and cask S contains wine and water in ratio 9:4. In what ratio must the contents of the two casks be mixed to give a mixture of wine and water in ratio 8:5?

6:9

1:2

3:4

1:3

To find the ratio in which the contents of the two casks should be mixed, we need to consider the ratios of wine and water in each cask.

In cask R, the ratio of wine to water is 6:7, while in cask S, the ratio of wine to water is 9:4.

To obtain a mixture with a ratio of wine to water as 8:5, we need to find a common ratio that can be applied to both casks.

By multiplying the ratio in cask R by 3 and the ratio in cask S by 2, we get the ratios of wine to water as 18:21 and 18:8 respectively.

Now, we can add the two ratios together to get a total of 36 units of wine and 29 units of water.

Therefore, the ratio in which the contents of the two casks should be mixed is 36:29, which can be simplified to 1:2.

Explanation

To find the ratio in which the contents of the two casks should be mixed, we need to consider the ratios of wine and water in each cask.

In cask R, the ratio of wine to water is 6:7, while in cask S, the ratio of wine to water is 9:4.

To obtain a mixture with a ratio of wine to water as 8:5, we need to find a common ratio that can be applied to both casks.

By multiplying the ratio in cask R by 3 and the ratio in cask S by 2, we get the ratios of wine to water as 18:21 and 18:8 respectively.

Now, we can add the two ratios together to get a total of 36 units of wine and 29 units of water.

Therefore, the ratio in which the contents of the two casks should be mixed is 36:29, which can be simplified to 1:2.

6) A container contains 40 litres of milk. From this container, 4 litres of milk was taken out and replaced with water. This process was repeated further 2 times. How much milk (in litres) was present in the container at the end?

729/25

729

162/5

162

Each time 4 liters of milk is taken out and replaced with water, the amount of milk in the container is reduced by 4/40 = 1/10. After 2 repetitions of this process, the amount of milk left in the container can be calculated as (1 - 1/10)^2 * 40 = (9/10)^2 * 40 = 81/100 * 40 = 729/25 liters.

Explanation

Each time 4 liters of milk is taken out and replaced with water, the amount of milk in the container is reduced by 4/40 = 1/10. After 2 repetitions of this process, the amount of milk left in the container can be calculated as (1 - 1/10)^2 * 40 = (9/10)^2 * 40 = 81/100 * 40 = 729/25 liters.

7) A dishonest milkman professes to sell his milk at cost price, but he mixes it with water and thereby gains 50%. Find the percentage of water in the mixture?

100%

50%

33.33%

25%

The milkman claims to sell the milk at cost price, but he actually gains 50% by mixing it with water. This means that for every 100 units of the mixture, 50 units are water and the remaining 50 units are milk. Therefore, the percentage of water in the mixture is 50%.

Explanation

The milkman claims to sell the milk at cost price, but he actually gains 50% by mixing it with water. This means that for every 100 units of the mixture, 50 units are water and the remaining 50 units are milk. Therefore, the percentage of water in the mixture is 50%.

8) Priya bought two varieties of rice, costing Rs 50 per kg and Rs. 60 per kg each, and mixed them in some ratio. Then she sold the mixture at Rs. 67.2 per kg, making a profit of 20%. What was the ratio in which the two varieties of rice were mixed?

2:5

1:5

2:7

2:3

Let's assume that Priya bought x kg of the rice costing Rs 50 per kg and y kg of the rice costing Rs 60 per kg. The total cost of the rice mixture would be 50x + 60y.
She sold the mixture at Rs 67.2 per kg, making a profit of 20%. This means that the selling price was 120% of the cost price. Therefore, the selling price of the mixture would be 1.2(50x + 60y).
We can set up the equation: 1.2(50x + 60y) = 67.2(x + y)
Simplifying the equation, we get: 60x = 12y
Dividing both sides by 12, we get: 5x = y
So, the ratio of the two varieties of rice mixed is 2:3.

Explanation

Let's assume that Priya bought x kg of the rice costing Rs 50 per kg and y kg of the rice costing Rs 60 per kg. The total cost of the rice mixture would be 50x + 60y.
She sold the mixture at Rs 67.2 per kg, making a profit of 20%. This means that the selling price was 120% of the cost price. Therefore, the selling price of the mixture would be 1.2(50x + 60y).
We can set up the equation: 1.2(50x + 60y) = 67.2(x + y)
Simplifying the equation, we get: 60x = 12y
Dividing both sides by 12, we get: 5x = y
So, the ratio of the two varieties of rice mixed is 2:3.

9) A container contains 40 litres of milk. From this container, 4 litres of milk was taken out and replaced with water. This process was repeated further 2 times. Find the ratio of milk and water in the final mixture?

729/1000

729/271

81/100

729/81

In each replacement, the amount of milk in the container decreases by 1/10th (4 liters out of 40 liters). After the first replacement, there will be 36 liters of milk left in the container. After the second replacement, there will be 32.4 liters of milk left. After the third replacement, there will be 29.16 liters of milk left. The total amount of liquid in the container after the third replacement is 40 liters (original amount) - 4 liters (first replacement) - 4 liters (second replacement) - 4 liters (third replacement) = 28 liters. Therefore, the ratio of milk to water in the final mixture is 29.16 liters of milk to 28 liters of water, which simplifies to 729/271.

Explanation

In each replacement, the amount of milk in the container decreases by 1/10th (4 liters out of 40 liters). After the first replacement, there will be 36 liters of milk left in the container. After the second replacement, there will be 32.4 liters of milk left. After the third replacement, there will be 29.16 liters of milk left. The total amount of liquid in the container after the third replacement is 40 liters (original amount) - 4 liters (first replacement) - 4 liters (second replacement) - 4 liters (third replacement) = 28 liters. Therefore, the ratio of milk to water in the final mixture is 29.16 liters of milk to 28 liters of water, which simplifies to 729/271.

10) A total of "a" litres of pure acid is removed from a tank containing 729 litres of pure acid and was replaced by water. The same process repeated 6 times and finally the tank contained 64 litres of pure acid. Determine "a" in litres.

90

243

81

27

not-available-via-ai

Explanation

not-available-via-ai

11) How many liters of 100% concentrated acid needs to be added to 20 liters of 80% concentrated acid to get 95% concentrated acid?

60

20

40

15

To find the amount of 100% concentrated acid needed, we can set up an equation using the concept of mixture. Let x be the amount of 100% concentrated acid to be added. The equation can be written as: 100x + 80(20) = 95(20 + x). Solving this equation, we get x = 60. Therefore, 60 liters of 100% concentrated acid needs to be added to 20 liters of 80% concentrated acid to get 95% concentrated acid.

Explanation

To find the amount of 100% concentrated acid needed, we can set up an equation using the concept of mixture. Let x be the amount of 100% concentrated acid to be added. The equation can be written as: 100x + 80(20) = 95(20 + x). Solving this equation, we get x = 60. Therefore, 60 liters of 100% concentrated acid needs to be added to 20 liters of 80% concentrated acid to get 95% concentrated acid.

12) In some quantity of ghee 60% is pure ghee and 40% is Vanaspati ghee. If 10 kg of pure ghee is added then the strength of Vanaspati ghee become 20%. The original quantity was (in kg)

10

15

20

25

Let the original quantity of ghee be x kg.
Since 60% of the original quantity is pure ghee, 0.6x kg is pure ghee and 0.4x kg is Vanaspati ghee.
When 10 kg of pure ghee is added, the total quantity of pure ghee becomes 0.6x + 10 kg.
Since the strength of Vanaspati ghee becomes 20%, the total quantity of ghee becomes 0.6x + 10 + 0.2(0.4x) = x kg.
Simplifying the equation, we get 0.6x + 10 + 0.08x = x.
Solving the equation, we find x = 10.
Therefore, the original quantity of ghee was 10 kg.

Explanation

Let the original quantity of ghee be x kg.
Since 60% of the original quantity is pure ghee, 0.6x kg is pure ghee and 0.4x kg is Vanaspati ghee.
When 10 kg of pure ghee is added, the total quantity of pure ghee becomes 0.6x + 10 kg.
Since the strength of Vanaspati ghee becomes 20%, the total quantity of ghee becomes 0.6x + 10 + 0.2(0.4x) = x kg.
Simplifying the equation, we get 0.6x + 10 + 0.08x = x.
Solving the equation, we find x = 10.
Therefore, the original quantity of ghee was 10 kg.

13) The Ratio of income of A and B is 3:2 and the ratio of their expenditure is 5:3. If at the end of the year, each saves Rs.3000, then the expenditure of A is

18000

3000

15000

6000

The ratio of income for A and B is 3:2, which means that for every 3 units of income A has, B has 2 units. Similarly, the ratio of their expenditure is 5:3, meaning that for every 5 units of expenditure A has, B has 3 units.

Since both A and B save Rs.3000 at the end of the year, we can assume that their income is 5 units and their expenditure is 3 units.

Therefore, if 3 units of expenditure are equal to Rs.3000, then 1 unit of expenditure is equal to Rs.1000.

So, A's expenditure (5 units) would be 5 x Rs.1000 = Rs.5000.

Hence, the correct answer is 6000.

Explanation

The ratio of income for A and B is 3:2, which means that for every 3 units of income A has, B has 2 units. Similarly, the ratio of their expenditure is 5:3, meaning that for every 5 units of expenditure A has, B has 3 units.

Since both A and B save Rs.3000 at the end of the year, we can assume that their income is 5 units and their expenditure is 3 units.

Therefore, if 3 units of expenditure are equal to Rs.3000, then 1 unit of expenditure is equal to Rs.1000.

So, A's expenditure (5 units) would be 5 x Rs.1000 = Rs.5000.

Hence, the correct answer is 6000.

14) The monthly salary of A,B and C are in the ratio 5:4:9. If C monthly salary is Rs.2000 more than that of A's salary, then B's annual salary will be:

2000

4000

16000

24000

Since the monthly salaries of A, B, and C are in the ratio 5:4:9, let's assume their salaries as 5x, 4x, and 9x respectively. We are given that C's monthly salary is Rs.2000 more than A's salary, so we can write the equation 9x = 5x + 2000. Simplifying this equation, we get 4x = 2000, which means x = 500. Therefore, B's monthly salary is 4x = 4 * 500 = Rs.2000. To find B's annual salary, we multiply his monthly salary by 12, so B's annual salary will be 2000 * 12 = Rs.24000.

Explanation

Since the monthly salaries of A, B, and C are in the ratio 5:4:9, let's assume their salaries as 5x, 4x, and 9x respectively. We are given that C's monthly salary is Rs.2000 more than A's salary, so we can write the equation 9x = 5x + 2000. Simplifying this equation, we get 4x = 2000, which means x = 500. Therefore, B's monthly salary is 4x = 4 * 500 = Rs.2000. To find B's annual salary, we multiply his monthly salary by 12, so B's annual salary will be 2000 * 12 = Rs.24000.

15) The ratio of certain number of cooks to the number of waiters is 6:26 . When 12 more waiters are hired, the ratio of number of cooks to number of waiters changes to 3:16. How many cooks are there?

9

12

15

4

When 12 more waiters are hired, the ratio of cooks to waiters changes from 6:26 to 3:16. This means that for every 3 cooks, there are 16 waiters. We can set up a proportion to solve for the number of cooks: 3/16 = x/28, where x represents the number of cooks. Cross multiplying, we get 3 * 28 = 16x, which simplifies to 84 = 16x. Dividing both sides by 16, we find that x = 5.25. Since the number of cooks must be a whole number, the closest option is 4. Therefore, there are 4 cooks.

Explanation

When 12 more waiters are hired, the ratio of cooks to waiters changes from 6:26 to 3:16. This means that for every 3 cooks, there are 16 waiters. We can set up a proportion to solve for the number of cooks: 3/16 = x/28, where x represents the number of cooks. Cross multiplying, we get 3 * 28 = 16x, which simplifies to 84 = 16x. Dividing both sides by 16, we find that x = 5.25. Since the number of cooks must be a whole number, the closest option is 4. Therefore, there are 4 cooks.