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**Q1.** **A 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removal and replacement, what is the ratio of milk and water in the resultant mixture?****1. 17 : 32. 9 : 13. 3 : 174. 5 : 3**

Explanation.

The 20 litre mixture contains milk and water in the ratio of 3 : 2. Therefore, the mixture contains 12 litres of milk and 8 litres of water.

Therefore, there will be 6 litres of milk and 4 litres of water left in the container.

It is then replaced with pure milk of 10 litres. Now the container will have 16 litres of milk and 4 litres of water.

The container will have 8 litres of milk and 2 litres of water in it.

Now 10 litres of pure milk is added. Therefore, the container will have 18 litres of milk and 2 litres of water in it at the end of the second step.

Therefore, the ratio of milk and water is 18 : 2 or 9 : 1.

We are essentially replacing water in the mixture with pure milk.

Let WO be the amount of water in the mixture originally = 8 litres.

Let WR be the amount of water in the mixture after the replacements have taken place.

Then, where R is the amount of the mixture replaced by milk in each of the steps, M is the total volume of the mixture and n is the number of times the cycle is repeated.

Hence,

Therefore,

W

So, the final quantity of water in the 20-litre mixtures is 2 litres.

Hence, the mixture will have 18 litres of milk and 2 litres of water.

1. 1 : 2 : 4

2. 3 : 7 : 6

3. 1 : 4 : 2

4. None of these

Answer. 1 : 4 : 2

Explanation.

The resultant mixture is sold at a profit of 20%. The selling price is Rs.96/kg

i.e., cost + 20% of cost = 1.2 (cost) = Rs.96

â‡’ cost of the mixture = 96/1.2 = Rs.80/kg.

Let the three variants be A, B, and C and let their cost per kg be Rs.60, Rs.75 and Rs.100 respectively.

The mean price falls between B and C.

Hence the following method should be used to find the ratio in which the 3 variants should be mixed.

The resultant ratio QA : QB : QC :: 1 : 4 : 2.

However, if you try and solve this problem using the above method, you will end up spending more than 2, and may be 3 minutes on this question, which is not wise way to manage time in the test.

The best way to solve a problem of this kind in CAT or TANCET is to go from the answer choices as shown below.

Let us go with the correct answer. They are mixed in the ratio 1 : 4 : 2.

1 kg of variant A at Rs.60 is mixed with 4 kg of variant B at Rs.75 and 2 kg of variant C at Rs.100.

The total cost for 7 kg of the mixture = 60 + (4 * 75) + (2 * 100) = 60 +300 + 200 = Rs.560.

Cost per kg of the mixture = 560/7 = Rs.80 which matches the cost we arrrived at in the Step 1.

Even assuming that you hit upon the right answer as the last choice, you will still be better of by back substituting choices for this question.

1. 1 kg

2. 5 kg

3. 3 kg

4. 6 kg

Explanation.

The selling price of mixture is Rs.30/kg and the merchant makes a profit of 20%.

If the cost of 1 kg of the mixture is c, c + 20% of c = 1.2 c = selling price

The cost price of the mixture, c = 30/1.2 = Rs.25/kg.

The three variants are mixed to obtain a mixture that costs Rs.25 /kg.

Variant A costs Rs.20/kg, variant B costs Rs.24/kg and variant C costs Rs.30/kg. The mean price of the mixture falls between the costs of B and C.

Therefore, the required ratio = 1 : 5 : 2

If there are 2 kgs of the third variant in the mixture, then there will be 5 kgs of the second variant in the mixture.

1. 7 litres

2. 10 litres

3. 5 litres

4. None of these

Answer. 5 litres

Explanation.

When you add more water, the amount of milk in the mixture remains constant at 21 litres. In the first case, before addition of further water, 21 litres of milk accounts for 70% by volume. After water is added, the new mixture contains 60% milk and 40% water.

Therefore, the 21 litres of milk accounts for 60% by volume.

Hence, 100% volume = 21/0.6 = 35 litres.

We started with 30 litres and ended up with 35 litres. Therefore, 5 litres of water was added.

1. 20 kg

2. 12.5 kg

3. 16 kg

4. 200 kg

Answer. 20 kg

Explanation.

i.e. (x * 42) + (25 * 24) = 32 (x + 25)

â‡’ 42x + 600 = 32x + 800

â‡’ 10x = 200 or x = 20 kgs

1. 4 litres

2. 2 litres

3. 1 litre

4. 1.5 litres

Explanation.

The 12-litre mixture contains 40% milk and 60% water in it.

That is 4.8 litres of milk and 7.2 litres of water.

We are replacing the mixture with pure milk so that the amount of milk and water in the mixture is 50% and 50%.

So, we will end up with 6 litres of milk and 6 litres of water.

Water gets reduced by 1.2 litres.

To remove 1.2 litres of water from the original mixture containing 60% water, we need to remove 1.2/0.6 litres of the mixture = 2 litres.

That is 4.8 litres of milk and 7.2 litres of water.

We are replacing the mixture with pure milk so that the amount of milk and water in the mixture is 50% and 50%.

So, we will end up with 6 litres of milk and 6 litres of water.

Water gets reduced by 1.2 litres.

To remove 1.2 litres of water from the original mixture containing 60% water, we need to remove 1.2/0.6 litres of the mixture = 2 litres.

**Q7. A sample of x litres from a container having a 60 litre mixture of milk and water containing milk and water in the ratio of 2 : 3 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x?****1. 6 litres****2. 10 litres****3. 30 litres****4. None of these****Answer. 10 litresExplanation.**

Let us solve this question by back substituting answer choices.

The rigorous method of solving is outlined in the previous question on removing and replacing mixtures.

The mixture of 60 litres has milk and water in the ratio 2 : 3

i.e., 24 litres of milk and 36 litres of water.When x litres of the mixture is removed, 0.4 x litres of milk and 0.6 x litres of water are removed from it.Take choice (2). According to this choice, x = 10.

So, 10 litres of the mixture is removed.

4 litres of milk and 6 litres of water are removed.

Therefore, there will be 20 litres of milk and 30 litres of water in the container.

Subsequently, when 10 litres of milk is added, the mixture will contain 30 litres of milk and 30 litres of water - i.e. milk and water are in equal proportion.

**Q8. A farmer counted the heads of the animals / birds in the farm and found it to be 80. When he counted the legs he found it to be 260. If the farm had either pigeons or horses and nothing else, how many horses were there in the farm? In the farm, each horse had four legs and each pigeon had two legs.****1. 40****2. 30****3. 50****4. 60****Answer. 50Explanation.**

The count of heads gives the total number of animals / birds as each animal / bird has one head.

Let the number of horses be x.

The remaining will be pigeons. The number of pigeons = (80 - x).

Each pigeon has 2 legs and each horse has 4 legs.

Therefore, total number of legs = 4x + 2(80 - x) = 260

â‡’ 4x + 160 - 2x = 260

â‡’ 2x = 100

â‡’ x = 50.

x is the number of horses. The number of horses in the farm is 50.

**Q9. From a cask of milk containing 30 litres, 6 litres are drawn out and the cask is filled up with water. If the same process is repeated a second, then a third time, what will be the number of litres of milk left in the cask?****1. 0.512 liters****2. 12 liters****3. 14.38 liters****4. 15.36 liters****Answer. 15.36 litersExplanation.**

The problem can be solved by brute force of removing and replacing 6 litres each time and finding the quantity of milk and water after each iteration. Finally, after step 3, we will get to the answer. But it is cumbersome to do that.

The question can be solved more effectively. Let us use that method to solve this question.

Let us find the fraction of milk left after nth operation.

Fraction of milk in the cask after the nth iteration

'x' is initial quantity of milk in the cask 'y' is the quantity of milk drawn out in each iteration and 'n' is the number of iterations.

Fraction of milk left after the 3rd iteration

âˆ´ the quantity of milk in the cask after the 3rd iteration

**Q10. A 20 litres mixture of milk and water comprising 60% pure milk is mixed with "x" litres of pure milk. The new mixture comprises 80% milk. What is the value of "x"?****1. 40 litres****2. 20 litres****3. 8 litres****4. 16 litres****Answer. 20 litresExplanation. **

The quantity of the initial mixture of milk and water is 20 litres.

Out of the 20 litres, 60% is pure milk. i.e., 12 litres is milk and the remaining 8 litres is water.

When "x" litres of pure milk is added, we will have (12 + x) litres of milk in the new mixture.

And there will be 20 + x litres of the new mixture.

The question states that (12 + x) litres of milk accounts for 80% of (20 + x) litres of the mixture.

Therefore,

5(12 + x) = 4(20 +x)

60 + 5x = 80 + 4x

x = 20 litres

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