# Cat Questions On Replacement PDF

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Cat Questions On Replacement PDF:

CAT Replacement questions and answers with solutions based on mixtures and alligations, previous year CAT questions. Download and practice important CAT Questions on replacement.

Question 1: From a tank containing 100 litres of milk, 5 litres of milk is taken out and replaced with an equal amount of water. 10 litres of the resulting mixture is again taken out and replaced with an equal amount of water. Again, some amount of the mixture is taken out and replaced with an equal amount of water. If the final quantity of milk left in the tank is 68.4 litres, how much milk was taken out in the third iteration?

a) 10
b) 15
c) 20
d) 25

Question 2: A vessel contains milk and water in the ratio 7 : 3. 3 litres of this mixture is taken out and replaced with water. Again, 2 litres of the mixture is taken out and replaced with water. What is the final ratio of milk and water in the vessel?

a) 79 : 87
b) 93 : 17
c) 43 : 69
d) 49 : 76

Question 3: Some ethanol from a container having 50 litres of ethanol is drawn out and replaced with an equal amount of water. This process is repeated thrice and after that, only 25.6 litres of ethanol is left in the container. What is the amount of ethanol removed in each iteration?

a) 20 l
b) 10 l
c) 30 l
d) 15 l

Question 4: From a can of 30 litres of milk, 10 litres are withdrawn and replaced with water. Again, 9 litres are withdrawn and replaced with water. In the similar manner, the quantity of mixture withdrawn and replaced by water is decreased by 1litre and repeated till 1 litre of mixture is withdrawn and replaced with water. What will be the quantity of milk left in final mixture?

a) $\frac{29!}{19!} * \frac{1}{30^{9} }$

b) $\frac{30!}{19!} * \frac{1}{30^{9} }$

c) $\frac{30!}{9!} * \frac{1}{30^{9} }$

d) $\frac{29!}{19!} * \frac{1}{30^{19} }$

Question 5: The quantity of apple juice to grape juice in a cocktail is in ratio 5 : 9. If X percent of juice is removed and replaced with grape juice, the concentration of apple juice in the resulting comes out to be in range 20% to 30%. What is the range of values of X?

a) 24 < x < 52
b) 18 < x < 40
c) 22 < x < 48
d) 16 < x < 44

Question 6: Ram prepares solutions of alcohol in water according to customers’ needs. This morning Ram has prepared 27 litres of a 12% alcohol solution and kept it ready in a 27 litre delivery container to be shipped to the customer. Just before delivery, he finds out that the customer had asked for 27 litres of 21% alcohol solution. To prepare what the customer wants, Ram replaces a portion of 12% solution by 39% solution. How many litres of 12% solution are replaced?

a) 5
b) 9
c) 10
d) 12
e) 15

Question 7: A mixture contains water and alcohol in the ratio 5:11. 11.11% of the mixture is drawn and replaced with water. Once this process is done, it is deemed that Round 1 has been performed on the mixture. After how many rounds does the concentration of alcohol in the mixture go below 50% for the first time?

a) Round 2
b) Round 3
c) Round 4
d) Round 5

Question 8: From a container containing 40l of alcohol, 4l of alcohol is replaced with water and then 5l of the resultant mixture is replaced with 6l of water. What is the ratio of the amount of alcohol to that of the amount of water in the final mixture?

a) 59:23
b) 61:21
c) 57:25
d) 63:19

We know that,
Final quantity = Initial quantity * $(1 – \frac{\text{Quantity removed}}{\text{Quantity after replacement}}) * (1 – \frac{\text{Quantity removed}}{\text{Quantity after replacement}})………$
Let the amount of milk taken out in the third iteration be $x$ litres.
The, 68.4 = 100 * $(1 – \frac{5}{100}) * (1 – \frac{10}{100}) * (1 – \frac{x}{100})$
On solving, we get $x$ = 20 litres.
Hence, 20 is the correct answer.

We know that Final quantity = Initial Quantity $(1 – \frac{\text{quantity removed}}{\text{total quantity}})$
In the first case, 3 litres are removed and replaced with water and in the second case 2 litres are removed and replaced with water
For milk,
=> Final proportion = Initial proportion $(1 – \frac{\text{quantity removed in the 1st case)}}{\text{total quantity}})$*$(1 – \frac{\text{quantity removed in the 2nd case}}{\text{total quantity}})$
=> Final proportion = $\frac{7}{10} * (1 – \frac{3}{10})* (1 – \frac{2}{10})$
=> Final proportion = $\frac{7}{10}$ * $\frac{7}{10}$ * $\frac{8}{10}$ = $\frac{49}{125}$
Therefore, ratio of milk and water = 49 : 76
Hence, option D is the correct answer.

We can use the formula
Quantity left = Quantity present initially * $(1-\frac{\text{Quantity removed}}{\text{Total Quantity}})^n$
where n is the number of iteration.
Let the quantity removed in each iteration be x litres
Putting all values we get
25.6 = 50 * $(1-\frac{x}{50})^3$
Solving this equation we can get x=10 litres
Hence, option B is correct.

Quantity of milk left after 1st iteration = 30 * $\frac{20}{30}$
Similarly, after 2nd iteration, Quantity = 30 * $\frac{20}{30}$$*$$\frac{21}{30}$
This will be repeated 10 times with quantity withdrawn decreasing by 1 each time.
So, Quantity of milk in final mixture = 30 * $\frac{20}{30} * \frac{21}{30} * \frac{22}{30}……..\frac{29}{30}$
= $\frac{29!}{19!} * \frac{1}{30^{9} }$
Hence, option A is correct.

Let the quantity of cocktail be 1400 units.
So the quantity of apple juice will be 500 units and quantity of grape juice will be 900 units.
X% of juice is removed i.e. 1400*X/100 = 14X units is removed
Out of which 5X would be apple juice and 9X would be grape juice.
New concentration of apple juice,
$\frac{500 – 5X}{1400}$%
According to the given condition we have,
$20 < \frac{500 – 5X}{1400} * 100 < 30$
$20 < \frac{500 – 5X}{14} < 30$
$280 < 500 – 5X < 420$
$-220 < – 5X < -80$
$16 < x < 44$
Hence, option D is the right answer.

Let Ram replaces $x$ litres of 12 % sol. with 39 % solution

Now, quality of 12 % solution in 27 litre = $\frac{12}{100} \times 27$

=> After replacing we have volume of 12 % solution

= $(\frac{12}{100} \times 27) – (\frac{12 x}{100}) + (\frac{39 x}{100})$

= $\frac{324 + 27 x}{100}$

This is equal to 27 litre of 21 % solution.

=> $\frac{324 + 27 x}{100} = \frac{21}{100} \times 27$

=> $27x = 567 – 324 = 243$

=> $x = \frac{243}{27} = 9$

Initially, the ratio of alcohol in the mixture is 11/16.
Now, 11.11% $\approx$ 1/9.
So, every time the process is performed, 1/9th of the alcohol is replaced with water. So, after each round, the concentration of alcohol becomes 1-1/9 = 8/9 of the previous concentration.
Concentration after round 1 = $\frac{8}{9} \times \frac{11}{16} > \frac{1}{2}$
Concentration after round 2 = $\frac{8}{9} \times \frac{8}{9} \times \frac{11}{16} > \frac{1}{2}$
Concentration after round 3 = $\frac{8}{9} \times\frac{8}{9} \times \frac{8}{9} \times \frac{11}{16} \approx 48$%.
Thus, Round 3 is the correct answer.

The amount of alcohol left after first replacement = $40(1- \frac{4}{40})$
The amount of alcohol left after second replacement = $40(1- \frac{4}{40})$*$(1- \frac{5}{40})$ = $31.5$