# Arithmetic - Mixture, Alligation, Removal & Replacement - Previous Year CAT/MBA Questions

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**CAT 2023 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A mixture P is formed by removing a certain amount of coffee from a coffee jar and replacing the same amount with cocoa powder. The same amount is again removed from mixture P and replaced with same amount of cocoa powder to form a new mixture Q. If the ratio of coffee and cocoa in the mixture Q is 16 : 9, then the ratio of cocoa in mixture P to that in mixture Q is

- (a)
4 : 9

- (b)
1 : 3

- (c)
1 : 2

- (d)
5 : 9

Answer: Option D

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**Text Explanation** :

Let the initial quantity of coffee in the jar be 100 kg and r kg is replaced each time.

Since r kg out of 100 kg is removed, fraction of coffee removed = r/100

∴ fraction of coffee remaining = $\left(1-\frac{\mathrm{r}}{100}\right)$

⇒ Quantity of coffee remaining after first replacement = 100 × $\left(1-\frac{\mathrm{r}}{100}\right)$

And quantity of cocoa after first replacement = r kgs.

⇒ Similarly, quantity of coffee remaining after second replacement = 100 × ${\left(1-\frac{\mathrm{r}}{100}\right)}^{2}$

Now, after 2nd replacement coffee and cocoa are in the ratio of 16 : 9

⇒ Quantity of coffee left after 2nd replacement = $\frac{16}{16+9}$ × 100 = 64 kg

⇒ 100 × ${\left(1-\frac{\mathrm{r}}{100}\right)}^{2}$ = 64

⇒ ${\left(1-\frac{\mathrm{r}}{100}\right)}^{2}$= $\frac{64}{100}$

⇒ $\left(1-\frac{\mathrm{r}}{100}\right)$ = $\frac{8}{10}$ = $\frac{4}{5}$

⇒ r = 20 kg

∴ 20 kg of cocoa is added after 1st replacement.

Also, quantity of cocoa after 2nd replacement = 100 – 64 = 36 kgs

⇒ Required ratio = 20 : 36 = 5 : 9.

Hence, option (d).

**Concept**:

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**CAT 2023 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A container has 40 liters of milk. Then, 4 liters are removed from the container and replaced with 4 liters of water. This process of replacing 4 liters of the liquid in the container with an equal volume of water is continued repeatedly. The smallest number of times of doing this process, after which the volume of milk in the container becomes less than that of water, is

Answer: 7

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**Text Explanation** :

4 liters out of 40 liters is remvoed, hence 4/40 = 1/10^{th} is removed every time.

∴ 1 - 1/10 = 9/10^{th} remains every time.

Quantity of milk will become less than that of water as soon as quantity of milk goes below 20 liters.

∴ Quantity of milk remaining after n replacements = 40 × (9/10)^{n} < 20

⇒ (0.9)^{n} < 0.5

Least value of n satisfying above inequality is 7.

Hence, 7.

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**CAT 2023 QA Slot 3 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

Anil mixes cocoa with sugar in the ratio 3 : 2 to prepare mixture A, and coffee with sugar in the ratio 7 : 3 to prepare mixture B. He combines mixtures A and B in the ratio 2 : 3 to make a new mixture C. If he mixes C with an equal amount of milk to make a drink, then the percentage of sugar in this drink will be

- (a)
21

- (b)
17

- (c)
16

- (d)
24

Answer: Option B

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**Text Explanation** :

Let 20 kgs and 30 kgs of A and B are mixed.

∴ Amount of sugar in C = 2/5 × 20 + 3/10 × 30 = 17 kgs

Now we have 50 kgs of C mixed with 50 kgs of milk i.e., 100 kgs of final solution.

⇒ Concentration of sugar in final solution = 17/(50 + 50) × 100% = 17%

Hence, option (b).

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**CAT 2022 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A mixture contains lemon juice and sugar syrup in equal proportion. If a new mixture is created by adding this mixture and sugar syrup in the ratio 1 : 3, then the ratio of lemon juice and sugar syrup in the new mixture is

- (a)
1 : 4

- (b)
1 : 6

- (c)
1 : 5

- (d)
1 : 7

Answer: Option D

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**Text Explanation** :

Let 10 units of mixture is mixed with 30 units of sugar syrup.

Amount of sugar syrup in 10 units of mixture = 5 units

Amount of lemon juice in 10 units of mixture = 5 units

∴ Total amount of lemon juice in the final mixture = 5 units, and

Total amount of sugar syrup in the final mixture = 5 + 30 = 35 units

Ratio of lemon juice and sugar syrup in final mixture = 5 : 35 = 1 : 7.

Hence, option (d).

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**CAT 2022 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

There are two containers of the same volume, first container half-filled with sugar syrup and the second container half-filled with milk. Half the content of the first container is transferred to the second container, and then the half of this mixture is transferred back to the first container. Next, half the content of the first container is transferred back to the second container. Then the ratio of sugar syrup and milk in the second container is

- (a)
5 : 6

- (b)
4 : 5

- (c)
5 : 4

- (d)
6 : 5

Answer: Option A

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**Text Explanation** :

Let container A initially have 100 liters of sugar while container B have 100 liters of milk.

Now the second container has sugar and milk in the ratio of 1 : 2.

When half i.e., 75 liters of it is transferred, 25 liters of sugar and 50 liters of milk will be transferred.

Now the first container has sugar and milk in the ratio of 3 : 2.

When half i.e., 62.5 liters of it is transferred, 37.5 liters of sugar and 25 liters of milk will be transferred.

∴ Ratio of sugar and milk in 2nd container = 62.5 : 75 = 625 : 750 = 25 : 30 = 5 : 6

Hence, option (a).

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**CAT 2022 QA Slot 3 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A glass contains 500 cc of milk and a cup contains 500 cc of water. From the glass, 150 cc of milk is transferred to the cup and mixed thoroughly. Next, 150 cc of this mixture is transferred from the cup to the glass. Now, the amount of water in the glass and the amount of milk in the cup are in the ratio

- (a)
10 : 13

- (b)
10 : 3

- (c)
3 : 10

- (d)
1 : 1

Answer: Option D

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**Text Explanation** :

When two containers contain equal amount of milk and water respectively and then equal amounts are transferred from 1^{st} to 2^{nd} and then from 2^{nd} to 1^{st}.

The amount of milk in 1^{st} becomes equal to amount of water in 2nd while

amount of water in 1^{st} becomes equal to amount of milk in 2^{nd}.

Hence, option (d).

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**CAT 2021 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is

Answer: 200

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**Text Explanation** :

Let x cc is removed from first bottle and replaced with solution from second bottle.

Using alligation rule

⇒ (800-x)/x = 4/12 = 1/3

⇒ 2400 – 3x = x

⇒ x = 600 cc.

∴ Solution left in second bottle = 800 – 600 = 200 cc.

Hence, 200.

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**CAT 2021 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

From a container filled with milk, 9 litres of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is

Answer: 45

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**Text Explanation** :

Initially the container is full of milk. Let the capacity of container be ‘V’ liters.

When 9 liters are drawn, fraction of quantity drawn = $\frac{9}{V}$.

Hence, fraction of milk drawn will also be $\frac{9}{V}$

∴ Fraction of milk remaining will be $\left(1-\frac{9}{V}\right)$

Only water is added.

⇒ Quantity of milk remaining after first replacement = $V\left(1-\frac{9}{V}\right)$

Similarly, after second replacement quantity of milk remaining = $V{\left(1-\frac{9}{V}\right)}^{2}$

This is $\frac{16}{25}$ ^{th} of the total volume.

∴ $V{\left(1-\frac{9}{V}\right)}^{2}$ = $\frac{16}{25}$V

⇒ 1 – $\frac{9}{V}$ = $\frac{4}{5}$

⇒ V = 45 liters.

Hence, 45.

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**CAT 2021 QA Slot 3 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is

- (a)
2.5

- (b)
3.5

- (c)
3

- (d)
4

Answer: Option C

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**Text Explanation** :

Let the weight of initial alloy of silver and copper be x kg and it has p% silver.

This alloy is mixed with 3 kg pure silver.

Using alligation:

⇒ $\frac{a}{3}$ = $\frac{100-90}{90-p}$ = $\frac{10}{90-p}$ …(1)

This alloy is not mixed with another alloy of 2kg containing 90% silver.

Using alligation:

⇒ $\frac{a}{2}$ = $\frac{90-84}{84-p}$ = $\frac{6}{84-p}$ …(2)

From (1) and (2), we get

$\frac{30}{90-p}$ = $\frac{12}{84-p}$

⇒ 2520 – 30p = 1080 – 12p

⇒ 18p = 1440

⇒ p = 80%

∴ a = $\frac{30}{90-p}$ = $\frac{30}{90-80}$ = 3 kg.

Hence, option (c).

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**CAT 2020 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one-fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?

Answer: 8

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**Text Explanation** :

40 litres solution had dye and water in the ratio 2 : 3

∴ dye = 16 l and water = 24 l.

After adding x liters of water, now proportion is 2 : 5

⇒ $\frac{16}{24+x}=\frac{2}{5}$

⇒ x = 16 liters

Dye = 16 l and water = 40 l now in the solution.

Now, 1/4^{th} is removed from the solution, hence dye left = ¾ × 16 = 12 l and water left = ¾ × 40 = 30 l

Now, After adding y liters of dye, proportion becomes 2 : 3.

⇒ $\frac{12+y}{30}=\frac{2}{3}$

⇒ y = 8 liters

Hence, 8.

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**CAT 2020 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A, B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg, of the metal C is

- (a)
70

- (b)
84

- (c)
96

- (d)
48

Answer: Option B

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**Text Explanation** :

Let 1 liter of A, B and C weight 5kg, 2kg and 6 kg respectively.

Now, let’s suppose 3, 4 and 7 liters of A, B and C are mixed.

∴ Total weight of 14 liters of this solution = 3 × 5 + 4 × 2 + 7 × 6 = 65 kgs.

⇒ 65 kgs of the solution contains 7 × 6 = 42 kgs of C

∴ 130 kgs of the solution contains 2 × 42 = 84 kgs of C

Hence, option (b).

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**CAT 2020 QA Slot 3 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

Two alcohol solutions, A and B, are mixed in the proportion 1 : 3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is

- (a)
92%

- (b)
90%

- (c)
89%

- (d)
94%

Answer: Option A

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**Text Explanation** :

Given, A and B are mixed in the ratio 1 : 3

Let 10 liters of A and 30 liters of B is mixed initially i.e., total 40 liters solution.

Now volume is double by adding 40 liters of A.

∴ In the solution A = 10 + 40 = 50 liters and B = 30 liters

Concentration of alcohol in this 80-liter solution = 72%

Let the alcohol concentration in B = b%

∴ 80 × 72% = 50 × 60% + 30 × b%

⇒ 576 = 300 + 30b

⇒ b = 92%

Hence, option (a).

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**CAT 2019 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

- (a)
70

- (b)
85

- (c)
80

- (d)
75

Answer: Option C

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**Text Explanation** :

Density of liquid 1 = 1000 g/L.

Density of liquid 2 = 800 g/L.

Half litre of the mixture weighs 480 gm, so 1 L of the mixture weighs 960 gm.

So, density of the mixture = 960 g/L.

Using the alligation cross;

$\frac{Liquid1}{Liquid2}=\frac{(960-800)}{(1000-960)}$ = $\frac{4}{1}$.

Percentage of liquid 1 in the mixture = (4/5) × 100 = 80%.

Hence, option (c).

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**CAT 2019 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A is

- (a)
15

- (b)
12

- (c)
13

- (d)
14

Answer: Option D

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**Text Explanation** :

Vessels A, B and C contains salt solution of strengths 10%, 22% and 32% respectively

It is also given that the amount of salt solution = 500 ml

So, Vessels A, B and C contains salt of 50 grams, 110 grams and 160 grams respectively

100 ml of Solution is transferred from A to B:

A would have 400 ml, B would have 600 ml of solution

Amount salt from A which is transferred to B = (Initial salt amount)/5 = 10 grams

So, Total salt in B = 110 + 10 = 120 grams (After first transfer)

Total salt in A = 40 grams (After first transfer)

Now, 100 ml from Vessel B is transferred to Vessel C

So similarly, 1/6th of salt would transfer from B to C

Total Salt in B = 120 - 20 = 100 grams (After second transfer)

Total Salt in C = 160 + 20 = 180 grams (After second transfer)

Now, 100 ml from Vessel C is transferred to Vessel A

So similarly, 1/6^{th} of salt would transfer from C to A

Total Salt in C = 160 - 30 = 130 grams (After third transfer)

Total Salt in A = 40 + 30 = 70 grams (After third transfer)

So, Vessel A contains 70 grams Salt in 500 ml solution

Strength of Salt Solution in Vessel A = 14%

Hence, option (d).

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**CAT 2018 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio

- (a)
17 : 25

- (b)
18 : 25

- (c)
21 : 25

- (d)
19 : 24

Answer: Option D

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**Text Explanation** :

Let cost of tea A and B be a and b respectively.

If 3 kg of tea A is mixed with 2 kg of tea B, (3a + 2b) × 1.1 = 40× 5 = 200

If 2 kg of tea A is mixed with 3 kg of tea B, (2a + 3b) × 1.05 = 40× 5 = 200

Therefore,

(3a + 2b) × 1.1 = (2a + 3b) × 1.05

∴ 3.3a + 2.2b = 2.1a + 3.15b

∴ a/b = 19/24

Hence, option (d).

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**CAT 2018 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is

- (a)
20

- (b)
26

- (c)
16

- (d)
22

Answer: Option A

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**Text Explanation** :

Let the 10 litres of mixture has ‘Y’ litres of A and (10 – Y) litres of B. Let cost of paint B be Rs. X and that of A be Rs. (X + 8).

We know that, Y ≥ (10 – Y) ⇒ Y ≥ 5

The trader makes 10% profit by selling this mixture at Rs. 264.

∴ Cost price of the mixture = $\frac{264}{1.1}$ = Rs. 240

∴ (X + 8) × Y + (10 – Y) × X = 240

∴ 10X + 8Y = 240

∴ X = 24 – 0.8Y

For maximum value of X, we need to consider minimum value of Y.

∴ X = 24 – (0.8 × 5) = Rs. 20

Hence, option (a).

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**CAT 2018 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now

- (a)
25.4

- (b)
20.5

- (c)
35.2

- (d)
30.3

Answer: Option C

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**Text Explanation** :

Originally, the volume of alcohol = 4 times the volume of water. Therefore, original percentage of alcohol = 80% and original volume of water = 20%.

When 10% of the mixture is removed and replaced with water, the percentage of alcohol remained in the mixture

= $80\%\times {\left(\frac{9}{10}\right)}^{2}$ = 64.8%

Therefore the percentage of water in the mixture = (100 - 64.8)% = 35.2%.

Hence, option (c).

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**CAT 2018 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is?

- (a)
1 : 3

- (b)
1 : 4

- (c)
2 : 5

- (d)
3 : 10

Answer: Option A

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**Text Explanation** :

Suppose the concentrations of the solutions A, B and C are ‘a’, ‘b’ and ‘c’ respectively.

If the three solutions are mixed in the ratio 1 : 2 : 3, the resultant solution has 20% concentration. Therefore, we have

$\frac{a+2b+3c}{1+2+3}=20$ or a + 2b + 3c = 120 ...(1)

If the three solutions are mixed in the ratio 3 : 2 : 1, the resultant solution has 30% concentration. Therefore, we have

$\frac{3a+2b+c}{3+2+1}=30$ or 3a + 2b + c = 180 ...(2)

Multiplying equation (1) with 3 and (2) with 2 and equating both, we get

3a + 6b + 9c = 6a + 4b + 2c

⇒ 2b + 7c = 3a

Now, concentration of solution D = $\frac{2b+7c}{9}$ = $\frac{3a}{9}$ = $\frac{a}{3}$

Therefore the ratio of D's concentration to A's concentration $=\frac{a/3}{a}=\frac{1}{3}$

Hence, option (a).

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**CAT 2018 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio

- (a)
220 : 149

- (b)
251 : 163

- (c)
229 : 141

- (d)
239 : 161

Answer: Option D

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**Text Explanation** :

Suppose the mixture from drum 1 = 300x litres and the mixture from drum 2 = 400x litres. Therefore, we have the following

Therefore the volume of Paint A in Drum B = 455x – 216x = 239x and the volume of Paint B in Drum B = 245x − 84x = 161x.

Therefore the required ratio is 239 : 161.

Hence, option (d).

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**CAT 2018 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1 : 3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is

- (a)
52%

- (b)
50%

- (c)
48%

- (d)
55%

Answer: Option B

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**Text Explanation** :

A 20% solution is mixed with x% solution in the ratio 1 : 3

Let 10 liters of 20% solution is mixed with 30 liters of x% solution.

Now, the total volume of the mixture = 40 liters.

This solution is mixed with 20% solution having equal volume.

∴ A total of 50 liters of 20% solutio is mixed with 30 liters of x% solution.

⇒ Final concentration = 31.25% = $\frac{20\%\times 50+x\times 30}{80}$

⇒ 2500 = 1000 + 30x

⇒ x = 50%

Hence, option (b).

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**CAT 2017 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is:

- (a)
9.5

- (b)
10

- (c)
4.5

- (d)
6

Answer: Option A

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**Text Explanation** :

Let the average score of the boys in the midsemester examination be b.

Average score of the girls = b + 5

Average of the class = $\frac{20\mathrm{b}+30(\mathrm{b}+5)}{50}$ = b + 3

New average of girls = (b + 5) - 3 = b + 2

Let new average of the boys = x

Average score of the entire class increased by 2 and is hence (b + 3) + 2 = b + 5

∴ $\frac{20\mathrm{x}+30(\mathrm{b}+2)}{50}$ = b + 5

⇒ 20x + 30b + 60 = 50b + 250

⇒ 20x = 20b + 190

⇒ x = b + 9.5

Increase in the average of boys is 9.5

Hence, option (a).

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**CAT 2017 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

Bottle 1 contains a mixture of milk and water in 7 : 2 ratio and Bottle 2 contains a mixture of milk and water in 9 : 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3 : 1 ratio?

- (a)
27 : 14

- (b)
27 : 13

- (c)
27 : 16

- (d)
27 : 18

Answer: Option B

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**Text Explanation** :

Bottle 1 contains 7/9 of milk

Bottle 1 contains 9/13 of milk

The final mixture contains 3/4 of milk

Now using the allegation rule, we get the ratio of quanities in bottles as:

⇒ $\frac{\mathrm{Quantity}\mathrm{in}\mathrm{bottle}1}{\mathrm{Quantity}\mathrm{in}\mathrm{bottle}2}$ = $\frac{{\displaystyle \frac{3}{4}}-{\displaystyle \frac{9}{13}}}{{\displaystyle \frac{7}{9}}-{\displaystyle \frac{3}{4}}}=\frac{{\displaystyle \frac{3}{52}}}{{\displaystyle \frac{1}{36}}}$

⇒ $\frac{\mathrm{Quantity}\mathrm{in}\mathrm{bottle}1}{\mathrm{Quantity}\mathrm{in}\mathrm{bottle}2}$ = $\frac{3}{52}\xf7\frac{1}{36}$ = $\frac{3}{52}\times \frac{36}{1}$ = $\frac{108}{52}$ = $\frac{27}{13}$ = 27 : 13

Hence, option (b).

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**CAT 2017 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

Consider three mixtures – the first having water and liquid A in the ratio 1 : 2, the second having water and liquid B in the ratio 1 : 3, and the third having water and liquid C in the ratio 1 : 4. These three mixtures of A, B and C respectively, are further mixed in the proportion 4 : 3 : 2. Then the resulting mixture has

- (a)
The same amount of water and liquid B

- (b)
The same amount of liquids B and C

- (c)
More water than liquid B

- (d)
More water than liquid A

Answer: Option C

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**Text Explanation** :

Proportion of water in three mixtures with A, B and C is $\frac{1}{3}$, $\frac{1}{4}$ and $\frac{1}{5}$ respectively.

[Sum of these proportions is 4, 5 and 6 and LCM(4, 5, 6) = 60]

Now as three mixture with A, B and C are mixed in the ratio 4 : 3 : 2.

Let the amount of three mixtures mixed be 4 × 60, 3 × 60 and 2 × 60 = 240, 180 and 120 liters respectively.

Quantity of water in liquid A, B and C is $\frac{1}{3}$ × 240 + $\frac{1}{4}$ × 180 + $\frac{1}{5}$ × 120 = 80 + 45 + 24 = 149

Quantity of liquid A in resultant mixture = $\frac{2}{3}$ × 240 = 160

Quantity of liquid B in resultant mixture = $\frac{3}{4}$ × 180 = 135

Quantity of liquid C in resultant mixture = $\frac{4}{5}$ × 120 = 96

Now let is examine the options individually

(a) The resultant mixture has 149 liters water and 160 of liquid B.

Hence, option (a) in incorrect.

(b) The resultant mixture has 135 liters of liquid B and 96 liters of liquid C.

Hence, option (b) is incorrect.

(c) Now quantity of water in resultant mixture 149 liters and quantity of liquid B is 135 liters.

Hence, option (c) is correct.

(d) Quantity of liquid A in resultant, mixture is 160 liters and quantity of water = 149 liters.

Hence, option (d) is incorrect.

Hence, option (c).

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**CAT 2004 QA | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he has sold. What is the current proportion of water to milk?

- (a)
2 : 3

- (b)
1 : 2

- (c)
1 : 3

- (d)
3 : 4

Answer: Option A

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**Text Explanation** :

The quantity of milk and water is as shown in the table.

∴ Current proportion of water and milk is 40 : 60 = 2 : 3

Hence, option (a).

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**CAT 2001 QA | Arithmetic - Mixture, Alligation, Removal & Replacement CAT Question**

Fresh grapes contain 90% water by weight while dry grapes contain 20% water by weight. What is the weight of dry grapes available from 20 kg of fresh grapes?

- (a)
2 kg

- (b)
2.4 kg

- (c)
2.5 kg

- (d)
None of these

Answer: Option C

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**Text Explanation** :

Total weight of fresh grapes = 20 kg

Weight of grape mass =

In dried grapes, water is 20%.

∴ Grape mass is 80%.

∴ Total weight of dried grapes =

Hence, option (c).

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