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

You can practice all previous year CAT questions from the topic Arithmetic - Mixture, Alligation, Removal & Replacement. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

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

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

**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**

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

**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**

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

**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**

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

**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**

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

**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**

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

**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 2 | Arithmetic - Mixture, Alligation, Removal & Replacement**

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

**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 (4).

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

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

**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 4.

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

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

**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 1.

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

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

**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

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

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

Hence, option 3.

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

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

**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

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

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

$\frac{3a+2b+c}{3+2+1}=30or3a+2b+c$ = 180 … (II)

Multiplying equation (I) with 3 and (II) 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 1.

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

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

**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 4.

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

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

**Explanation** :

When a 20% solution is mixed with x% solution in the ratio 1 : 3, the resultant concentration (in %) of the solution

$=\frac{20\left(1\right)+x\left(3\right)}{1+3}=\frac{3x+20}{4}=0.75x+5$

When this solution is mixed with 20% solution having equal volume, the resultant concentration (in %) of the solution

$=\frac{0.75x+5+20}{2}=\frac{3}{8}x+12.5\phantom{\rule{0ex}{0ex}}\therefore \frac{3}{8}x+12.5=31.25$

Solving for x, x = 50.

Hence, option 2.

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

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

**Explanation** :

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

Average score of the girls = b + 5

In the final exam, average score of the girls = b + 5 – 3 = b + 2.

Average score of the entire class increased by 2 and is hence $\frac{50(b+5)-30(b+2)}{50}+2$ i.e. b + 5.

Average score of the boys = $\frac{50(b+5)-30(b+2)}{20}$ = b + 9.5

Increases in the average of boys is 9.5

Hence, option 1.

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

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

**Explanation** :

Bottle 1 contains 7/9 of mil

Bottle 1 contains 9/13 of mil

The final mixture contains 3/4 of milk

Now using the allegation rule, we get the ratio of milk to water as follows

The final mixture contains 3/4 of milk

$\frac{{\displaystyle \frac{3}{4}}-{\displaystyle \frac{9}{13}}}{{\displaystyle \frac{7}{9}}-{\displaystyle \frac{3}{4}}}=\frac{{\displaystyle \frac{3}{52}}}{{\displaystyle \frac{1}{36}}}$

$\Rightarrow \frac{3}{52}+\frac{1}{36}=\frac{108}{52}or\frac{27}{13}$ = 27 : 13

Hence, option 2.

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

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

**Explanation** :

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

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

Quantity of water in liquid A, B and C is $\frac{4}{9}\left(\frac{1}{3}\right),\frac{3}{9}\left(\frac{1}{4}\right)$ and $\frac{2}{9}\left(\frac{1}{5}\right)$

i.e., $\frac{4}{27},\frac{1}{12}$ and $\frac{2}{45}$

So quantity of water = $\frac{4}{27}+\frac{1}{12}+\frac{2}{45}$

$\Rightarrow \frac{80+45+24}{540}=\frac{149}{540}$

Now proportion of liquid A in resultant mixture

$=\frac{4}{9}\times \frac{2}{3}=\frac{8}{27}$

Similarly proportion of liquid B and liquid C in the resultant mixture is

$\left(\frac{3}{9}\times \frac{3}{4}=\frac{1}{4}\right)$ and $\left(\frac{2}{9}\times \frac{4}{5}=\frac{8}{45}\right)$

Now let is examine the statement in each option individually

1] The resultant mixture has $\frac{149}{540}$ of water and $\frac{1}{4}$ of liquid B.

Now $\frac{149}{540}>\frac{1}{4}$

Hence, statement (1) is false.

2] The resultant mixture has $\frac{1}{4}$ of liquid B and $\frac{8}{45}$ of liquid C. As $\frac{1}{4}>\frac{8}{45}.$

So statement (2) is false.

3] Now quantity of water in resultant mixture

$=\frac{149}{540}$ and quantity of liquid B is $\frac{1}{4}.$

As $\frac{149}{540}>\frac{1}{4}$

Statement (3) is true.

4] Quantity of liquid A in resultant, mixture is

$\frac{8}{27}=\frac{160}{540}>\frac{149}{540}$

So quantity of water is greater than the quantity of the resultant mixture. So statement (4) is false.

Hence, option 3.

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

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

**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 1.

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

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

**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 3.

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**Directions: Answer the questions based on the following information.**

The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.

**CAT 1999 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

What is the minimum amount of sucrose (to the nearest gram) that must be added to one gram of saccharin to make a mixture that will be at least 100 times as sweet as glucose?

- A.
7

- B.
8

- C.
9

- D.
100

Answer: Option C

**Explanation** :

If the mixture is to be made 100 times as sweet as glucose, its sweetness should be 74. The ratio in which saccharin and sucrose be mixed to get the above level of sweetness is given by the following alligation table.

In other words, it means to achieve the given level of sweetness, you need to add 601 g of sucrose to 73 g of saccharin. Hence to 1 g of saccharin, the amount of sucrose to be added is $\frac{601}{73}$ = 8.23.g.

Hence, option (c).

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**CAT 1999 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Approximately how many times sweeter than sucrose is a mixture consisting of glucose, sucrose and fructose in the ratio of 1 : 2 : 3?

- A.
1.3

- B.
1.0

- C.
0.6

- D.
2.3

Answer: Option A

**Explanation** :

$\frac{\left[\right(0.74)+(1.000)2+(1.7\left)3\right]}{6}$ = 1.31.

Hence, option (a).

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**CAT 1998 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the

second container. Then

- A.
A > B

- B.
A < B

- C.
A = B

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Let the capacity of each cup be 100 ml. So 300 ml of alcohol is taken out from the first container and poured into the second one. So the first vessel will have 200 ml of alcohol and the second one will have 500 ml of water and 300 ml of alcohol. So the ratio of water to alcohol in the second vessel is 5 : 3.

Hence, proportion of alcohol in B = 3 : 8

Now if 300 ml of mixture is removed from the second container, it will have $\left(300\times \frac{5}{8}\right)$ = 187.5 ml of water and $\left(300\times \frac{3}{8}\right)$ = 112.5 ml of alcohol. Now if this mixture is poured in the second vessel, that vessel would have (200 + 112.5) = 312.5 ml of alcohol and 187.5 ml of water. Hence, ratio of alcohol to water in this container = 312.5 : 187.5 = 5 : 3

Hence, proportion of water = A = 3 : 8

Hence, we find that A = B

**Note: **This result will be independent of the capacity of the cup.

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**CAT 1998 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

In a group of 150 students, find the number of girls.

I. Each girl was given 50 paise, while each boy was given 25 paise to purchase goods totalling Rs. 49.

II. Girls and boys were given 30 paise each to buy goods totalling Rs. 45.

Answer: 1

**Explanation** :

Using the first statement alone, we can alligate and find the ratio of boys to girls and hence the number of girls, i.e. as shown in the adjacent diagram, 150 students when divided in the ratio 115 : 260, give 46 girls and 104 boys. The second statement, however, does not throw any further light on the data given in the question as it simply suggests 0.3B + 0.3G = 45 or B + G = 150, which is already known. Hence, only statement I is required to answer the question.

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**CAT 1997 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

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

- A.
2 kg

- B.
2.5 kg

- C.
2.4 kg

- D.
None of these

Answer: Option B

**Explanation** :

20 kg fresh grapes will contain (0.9 × 20) = 18 kg water and 2 kg mass. If the dry grape has to contain 2 kg mass, it should constitute 80% of that. Hence, if 80% of dry grapes corresponds to 2 kg, its total weight will be $\left(\frac{2}{0.8}\right)$ = 2.5 kg

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**CAT 1996 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?

- A.
2 : 3

- B.
4 : 3

- C.
3 : 2

- D.
3 : 4

Answer: Option D

**Explanation** :

We can solve this by alligation. But while we alligate, we have to be careful that it has to be done with respect to any one of the two liquids, viz. either A or B. We can verify that in both cases, we get the same result. e.g. the proportion of A in the first vessel is $\frac{5}{6}$ and that in the second vessel is $\frac{1}{4},$ and we finally require $\frac{1}{2}$ parts of

A. Similarly, the proportion of B in the first vessel is $\frac{1}{6},$ that in the second vessel is $\frac{3}{4}$ and finally we want it to be $\frac{1}{2}.$ With respect to liquid A

With respect to liquid B

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**CAT 1994 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A man buys spirit at Rs. 60 per litre, adds water to it and then sells it at Rs. 75 per litre. What is the ratio of spirit to water if his profit in the deal is 37.5%?

- A.
9 : 1

- B.
10 : 1

- C.
11 : 1

- D.
None of these

Answer: Option B

**Explanation** :

Since SP of spirit and solution water = Rs.75/l and there is a profit of 37.5%, CP of spirit and water solution

$=\frac{75}{1.375}$ = Rs. 54.54/L

This should indeed be the weighted average of the costs of spirit and water. So if we alligate, we can get the ratio of spirit : water (assuming that cost of water is 0).

i.e. 10 : 1.

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