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

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

**XAT 2018 QADI | Arithmetic - Mixture, Alligation, Removal & Replacement**

A mixture comprises water and liquids A and B. The volume of water is 1/3rd of the total mixture and the volume of liquids A and B are in the ratio 5:3. To remove the water, the mixture is passed through a porous medium which completely absorbs the water and partially absorbs liquid A. Altogether this porous medium absorbs 200 ml of the initial mixture. If the ratio of volume of liquids A and B in the residual concentrated mixture becomes 7:9 then find the volume of water absorbed by the porous medium.

- A.
60 ml

- B.
200/3 ml

- C.
80 ml

- D.
100 ml

- E.
120 ml

Answer: Option E

**Explanation** :

Let the volume of liquid A and B in the initial mixture be 5x and 3x ml.

∴ Total volume of A and B = 8x ml, this is 2/3^{rd} of the total mixture.

⇒ Total volume of the mixture = 3/2 × 8x = 12x ml and volume of water = 1/3^{rd} of 12x i.e., 4x ml.

∴ Volume of water, liquid A and liquid B in the mixture will be 4x, 5x and 3x respectively.

When this mixture passes through the porous material whole of water and part of liquid A is absorbed. B is not absorbed by the porous material.

Now, since the ratio of volume of liquids A and B in the residual concentrated mixture becomes 7:9

⇒ (Remaining volume of A) : (3x) = 7 : 9

⇒ Remaining volume of A = 7x/3 ml.

∴ Volume of A absorbed = 5x – 7x/3 = 8x/3

Volume of water absorbed = 4x ml

⇒ 4x + 8x/3 = 200

⇒ x = 30.

∴ Volume of water absorbed = 4x = 120 ml.

Hence, option (d).

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**IIFT 2017 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A pharmaceutical company produces two chemicals X and Y, such that X consists of 5% salt A and 10% salt B and Y consists of 10% salt A and 6% salt B. For producing the chemicals X and Y, the company requires at least 7 gm of Salt A and at least 7 gm of salt B. If chemicals X costs Rs. 10.50 per gm and chemical Y costs Rs. 7.80 per gm, what is the minimum cost at which the company can meet the requirement by using a combination of both types of chemicals?

- A.
Rs. 810

- B.
Rs. 850

- C.
Rs. 537

- D.
None

Answer: Option A

**Explanation** :

Let the total quantity of chemicals A and B be a and b respectively.

∴ Quantity of salt A required = (0.05a + 0.1b) and Quantity of salt B required = (0.1a + 0.06b)

Atleast 7 gms of each salt need to be present.

∴ 0.05a + 0.1b ≥ 7 and 0.1a + 0.06b ≥ 7

On solving, b ≥ 50 and a ≥ 40

Consider the minimum quantities required of A and B as cost is to be minimized.

∴ Minimum cost = (10.5)(40) + (7.8)(50)

= 420 + 390 = Rs. 810

Hence, option 1.

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**IIFT 2017 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

The Drizzle Pvt. Ltd., a squash company has 2 cans of juice. The first contains 25% water and the rest is fruit pulp. The second contains 50% water and rest is fruit pulp. How much juice should be mixed from each of the containers so as to get 12 litres of juice such that the ratio of water to fruit pulp is 3 : 5?

- A.
6 litres, 6 litres

- B.
4 litres, 8 litres

- C.
5 litres, 7 litres

- D.
9 litres, 3 litres

Answer: Option A

**Explanation** :

Quantity of water in 12 litres of juice = (3/8) × 12 = 4.5 litres

Quantity of fruit pulp in 12 litres = 12 − 4.5 = 7.5 litres

Let x litres of the first can and y litres of the second can be used.

∴ Quantity of water used in mixture = (0.25x + 0.5y) and quantity of pulp used = (0.75x + 0.5y)

Comparing the expressions for quantity of water and pulp in the mixtures:

0.25x + 0.5y = 4.5 ... (i)

0.75x + 0.5y = 7.5 ... (ii)

On solving, x = 6 and y = 6

Hence, option 1.

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**IIFT 2015 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

There are two alloys P and Q made up of silver, copper and aluminium. Alloy P contains 45% silver and rest aluminum. Alloy Q contains 30% silver, 35% copper and rest aluminium. Alloys P and Q are mixed in the ratio of 1:4.5. The approximate percentages of silver and copper in the newly formed alloy is:

- A.
33% and 29%

- B.
29% and 26%

- C.
35% and 30%

- D.
None of the above

Answer: Option A

**Explanation** :

P and Q are mixed in the ratio 1 : 4.5 i.e., 2 : 9.

Assume that 1 kg of P is mixed with 4.5 kg of Q.

1 kg of P has 0.45 kg of silver.

4.5 kg of Q has (0.3 × 4.5 =) 1.35 kg of silver and (0.35 × 4.5 =) 1.575 kg of copper.

Thus, the newly formed alloy of 5.5 kg has 1.8 kg of silver and 1.575 kg of copper.

∴ % of silver ≈ 33 and % of copper = 29

Hence, option 1.

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**XAT 2015 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Product M is produced by mixing chemical X and chemical Y in the ratio of 5 : 4. Chemical X is prepared by mixing two raw materials, A and B, in the ratio of 1 : 3. Chemical Y is prepared by mixing raw materials, B and C, in the ratio of 2 : 1. Then the final mixture is prepared by mixing 864 units of product M with water. If the concentration of the raw material B in the final mixture is 50%, how much water had been added to product M?

- A.
328 units

- B.
368 units

- C.
392 units

- D.
392 units

- E.
616 units

Answer: Option B

**Explanation** :

In 864 units of M,

X = 5/9 × 864 = 480 units

Y = 864 – 480 = 384 units

B in 480 units of X = 3/4 × 480 = 360 units

B in 384 units of Y = 2/3 × 384 = 256 units

Total units of B = 360 + 256 = 616

Concentration of B in the final mixture is 50%

Thus, water in the final mixture = (2 × 616) – 864 = 368 units.

Hence, option 2.

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**IIFT 2014 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

X and Y are the two alloys which were made by mixing Zinc and Copper in the ratio 6 : 9 and 7 : 11 respectively. If 40 grams of alloy X and 60 grams of alloy Y are melted and mixed to form another alloy Z, what is the ratio of Zinc and Copper in the new alloy Z?

- A.
6 : 9

- B.
59 : 91

- C.
5 : 9

- D.
59 : 90

Answer: Option B

**Explanation** :

Zinc in alloy Z = 40 × $\frac{6}{15}$ + 60 × $\frac{7}{18}$

= $\frac{118}{3}$

Copper in alloy Z = 40 × $\frac{9}{15}$ + 60 × $\frac{11}{18}$

= $\frac{182}{3}$

Ratio of Zinc and Copper in Z = 118 : 182 = 59 : 91

Hence, option 2.

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**IIFT 2014 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A milk vendor sells 10 litres of milk from a can containing 40 litres of pure milk to the 1^{st} customer. He then adds 10 litres of water to the milk can. He again sells 10 litres of mixture to the 2^{nd} customer and then adds 10 litres of water to the can. Again he sells 10 litres of mixture to the 3^{rd} customer and then adds 10 litres of water to the can and so on. What amount of pure milk will the 5^{th} customer receive?

- A.
$\frac{510}{128}$ litres

- B.
$\frac{505}{128}$ litres

- C.
$\frac{410}{128}$ litres

- D.
$\frac{405}{128}$ litres

Answer: Option D

**Explanation** :

Total volume, V = 40 litres

Volume replaced, v = 10 litres

Number of replacements = 4

Pure milk in the mixture after 4 replacements = [(40 – 10)/ 40]^{4}

Pure milk in 10 litres of the mixture

= 10 × (3/4)^{4} = 405/128

Hence, option 4.

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**IIFT 2013 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Two alloys of aluminium have different percentages of aluminium in them. The first one weighs 8 kg and the second one weighs 16 kg. One piece each of equal weight was cut off from both the alloys and first piece was alloyed with the second alloy and the second piece alloyed with the first one. As a result, the percentage of aluminium became the same in the resulting two new alloys. What was the weight of each cut-off piece?

- A.
3.33 kg

- B.
4.67 kg

- C.
5.33 kg

- D.
None of the above

Answer: Option C

**Explanation** :

Let weight of the cut-off piece = X kg

Let percentage of aluminium in 8 kg and 16 kg alloy be a and b respectively.

∴ $\frac{(8-X)a+Xb}{8\times 100}$ = $\frac{(16-X)b+Xa}{16\times 100}$

∴ 16a – 2Xa + 2Xb = 16b – Xb + Xa

∴ 16 – 2X = X and 2X = 16 – X

∴ X = $\frac{16}{3}$ = 5.33 kg

Hence, option 3.

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**XAT 2012 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Gopal sells fruit juice mixture using orange juice and pineapple juice. Gopal prepares this mixture by drawing out a jug of orange juice from a 10 litre container filled with orange juice, and replacing it with pineapple juice. If Gopal draws out another jug of the resultant mixture and replaces it with pineapple juice, the container will have equal volumes of orange juice and pineapple juice. The volume of the jug in litres, is

- A.
2

- B.
>2 and ≤2.5

- C.
2.5

- D.
>2.5 and ≤3

- E.
≥3

Answer: Option D

**Explanation** :

Assume that the volume of the jug is l liters.

Hence, after first replacement, the juice mixture contains l liters of pineapple juice.

When the juice mixture is drawn out for the second time using the jug, the amount of pineapple juice in the jug = l × ( l / 10)

This is replaced by l litres of pineapple juice.

Hence, amount of pineapple juice after two replacements = l + ( l – l × ( l / 10)) = 5

Hence, we get,

l^{2} – 20l + 50 = 0

Solving the above quadratic we get,

l = 5(2 - √2) or l = 5(2 + √2)

As, 5(2 + √2) > 10, l = 5(2 - √2) ≈ 2.92

Hence, option 4.

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**XAT 2012 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A medical practitioner has created different potencies of a commonly used medicine by dissolving tables in water and using the resultant solution.

*Potency 1 solution: When 1 tablet is dissolved in 50 ml, the entire 50 ml is equivalent to one dose.Potency 2 solution: When 2 tablets are dissolved in 50 ml, the entire 50 ml of this solution is equivalent to 2 doses,… and so on.*

This way he can give fractions of tablets based on the intensity of infection and the age of the patient.

For particular patient, he administers 10 ml of potency 1, 15 ml of potency 2 and 30 ml of potency 4. The dosage administered to the patient is equivalent to

- A.
> 2 and ≤ 3 tablets

- B.
> 3 and ≤ 3.25 tablets

- C.
> 3.25 and ≤ 3.5 tablets

- D.
> 3.5 and ≤ 3.75 tablets

- E.
> 3.75 and ≤ 4 tablets

Answer: Option B

**Explanation** :

50 ml of potency 1 solution is equivalent to 1 tablet; 50 ml of potency 2 solution is equivalent to 2 tablets and so on.

Hence, 10 ml of potency 1 solution is equivalent to 10/50 = 1/5 tablet.

Similarly, 15 ml of potency 2 and 30 ml of potency 4 corresponds to 15/50 × 2 and 30/50 × 4 tablets respectively.

Hence, the dosage administered is equivalent to

1/5 + 3/5 + 12/5 = 16/5 = 3.2 tablets

Hence, option 2.

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**XAT 2012 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

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

Answer: Option B

**Explanation** :

Let Ram replace x litres of 12% solution with 39% solution.

Hence, amount of alcohol in new solution = (27 – x) × 0.12 + x × 0.39 = 27 × 0.12 + x × 0.27

Now, new concentration of the solution is 21%, hence, volume of alcohol = 27 × .21

Hence, 27 × 0.12 + x × 0.27 = 27 × 0.21

Hence, 0.12 + x/100 = 0.21

Hence, x = 9.

Hence, option 2.

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**IIFT 2012 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A 10 litre cylinder contains a mixture of water and sugar, the volume of sugar being 15% of total volume. A few litres of the mixture is released and an equal amount of water is added. Then the same amount of the mixture as before is released and replaced with water for a second time. As a result, the sugar content becomes 10% of total volume. What is the approximate quantity of mixture released each time?

- A.
1 litres

- B.
1.2 litres

- C.
1.5 litres

- D.
2 litres

Answer: Option D

**Explanation** :

The amount of sugar in the cylinder = 1.5 litres

Now, 1.5$\left\{1-{\left(\frac{a}{10}\right)}^{2}\right\}$ is the amount of sugar

left after a litres has been replaced twice.

Now after replacement, sugar left is 10% of the total solution i.e.10% of 10 litres = 1 litre

∴ 1 = 1.5$\left\{1-{\left(\frac{a}{10}\right)}^{2}\right\}$

∴* *0.66 = $\left\{1-{\left(\frac{a}{10}\right)}^{2}\right\}$

$\sqrt{0.66}$ ≈ 0.8

∴ 0.8 = 1 - $\left(\frac{a}{10}\right)$

∴ $\left(\frac{a}{10}\right)$ = 0.2

∴ a = 2

Hence, option 4.

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**IIFT 2011 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

The ratio of ‘metal 1’ and ‘metal 2’ in Alloy ‘A’ is 3:4. In Alloy ‘B’ same metals are mixed in the ratio 5:8. If 26 kg of Alloy ‘B’ and 14 kg of Alloy ‘A’ are mixed then find out the ratio of ‘metal 1’ and ‘metal 2’ in the new Alloy.

- A.
3 : 2

- B.
2 : 5

- C.
2 : 3

- D.
None of the above

Answer: Option C

**Explanation** :

Ratio of metal 1 and metal 2 in alloy A = 3 : 4

In 14 kg, there will be 3/7 × 14 = 6 kg of metal 1 and 4/7 × 14 = 8 kg of metal 2.

Similarly, alloy B contains 10 kg and 16 kg of metal 1 and metal 2 respectively.

∴ Metal 1 content in final mixture = 6 + 10 = 16 kg

Metal 2 content in final mixture = 8 + 16 = 24 kg

∴ Required ratio = 16 : 24 ≡ 2 : 3

Hence, option 3.

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**IIFT 2009 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A petrol tank at a filling station has a capacity of 400 litres. The attendant sells 40 litres of petrol from the tank to one customer and then replenishes it with kerosene oil. This process is repeated with six customers. What quantity of pure petrol will the seventh customer get when he purchases 40 litres of petrol?

- A.
20.50 litres

- B.
21.25 litres

- C.
24.75 litres

- D.
22.40 litres

Answer: Option B

**Explanation** :

$\frac{\mathrm{Quantity}\mathrm{of}\mathrm{petrol}\mathrm{remaining}\mathrm{after}{\mathrm{n}}^{\mathrm{th}}\mathrm{replacement}}{\mathrm{Total}\mathrm{Quantity}}$ = ${\left(\frac{x-y}{x}\right)}^{n}$

where x is the original quantity,

y is the quantity that is replaced, and

n is the number of times the replacement process is repeated.

∴ At the end of 6 replacements, each litre contains

${\left(\frac{400-40}{400}\right)}^{6}$ = ${\left(\frac{360}{400}\right)}^{6}$ = (0.9)^{6} ≈ 0.53 litres of petrol

∴ The customer who purchases 40 litres of petrol gets 40 × 0.53 ≈ 21.2 litres of pure petrol.

Hence, option 2.

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