Arithmetic - Mixture, Alligation, Removal & Replacement - Previous Year CAT/MBA Questions
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A small jar contained water, lime and sugar in the ratio of 90 : 7 : 3. A glass contained only water and sugar in it. Contents of both (small jar and glass) were mixed in a bigger jar and the ratio of contents in the bigger jar was 85 : 5 : 10 (water, lime and sugar respectively). Find the percentage of water in the bigger jar?
- (a)
70
- (b)
75
- (c)
80
- (d)
72.5
- (e)
85
Answer: Option E
Text Explanation :
Let the quantity of water, line and sugar in the bigger jar be 85x, 5x and 10x respectively.
Now, percentage of water in bigger jar = = 85%
Hence, option (e).
Workspace:
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
Text 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 2/3rd of the total mixture.
⇒ Total volume of the mixture = 3/2 × 8x = 12x ml and volume of water = 1/3rd 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).
Workspace:
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
Text 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 (b).
Workspace:
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
Text 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,
l2 – 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 (d).
Workspace:
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
Text 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 (b).
Workspace:
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
Text 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 (b).
Workspace:
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