CAT — Mixture & Alligations
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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
1 : 3
1 : 2
5 : 9
4 : 9
5 : 9
Let the initial quantity of coffee in the jar be 1 unit.
Suppose a fraction x of the mixture is removed and replaced each time.
After the first replacement (Mixture P)
Coffee remaining = 1 − x
Cocoa added = x
So, Coffee : Cocoa = (1 − x) : x
After the second replacement (Mixture Q)
Initially in mixture P, Coffee = 1 − x, Cocoa = x
A fraction x of the mixture is removed.
Coffee removed = x(1 − x)
Coffee remaining = (1 − x) − x(1 − x) = (1 − x)²
Hence, coffee in mixture Q is (1 − x)²
Given, Coffee : Cocoa = 16 : 9
Therefore, Coffee fraction = 16/(16 + 9) = 16/25
So, (1 − x)² = 16/25 → 1 − x = 4/5 → x = 1/5
Cocoa in Mixture P = x = 1/5
Cocoa in Mixture Q = 1 − Coffee = 1 − 16/25 = 9/25
Required ratio = (1/5) : (9/25) = (5/25) : (9/25) = 5 : 9
Answer:
C. 5 : 9
A container holds 200 litres of a solution of acid and water, having 30% acid by volume. Atul replaces 20% of this solution with water, then replaces 10% of the resulting solution with acid, and finally replaces 15% of the solution thus obtained, with water. The percentage of acid by volume in the final solution obtained after these three replacements, is nearest to
23
25
29
27
27
Initial acid = 60 litres.
Step 1 (replace 20% with water): Acid removed = 30% of 40 = 12 L. Acid left = 48 L.
Step 2 (replace 10% with acid): Acid percentage = 48/200 = 24%. Acid removed = 24% of 20 = 4.8 L. Acid left = 43.2 L. Add 20 L pure acid → acid = 63.2 L.
Step 3 (replace 15% with water): Acid percentage = 63.2/200 = 31.6%. Acid removed = 31.6% of 30 = 9.48 L. Acid left = 53.72 L.
Final acid % = (53.72/200) × 100 = 26.86% ≈ 27%.
− A certain amount of water was poured into a 300 litre container and the remaining portion of the container was filled with milk. Then an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains 72% milk, then the amount of water, in litres, that was initially poured into the container was
Given:
A 300-litre container is partially filled with water.
The remaining part is filled with milk.
Then, a quantity of the mixture equal to twice the initial amount of water is removed and replaced with water.
The final solution contains 72% milk.
Find the amount of water initially poured.
Step 1: Let the initial amount of water be x litres.
Then,
Initial milk = 300 − x litres.
Step 2: Amount of mixture removed
The amount removed is
2x litres.
Since the mixture is uniform,
Milk removed
= (300 − x)/300 × 2x
Step 3: Milk remaining
Milk remaining
= (300 − x) − (300 − x)/300 × 2x
After adding water, only the amount of water changes.
Since the final solution contains 72% milk,
Milk remaining
= 72% of 300
= 216 litres.
Hence,
(300 − x) − (300 − x)/300 × 2x = 216
Step 4: Solve the equation
Factor out (300 − x),
(300 − x)(1 − 2x/300) = 216
(300 − x)(300 − 2x) = 64800
(300 − x)(150 − x) = 32400
Expanding,
45000 − 450x + x² = 32400
x² − 450x + 12600 = 0
(x − 30)(x − 420) = 0
Since x cannot exceed 300,
x = 30
Final Answer
The amount of water initially poured into the container was
30 litres
A glass is filled with milk. Two-thirds of its content is poured out and replaced with water. If this process of pouring out two-thirds the content and replacing with water is repeated three more times, then the final ratio of milk to water in the glass, is
1 : 27
1 : 80
1 : 81
1 : 26
1 : 80
Step 1: Find the fraction of milk left after each operation
Since two-thirds is removed, one-third of the milk remains after each operation.
After 1st operation: Milk = 1/3
After 2nd operation: Milk = (1/3)² = 1/9
After 3rd operation: Milk = (1/3)³ = 1/27
After 4th operation: Milk = (1/3)⁴ = 1/81
Step 2: Find the fraction of water
Water = 1 − 1/81 = 80/81
Step 3: Find the ratio
Milk : Water = (1/81) : (80/81) = 1 : 80
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
After n replacements, milk remaining = 40 × (9/10)ⁿ We need milk < water → milk < 20 → (9/10)ⁿ < 1/2 n = 6: (0.9)⁶ = 0.531441 > 0.5 → Milk ≈ 21.26 L > Water. Not yet. n = 7: (0.9)⁷ = 0.4782969 < 0.5 → Milk ≈ 19.13 L < Water ≈ 20.87 L ✓ Smallest required number of replacements = 7
A mixture of coffee and cocoa, 16% of which is coffee, costs Rs 240 per kg. Another mixture of coffee and cocoa, of which 36% is coffee, costs Rs 320 per kg. If a new mixture of coffee and cocoa costs Rs 376 per kg, then the quantity, in kg, of coffee in 10 kg of this new mixture is
5
4
2.5
6
5
Step 1: Let the price of pure coffee be C Rs/kg and pure cocoa be K Rs/kg. From the first mixture: 0.16C + 0.84K = 240 → 16C + 84K = 24000 → 4C + 21K = 6000. From the second mixture: 0.36C + 0.64K = 320 → 36C + 64K = 32000 → 9C + 16K = 8000.
Step 2: Find the prices of pure coffee and cocoa. Multiply first equation by 9: 36C + 189K = 54000. Multiply second equation by 4: 36C + 64K = 32000. Subtract: 125K = 22000 → K = 176. Substitute into 9C + 16K = 8000: 9C + 2816 = 8000 → 9C = 5184 → C = 576.
Step 3: Let the percentage of coffee in the new mixture be x. Since the new mixture costs Rs. 376 per kg: 576x + 176(1 − x) = 376 → 576x + 176 − 176x = 376 → 400x = 200 → x = 0.5. Thus, the new mixture contains 50% coffee.
Step 4: Find the quantity of coffee in 10 kg. Coffee = 50% of 10 kg = 5 kg.
− Vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively. A certain volume is taken out from A and poured into B. After stirring, the same volume is taken out from B and poured into A. If the resultant ratio of alcohol and water in A is 15 : 4, then the volume, in litres, initially taken out from A is
Step 1: Let x litres be transferred from A to
B.
After transfer: Alcohol in A = 60−x; Vessel B has x alcohol + 60 water (total 60+x).
Step 2: Transfer x litres back from B to
A.
Alcohol returned = x²/(60+x)
Water returned = 60x/(60+x)
Step 3: Final contents of vessel A
Alcohol in A = (60−x) + x²/(60+x) = 3600/(60+x)
Water in A = 60x/(60+x)
Step 4: Ratio condition
3600/(60+x) : 60x/(60+x) = 15 : 4
3600 : 60x = 15 : 4
3600 × 4 = 60x × 15 → 14400 = 900x → x = 16
A vessel contained a certain amount of a solution of acid and water. When 2 litres of water was added to it, the new solution had 50% acid concentration. When 15 litres of acid was further added to this new solution, the final solution had 80% acid concentration. The ratio of water and acid in the original solution was
5 : 3
3 : 5
5 : 4
4 : 5
3 : 5
Given:
A vessel contains a solution of acid and water.
● After adding 2 litres of water, the solution becomes 50% acid.
● Then, 15 litres of acid is added, making the solution 80% acid.
Find the ratio of water to acid in the original solution.
Step 1: Let the original quantities be
Acid = A litres
Water = W litres
Step 2: Use the first condition
After adding 2 litres of water,
Acid = A
Water = W + 2
Since the solution is 50% acid,
A = W + 2
Step 3: Use the second condition
After adding 15 litres of acid,
Acid = A + 15
Water = W + 2
The acid concentration becomes 80%.
Therefore,
(A + 15)/(A + W + 17) = 4/5
Cross-multiplying,
5(A + 15) = 4(A + W + 17)
5A + 75 = 4A + 4W + 68
A = 4W − 7
Step 4: Solve the equations
From Step 2,
A = W + 2
Substitute into
A = 4W − 7
W + 2 = 4W − 7
3W = 9
W = 3
Hence,
A = 5
Step 5: Find the required ratio
Water : Acid
= 3 : 5
Answer:
B. 3 : 5
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