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- Question
There are two mixtures of honey and water in which the ratio of honey and water are as 1:3 and 3:1 respectively. Two litres are drawn from first mixture and 3 litres from second mixture, are mixed to form another mixture. What is the ratio of honey and water in it ?
Options- A. 111:108
- B. 11:9
- C. 103:72
- D. None
- Correct Answer
- 11:9
ExplanationFrom the given data,
The part of honey in the first mixture = 1/4
The part of honey in the second mixture = 3/4
Let the part of honey in the third mixture = x
Then,
1/4 3/4
x
(3/4)-x x-(1/4)
Given from mixtures 1 & 2 the ratio of mixture taken out is 2 : 3
=>
=> Solving we get the part of honey in the third mixture as 11/20
=> the remaining part of the mixture is water = 9/20
Hence, the ratio of the mixture of honey and water in the third mixture is 11 : 9 .
- 1. A milkman has 20 liters of milk. If he mixes 5 liters of water, which is freely available, in 20 liters of pure milk. If the cost of pure milk is Rs. 18 per litre, then the profit of the milkman, when he sells all the mixture at cost price, the profit % ?
Options- A. 22.5%
- B. 25%
- C. 20%
- D. 33.33% Discuss
- 2. In what ratio must wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg so that the mixture be worth Rs. 3/kg?
Options- A. 1:2
- B. 3:2
- C. 2:1
- D. 2:3 Discuss
- 3. An alloy of copper and bronze weight 50g. It contains 80% Copper. How much copper should be added to the alloy so that percentage of copper is increased to 90%?
Options- A. 45 gm
- B. 50 gm
- C. 55 gm
- D. 60 gm Discuss
- 4. The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?
Options- A. 8:3
- B. 6:7
- C. 7:5
- D. 11:7 Discuss
- 5. Equal quantities of three mixtures of milk and water are mixed in the ratio 1:2, 2:3 and 3:4. The ratio of water and milk in the mixture is ?
Options- A. 193 : 122
- B. 97 : 102
- C. 115 : 201
- D. 147 : 185 Discuss
- 6. From a tank of petrol , which contains 200 liters of petrol, the seller replaces each time with kerosene when he sells 40 liters of petrol(or mixture). Everytime he sells out only 40 liters of petrol (pure or impure). After replacing the petrol with kerosene 4th time, the total amount of kerosene in the mixture is
Options- A. 81.92L
- B. 96L
- C. 118.08L
- D. None of these Discuss
- 7. In what ratio must a merchant mix two varieties of oils worth Rs. 60/kg and Rs. 65/kg, so that by selling the mixture at Rs. 68.20/kg, he may gain 10% ?
Options- A. 2:3
- B. 4:3
- C. 3:4
- D. 3:2 Discuss
- 8. From the 50 liters of milk, 5 liters of milk is taken out and after it 5 liters of water is added to the rest amount of milk. Again 5 liters of milk and water is drawn out and it was replaced by 5 liters of water. If this process is continued similarly for the third time, the amount of milk left after the third replacement:
Options- A. 45L
- B. 36.45L
- C. 40.5L
- D. 42.5L Discuss
- 9. In a mixture of 240 lt. water is 20% and rest is Milk. What quantity of mixture should be taken out and replaced with water so that water becomes 40%?
Options- A. 60 lit
- B. 55 lit
- C. 45 lit
- D. 50 lit Discuss
- 10. A container contains 120 lit of Diesel. From this container, 12 lit of Diesel was taken out and replaced by kerosene. This process was further repeated for two times. How much diesel is now there in the container ?
Options- A. 88.01 lit
- B. 87.48 lit
- C. 87.51 lit
- D. 87.62 lit Discuss
Alligation or Mixture problems
Search Results
Correct Answer: 25%
Explanation:
As water costs free, water sold at cost price of milk gives the profit.
Required profit % = 5/20 x 100 = 5 x 5 = 25%.
Correct Answer: 2:1
Explanation:
Given rate of wheat at cheap = Rs. 2.90/kg
Rate of wheat at cost = Rs. 3.20/kg
Mixture rate = Rs. 3/kg
Ratio of mixture =
2.90 3.20
3
(3.20 - 3 = 0.20) (3 - 2.90 = 0.10)
0.20 : 0.10 = 2:1
Hence, wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg in the ratio of 2:1, so that the mixture be worth Rs. 3/kg.
Correct Answer: 50 gm
Explanation:
Initial quantity of copper = = 40 g
And that of Bronze = 50 - 40 = 10 g
Let 'p' gm of copper is added to the mixture
=> = 40 + p
=> 45 + 0.9p = 40 + p
=> p = 50 g
Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.
Correct Answer: 7:5
Explanation:
Milk in 1-litre mixture of A = 4/7 litre.
Milk in 1-litre mixture of B = 2/5 litre.
Milk in 1-litre mixture of C = 1/2 litre.
By rule of alligation we have required ratio X:Y
X : Y
4/7 2/5
\ /
(Mean ratio)
(1/2)
/ \
(1/2 ? 2/5) : (4/7 ? 1/2)
1/10 1/1 4
So Required ratio = X : Y = 1/10 : 1/14 = 7:5
Correct Answer: 193 : 122
Explanation:
Given the three mixtures ratio as (1:2),(2:3),(3:4)
(1+2),(2+3),(3+4)
Total content = 3,5,7
Given equal quantities of the three mixtures are mixed, then LCM of 3, 5, 7 = 105
105/3 = 35 , 105/5 = 21 , 105/7 = 15
Now, the individual equal quantity ratios are (35x1, 35x2), (21x2, 21x3), (15x3, 15x4)
i.e (35,70), (42,63), (45,60)
So overall mixture ratio of milk and water is
35+42+45 : 70+63+60
122:193
But in the question asked the ratio of water to milk = 193 : 122
Correct Answer: 118.08L
Explanation:
The amount of petrol left after 4 operations = 81.92
Hence the amount of kerosene = 200 - 81.92 = 118.08 liters
Correct Answer: 3:2
Explanation:
Let he mixes the oils in the ratio = x : y
Then, the cost price of the oils = 60x + 65y
Given selling price = Rs. 68.20
=> Selling price = 68.20(x+y)
Given profit = 10% = SP - CP
=> 10/100 (60x + 65y) = 68.20(x+y)-(60x + 65y)
=> 6x + 6.5y = 8.20x + 3.20y
=>2.2x = 3.3y
=> x : y = 3 : 2
Correct Answer: 36.45L
Explanation:
General Formula:
Final or reduced concentration = initial concentration x
where n is the number of times the same operation is being repeated. The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture. Here amount being replaced denotes the quantity which is to be withdrawn in each time.
Therefore,
= 36.45 L
Correct Answer: 60 lit
Correct Answer: 87.48 lit
Explanation:
For these type of problems,
Quantity of Diesel remained =
Here p = 12 , q = 120
=> = 120 x 0.9 x 0.9 x 0.9 = 87.48 lit.
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