## Allegations and Mixtures Practice Questions – Set 4

Allegations and Mixtures Practice Questions – Set 4
1. 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?
a) 106 : 69
b) 103 : 72
c) 89 : 86
d) 11 : 9
e) None of these
Explanation: The percentage of honey in the first mixture = ¼ × 100 = 25%
The percentage of honey in the second mixture = ¾ × 100 = 75%
Given that, 2 litres are drawn from first mixture and 3 litres from second mixture, then
Part of honey in the new mixture = (2 × 25 + 3 × 75)/5 = 275/5 = 55%
Then the ratio of honey and water in the new mixture would be 55 : (100 – 55) = 55 : 45 = 11 : 9

2. Jagtap purchases 30 kg of wheat at the rate of 11.50 per kg and 20 kg of wheat at the rate of Rs. 14.25 per kg. He mixed the two and sold the mixture. Approximately at what price per kg should he sell the mixture to make 30 per cent profit?
a) Rs.16.38
b) Rs.18.20
c) Rs.15.60
d) Rs.14.80
e) None of these
Explanation: Given, Jagtap purchases 30 kg of wheat at the rate of 11.50 per kg and 20 kg of wheat at the rate of Rs. 14.25 per kg.
Total cost of the mixture = 30 × 11.50 + 20 × 14.25
Total cost of the mixture = Rs. 630
Total kg of mixture = 30 + 20 = 50 kg
Cost of mixture per kg = 630/50 = Rs 12.6 per kg
Given, he sells the mixture to make 30 per cent profit.
Selling price of mixture per kg = 12.6 + 30% of 12.6
Selling price of mixture per kg = RS. 16.38

3. A and B are two alloys of bronze and platinum prepared by mixing metals in the ratio of 4 : 3 and 9 : 5, respectively. If equal quantities of alloys A and B are melted to form a third alloy C, then find the ratio of bronze and platinum in C.
a) 19 : 25
b) 17 : 11
c) 9 : 7
d) 22 : 13
e) 12 : 29
Explanation: Let x be the quantity of each melted alloy A and B.
Quantity of bronze in A = 4x/7
Quantity of platinum in A = 3x/7
Quantity of bronze in B = 9x/14
Quantity of platinum in B = 5x/14
Total quantity of bronze in alloy C = 4x/7 + 9x/14 = 17x/14
Total quantity of platinum in alloy C = 3x/7 + 5x/14 = 11x/14
Requried ratio = 17x/14 : 11x/14 = 17: 11

4. A liuid P is 2 ½  times as heavy as water and water is 1 3/5 times as heavy as another liquid ‘Q’. The amount of liquid ‘P’ that must be added to 9 litres of the liquid ‘Q’ so that the mixture may weigh as much as an equal volume of water, will be
a) 7 litres
b) 5 1/6 litres
c) 5 litres
d) 2 ¼ litres
e) 3 litres
Explanation: Let the weight of liquid Q be w kg.
Thus, weight of water = 1 3/5 times as heavy as liquid ‘Q’ = 8w/5
Weight of liquid P = 2 1/2 times as heavy as water = 5/2 × 8w/5=4w
Let the amount of liquid P to be added to 9 litres of Q be x litres.
Volume of mixture of P & Q = (9 + x) litres
Now, according to condition given in the problem, the weight of the mixture should be equal to the weight of equal amount of water.
(4w × x) + (w × 9) = (8w/5) × (9 + x)
4x + 9 = 8/5 × (9+x)
5 × (4x + 9) = 8 × (9 + x)
20x + 45 = 72 + 8x
20x 8x = 72 45
12x = 27

5. In what proportion must water be mixed with milk to gain 33 1/3 per cent by selling the mixture at the cost price?
a) 2 : 3
b) 1 : 3
c) 1 : 4
d) 3 : 4
e) None of these
Explanation: Let C.P of 1 litre milk = Rs 1
So, S.P of 1 litre mixture = CP of 1L milk = Rs 1
Gain = 33 1/3% = 100/3%
CP of 1L mix = 100/(100 + gain%) × SP
= [100/(100 + 100/3)] × 1 = 100/(400/3) = 3/4
By rule of allegation
CP of 1L water (0)                                                   CP of 1L milk (1)
CP of 1L mixture (3/4)
1 – (3/4) = 1/4                                                         (3/4) – 0 = ¾
Quantity of water : quantity of milk = ¼ : ¾ = 1:3

6. A vessel is full of a mixture of spirit and water in which there is found to be 17% of spirit by measure. Ten litres are drawn off and the vessel is filled up with water. The proportion of spirit is now found to be 15 1/9%. How much does the vessel hold?
a) 70 litres
b) 75 litres
c) 80 litres
d) 90 litres
e) 100 litres
Explanation: Let the initial quantity of mixture be 100x litres
Quantity of spirit in original mixture = 17x litres
Quantity of water in original mixture = 83x litres
10 litres of mixture is drawn off and replaced by 10 litres of water.
Thus, quantity of spirit in new mixture = 17x - (17% of 10 litres) = 17x - 1.7 litres
Quantity of water in new mixture = 83x - (83% of 10 litres) + 10 litres = 83x - 8.3 + 10 litres
= 83x - 1.7 litres
Given: proportion of spirit in new mixture = 15 1/9% = 136/9%
Proportion of water in new mixture = 764/9%
Thus, [(17x – 1.7)/(83x – 1.7)]litres = [(136/9)/(764/9)]
On solving, we get
1700x = 1530
x = 1530/1700 = 0.9 litres
Thus, the initial quantity of mixture will be 100x litres = 100 × 0.9 litres = 90 litres.

7. How much water must be added to a cask containing 40 1/2 litres of spirit worth Rs. 3.92 a litre to reduce the price to Rs. 3.24 a litre?
a) 9 1/3 litre
b) 9 litres
c) 8 ½ litres
d) 8 litres
e) None of these
Explanation: Quantity of sprit = 40 1/2 litres
Price of sprit = Rs.3.92 per litre (dearer)
Price of water = Rs. 0 (cheaper)
Now if we mix spirit and water, we have to mix in such a proportion that mixture would cost Rs.3.24 per litre
Consider the price at which mixture to be made be mean price i.e. 3.24 per litre.
Proportion in terms of quantity must be given as
Quantity of cheaper/Quantity of dearer = (dearer price – mean price)/(mean price – cheaper price)
=> Quantity of cheaper/(40 1/3) L = (3.92 – 3.24)/3.24 = 0.68/3.24 = 68/324
Quantity of cheaper = 68/324 × 81/2 = 17/2 = 8 ½ L

8. In a mixture of milk and water the proportion of water by weight was 75%. If in the 60 gms mixture 15 gms. Water was added, what would be the percentage of water in the new mixture?
a) 75%
b) 88%
c) 90%
d) 100%
e) None of these
Explanation: Given in the question, in the mixture of milk and water the proportion of water by weight was 75%.
Initially we have taken 60 gram mixture of milk & water.
Amount of water in this mixture = 75% of 60 gram = 45 gram
Amount of milk in the mixture = (60 45) gram = 15 gram
Externally we have added 15 gram water.
New amount of water = (45 + 15) gram = 60 gram
Now, new volume of mixture = (60 + 15) gram = 75 gram
[Due to addition of 15 gram water]
Percentage of water in new mixture = (Amount of water)/(Amount of Mixture) × 100%
Percentage of water in new mixture = 60/75 × 100% = 80%

9. 80% milk solution is mixed with 30% milk solution to produce 45 liter of 50% milk solution. If the price of 45 liter of 50% milk solution is Rs.1,080 and the price of 30% milk solution is Rs. 20 per liter, then what is the price of 80% milk solution in the final mixture?
a) Rs.640
b) Rs.650
c) Rs.400
d) Rs.250
e) Rs.540
Explanation: Say there is X liters of 30% milk solution and remaining (45 – X) liters of 80% milk.
Amount of milk in the solution = 0.30X + 0.80(45 – X)
Amount of milk in 45 liters of 50% solution = 0.50 × 45 = 22.5
0.30X + 36 – 0.8X = 22.5
0.5X = 13.5
X = 27 liters
Volume of 80% milk in the mixture = 45 – X = 18 liters.
Price of 30% milk at Rs.20 per liter = 20 × 27 = Rs.540
Price of 80% milk = 1080 – 540 = Rs.540

10. In a 729 litres mixture of milk and water, the ratio of milk to water is 7 : 2. To get a new mixture containing milk and water in the ratio 7 : 3, the amount of water to be added is–
a) 81 litres
b) 71 litres
c) 56 litres
d) 50 litres
e) None of these