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
Answer: D)
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
Answer: A)
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
Answer: B)
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
Answer: D)
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
Answer: B)
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
Answer: D)
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
Answer: C)
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
Answer: E)
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
Answer: E)
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
Answer: A)
Explanation: Quantity of milk in 729 litres of
mixture,
= 7/9 × 729 = 567 litres
⇒ Quantity of water = (729 – 567) litres = 162 litres
Let x litres of water is
mixed to get the required ratio of 7 : 3
∴ 567/(162 + x) = 7/3 => 7x = 567 => x = 81 liters
