# Quantitative Aptitude: Mixture and Alligation Questions Set 8

Mixture and Alligation Questions for NICL, SBI PO, IBPS PO, NIACL, LIC, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams

1. A chemist has 10L of a solution that is 10% nitric acid by volume. He wants to dilute the solution to 4% strength by adding water. How many litres of water must be add?
A) 40L
B) 33 L
C) 25L
D) 15 L
E) 20L
Option D
Solution:
Quantity of nitric acid = 10 *(1/10) = 1 L
Water = 10 – 1 = 9 L
Let x litre of water be added,
(10 + x ) * (4/100) = 1
=> x = 15 L
2. A bottle contains (3/4) of milk and the rest water. How much of the mixture must be taken away and replaced by an equal quantity of water so that the nixtude has half milk and half water?
A) 42(1/4)%
B) 33(1/3)%
C)22(1/3)%
D) 18(1/2)%
E) 21(1/2)%
Option B
Solution:
Ratio of milk : water = 3 : 1
water = (1/4)*100 = 25
Let x L is taken out , then
qty. of milk left= (3 – 3x/4)
water left = (1 – x/4) + x
Now , 3 – 3x/4 = (1 – x/4) + x
=> x = 4/3
Required % = 4/(3*4)*100 = 33(1/3)%

3. P and Q are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 : 11 resp. If equal quantities of the alloys are melted to form a third alloy R, Find the ratio of gold and copper.
A) 6 : 7
B) 7 : 5
C) 4 : 3
D) 5 : 6
E) 3 : 2
Option B
Solution:
In 1 kg of alloy P, Gold = 7/9
Copper = 2/9
In 1 kg of alloy Q, Gold = 7/18
Copper = 11/18
Therefore, Ratio of Gold and Copper in alloy R
= 7/9 + 7/18 : 2/9 + 11/18
= 21 : 15 = 7 : 5

4. A container has 30 L of water. If 3 L of water is replaced by 3 L of spirit and this operation is repeated twice , what will be the quantity of water in the new mixture?
A) 27.1 L
B) 25.5 L
C) 14.4 L
D) 24.3 L
E) 22 L
Option D
Solution:
Suppose a container contains x units of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid.
= x(1 – y/x)^n units
= Remaining water = 30(1 – 3/30)^2 = 24.3 L
5. Two barrels contain a mixture of ethanol and gasoline. The content of the ethanol is 60% in the first barrel and 30% in the second barrel. In what ratio must the mixtures from the first and the second barrels be taken to form a mixture containing 50% ethanol?
A) 2 : 1
B) 2 : 5
C) 1 : 3
D) 3 : 2
E) 4 : 5
Option A
Solution:
Mixture I ————– Mixture II
Ethanol – (3/5)———Ethanol- (3/10)
———————(1/2)
(1/5)—————————(1/10)
= 2 : 1
6. A solution of sugar syrup has 15% sugar. Another solution has 5% sugar. How many litres of the second solution must be added to 20 L of the first solution to make a solution of 20% sugar.
A) 60 L
B) 45 L
C) 50 L
D) 30 L
E) 20 L
Option E
Solution:
Let x L of second solution must be added.
Then, [15*20 + 5*x]/(20 + x) = 10
=> x = 20 L
7. A person has a chemical of Rs. 25 per litre. In what ratio should water be mixed in that chemical, so that after selling the mixture at Rs. 20 per litre he may get a profit of 25%?
A) 9 : 15
B) 10 : 13
C) 16 : 9
D) 15 : 22
E) 21 : 17
Option C
Solution:
Selling price of mixture = Rs. 20
Cost price of mixture = (100/125)*20 = Rs.16
Rule of mixture
25————-0
——–16
16————-9
So, the required ratio = 16 : 9
8. Three containers X, Y and Z are having mixtures of milk and water in the ratio 1 : 5 , 3 : 5 and 5 : 7 resp. If the capacities of the containers are in the ratio 5 : 4 : 5, then find the ratio of the milk to the water, if the mixtures of all the three containers are mixed together.
A) 44 : 119
B) 24 : 111
C) 46 : 143
D) 53 : 115
E) 55 : 157
Option D
Solution:
Ratio of milk and water
= [(1/6)*5 + (3/8)*4 + (5/12)*5] : [(5/6)*5 + (5/8)*4 + (7/12)*5] = 53 : 115
9. How many kg of sugar costing Rs. 5.75 per kg should be mixed with 75 kg of cheaper sugar costing Rs. 4.50 per kg so that the mixture is worth Rs. 5.50 per kg ?
A) 440 kg
B) 300 kg
C) 112 kg
D) 225 kg
E) 320 kg
Option B
Solution:
Sugar I —————- Sugar II
5.75 ————————-4.50
—————–5.50
1——————————0.25
ratio = 4 : 1
The required qty. of sugar I = (75/1)*4 = 300 kg
10. One test tube contains some acid and another test tube contains an equal quantity of water. To prepare a solution, 20 g of the acid is poured into the second test tube. Then, two-thirds of the so formed solution is poured from the second test tube into the first. If the fluid in the first test tube is four times that in second, what quantity of water was taken initially.
A) 150 g
B) 120 g
C) 90 g
D) 100 g
E) 150 g
Option D
Solution:
Initially, let x g of water and Acid was
taken. Initially 1st process
First test tube = (x - 20) g
Second test tube = (x + 20) g
2nd process
First test tube = (x - 20) + (x + 20)*(2/3)
Second test tube = (x + 20)*(1/3)
Now,
(x -20) + (2/3)(x +20) = 4*(1/3)(x + 20)
=> x = 100 g