__Question No. 01__**A cup contains milk and water in the ratio 3:1. How much mixture should be taken out and water added to make the ratio 1:1?**

(A) 1/3 part
of mixture

(B) 1/2 part
of mixture

(C) 1/4 part
of mixture

(D) 1/5 part
of mixture

Answer:
Option A

__Explanation:__
Let

*y*part of mixture is replaced by water
Therefore,
(4

*x*-*y*)/4*x*= 2*x*/3*x*
=> 12

*x*² - 3*xy*= 8*x*²
=>

*y*= 4*x*/3
I.e. 1/3
part of mixture.

__Question No. 02__**A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 liters of milk such that the ratio of water to milk is 3:5?**

(A) 4
liters, 8 liters

(B) 6
liters, 6 liters

(C) 5
liters, 7 liters

(D) 7
liters, 5 liters

Answer:
Option B

__Explanation:__
Let the cost
of 1 liter milk be Rs. 1

Milk in 1
liter mix in 1st can = 3/4 liter, C.P of 1 liter mix in 1st can Rs. 3/4

Milk in 1
liter mix in 2nd can = 1/2 liter, C.P of 1 liter mix in 2nd can Rs. 1/2

Milk in 1
liter of final mix = 5/8 liter, Mean price = Rs. 5/8

By the rule
of alligation, we have:
∴ Ratio
of two mixtures = (1/8) : (1/8) = 1 : 1

So, quantity
of mixture taken from each can = (½ × 12) = 6 liters

__Question No. 03__**In 2 gallons mixture of spirit and water, the percentage of water is 12. In another 3 gallons mixture of spirit and water, the percentage of water is 5. These two mixtures are poured in a third pot and ½ gallon more water is added to it. Find the percentage of water in this new mixture.**

(A) 17

^{1}*/*_{11}%
(B) 17

^{3}*/*_{11}%
(C) 16

^{2}*/*_{11}%
(D) 16

^{13}*/*_{27}%
Answer:
Option C

__Explanation:__
W

_{1 }= 12% of 2 = 24/100 = 0.24
∴ W

_{2}= 5% of 3 = 0.15
W

_{3}= 0.5
∴ W

_{1}+ W_{2}+ W_{3}= 0.24 + 0.15 + 0.5 = 0.89
∴ New
W% = (0.89/5.5) × 100 = 16

^{2}*/*_{11}%

__Question No. 04__**How many kg of sugar costing Rs. 9 per kg. must be mixed with 27 kg. of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg.?**

(A) 36 kg.

(B) 42 kg.

(C) 54 kg.

(D) 63 kg.

Answer:
Option D

__Explanation:__
S.P of 1 kg
of mixture = Rs. 9.24, Gain 10%

∴ C.P
of 1 kg of mixture = Rs. [(100/110) × 9.24] = Rs. 8.40

By the rule
of alligation, we have:

∴ Ratio
of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3

Let,

*x*kg of sugar of 1st kind be mixed with 27 kg of 2nd kind
Then, 7 : 3
=

*x*: 27
=>

*x*= (7 × 27)/3 = 63 kg.

__Question No. 05__**Three identical vessels contain the mixture of spirit and water. The ratio of spirit and water in each glass is 2:3, 3:4 and 4:5 respectively. The mixtures of all the three vessels are poured into a big pot. The ratio of spirit and water in the new mixture is**

(A) 401/544

(B) 27/37

(C) 19/37

(D) 13/37

Answer:
Option A

__Explanation:__
Net spirit =
(2/5) + (3/7) + (4/9) = 401/315

∴ Net
water = 3 - (401/315) = 544/325

Hence
required ratio = 401/544

__Question No. 06__**A container contains 40 liters of milk. From this container 4 liters of milk was taken out and replaced with water. This process was repeated further 2 times. How much milk is now contained by the container?**

(A) 26.34
liters

(B) 27.36
liters

(C) 28
liters

(D) 29.16
liters

Answer:
Option D

__Explanation:__
Amount of
milk left after 3 operations = [40 × {1 - (4/40)}

^{3}] liters
= [40 × (9/10) × (9/10) × (9/10)] = 29.16 liters

__Question No. 07__**In what ratio must water be mixed with milk to gain 16⅔ % on selling the mixture at cost price?**

(A) 1 : 6

(B) 6 : 1

(C) 2 : 3

(D) 4 : 3

Answer:
Option A

__Explanation:__
Let C.P of 1
liter milk be Rs.1

S.P of 1
liter of mixture = Rs. 1, gain = (50/3) %

∴ C.P
of 1 liter of mixture = [100 × (3/350) × 1] = 6/7

By the rule
of alligation, we have:

∴ Ratio
of water and milk = (1/7) : (6/7) = 1 : 6

__Question No. 08__**Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.**

(A) 1 : 3

(B) 2 : 3

(C) 3 : 4

(D) 4 : 5

Answer:
Option B

__Explanation:__
By the rule
of alligation, we have:

∴ Required
ratio = 60 : 90 = 2 : 3

__Question No. 09__**In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?**

(A) 3 : 2

(B) 3 : 4

(C) 3 : 5

(D) 4 : 5

Answer:
Option A

__Explanation:__
S.P of 1 kg
of mixture = 68.20, Gain = 10%

C.P of 1 kg
of the mixture = Rs [(100/110) × 68.20] = Rs. 62

By the rule
of alligation, we have:

∴ Required
ratio = 3 : 2

__Question No. 10__**A can contains a mixture of two liquids ‘A’ and ‘B’ in the ratio 7 : 5. When 9 liters of mixture are drawn off and the can is filled with ‘B’, the ratio of ‘A’ and ‘B’ becomes 7 : 9. how many liters of liquid ‘A’ was contained by the can initially?**

(A) 10

(B) 20

(C) 21

(D) 25

Answer:
Option C

__Explanation:__
Suppose the
can initially contains 7

*x*and 5*x*of mixtures*A*and*B*respectively
Quantity of
A in mixture left = [7

*x*- (^{7}/_{12}× 9)] liters = [7*x*- (21/4)] liters
Quantity of
B in the mixture left = [5

*x*- (^{5}/_{12}× 9)] liters = [5*x*- (15/4)] liters
∴ [7

*x*- (21/4)]/ [5*x*- (15/4)] = 7/9
=> (28

*x*- 21)/(20*x*+ 21) = 7/9
=> 252

*x*- 189 = 140*x*+ 147
=> 112

*x*= 336
=>

*x*= 3
So, the can
contained 21 liters of A.

__Question No. 11__**8 liters are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16:65. How much wine did the cask hold originally?**

(A) 18
liters

(B) 24
liters

(C) 32
liters

(D) 42
liters

Answer:
Option B

__Explanation:__
Let the
quantity of the wine in the cask originally be

*x*liters
Then,
quantity of wine left in the cask after 4 operations = [

*x*{1 - (8/*x*)}^{4}] liters
[

*x*{1 - (8/*x*)}^{4}]/*x*= 16/81
=> {1 -
(8/

*x*)}^{4}= (2/3)^{4}
=> [(

*x*- 8)/*x*] = 2/3
=> 3

*x*- 24 = 2*x*
=>

*x*= 24**Alligation and Mixtures: Next Tests:**