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## Arithmetic: Alligation and Mixture Test-2

Congratulations - you have completed Arithmetic: Alligation and Mixture Test-2.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
 Question 1
Given that 24 carat gold is pure gold, 18 carat gold is  3/4 gold and 20 carat gold is 5/6 gold, the ratio of the pure gold in 18 carat gold to the pure gold in 20 carat gold is
 A 5 : 8 B 9 : 10 C 15 : 24 D 8 : 5
Question 1 Explanation:
18 carat gold = ¾ pure gold = ¾ x 24 = 18
20 carrot = 5/6 pure gold = 5/6 x 24 = 20
Therefore the required ratio =  18 :20 = 9:10
 Question 2
A sink contains exactly 12 L of water. If water is drained from the sink until it holds exactly 6 L of water less than the quantity drained away, how many liters of water were drained away?
 A 2 L B 6 L C 3 L D 9 L
Question 2 Explanation:
Sink Capacity = 12 L
Effectively this means:
Remaining water (R)+ Drained water (D) = 12 L….(i)
Remaining Water (R)  = D – 6
Using (i) and (ii)
D – 6 +D =12
D = 9
 Question 3
Several liters of acid were drawn off a 54 L vessel full of acid and an equal amount of water -added. Again, the same volume of the mixture was drawn off and replaced by water. As a result the vessel contained 24 L of pure acid. How much of the acid was drawn off initially?
 A 12 L B 16 L C 18 L D 24 L
Question 3 Explanation:
Let the container contain a volume of v L (54 L in the given case)
And also let us suppose that the liquid which is drained out = u L
So the final quantity will be = v(1-u/v)n
Where n is the number of time of operation is carried out
So from the given information we can say that
24 = 54(1-u/54)2
(1-u/54)2= 24/54 = 4/9 (1-u/54) = 2/3
u/54 = 1/3
u =18L
 Question 4
A tin of oil was four- fifths full. When six bottles of oil were taken out and four bottles of oil were poured into it, it was three-fourths full. How many bottles of oil were contained by the tin?
 A 10 B 20 C 30 D 40
Question 4 Explanation:
Let us assume the total capacity of the tin = P bottles
In this case we assume that a single bottle represents one unit of tin.
So according to the question
(4/5P) – 6 +4=3/4P
(4/5P) – 3/4P = 2
P=40
 Question 5
A vessel contains 5 liters of a mixture of milk and water, the quantity of milk was 36% of the total mixture. A few liters of mixture was taken out and replaced by the same amount of water. This process is repeated 2 times and the quantity of milk becomes 16% only. How many liters of mixture was taken out every time?
 A 16 B 20 C 1.6 D 1.98
Question 5 Explanation:
Total amount of the mixture = 5L
Number of operations = 2
Amount of ingredient left/actual amount of milk =   {1 – x/amount of mixture}n
16%/36% = (1 – x/5)2
4/6 = 1-x/5
X = approximately 1.66L
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