ExamVeda
Login
Home

Alligation - Aptitude MCQ Questions and Solutions with Explanations

Alligation MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Alligation with easy and logical explanations. Alligation MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-8 section-1

71
The acid and water in two vessels A and B are in the ratio 4 : 3 and 2 : 3. In what ratio should the liquid in both the vessels be mixed to obtain a new mixture in vessel C containing half acid and half water ?
Discuss
Answer & Solution
Answer: Option A
Solution:
According to the question,
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Acid}}\,\,\,\,\,\,{\text{Water}} \cr & {\text{Vessel A}}\,\,{\text{4}}\,\,\,:\,\,\,\,{\text{3}} \cr & {\text{Vessel B}}\,\,\,2\,\,\,\,{\text{:}}\,\,\,\,\,{\text{3}} \cr} $$
Now using alligation,
Alligation mcq solution image
72
A vessel contained a solution of acid and water, in which water was 64%. Four litres of the solution was taken out of the vessel and the same quantity of water was added. If the resulting solution contains 30% acid, the quantity (in litres) of the water in the solution, at the beginning in the vessel, was:
Discuss
Answer & Solution
Answer: Option B
Solution:
\[\begin{array}{*{20}{c}} {}&{{\text{Acid}}}&{{\text{Water}}} \\ {{\text{Old}} \to }&{36\% }&{64\% } \\ {{\text{New}} \to }&{30\% }&{70\% } \end{array}\]
Alligation mcq question image
$$\eqalign{ & {\text{Old mixture}} = {\text{Remaining}} + {\text{Taken out}} \cr & = 20{\text{L}} + 4{\text{L}} \cr & = 24{\text{L}} \cr & {\text{Water}} = 24 \times \frac{{64}}{{100}} \Rightarrow 15.36{\text{L}} \cr} $$
73
A milkman borrowed a sum of Rs. 2,500 from two moneylenders for one loan on simple interest at the rate of 5% and 7%, respectively, for a period of 2 years. If the total interest paid by him was Rs. 276, then how much did he borrow at the rate of 7% p.a.?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Let, loan taken at rate of }}7\% = x \cr & {\text{then, }}\frac{{\left( {2500 - x} \right) \times 5}}{{100}} + \frac{{x \times 7}}{{100}} = \frac{{276}}{2} \cr & 12500 - 5x + 7x = 13800 \cr & 2x = 3800 - 12500 \cr & 2x = 1300 \cr & x = 650\,{\text{Answer}} \cr} $$
74
An alloy contains copper and tin in the ratio 3 : 2. If 250 gm of copper is added to this alloy then copper in it becomes double the quantity of tin in it. What is the amount (in gm) of tin in the alloy?
Discuss
Answer & Solution
Answer: Option D
Solution:
The ratio of copper and tin = 3x : 2x
According the question,
3x + 250 = 2(2x)
3x + 250 = 4x
x = 250
∴ Quantity of tin in the alloy = 2x
= 2 × 250
= 500 gm
75
A vessel contains a 32 litre solution of acid and water in which the ratio of acid and water is 5 : 3. If 12 litres of the solution are taken out and $$7\frac{1}{2}$$ litres of water are added to it, then what is the ratio of acid and water in the resulting solution?
Discuss
Answer & Solution
Answer: Option A
Solution:
Alligation mcq question image
$$\eqalign{ & {\text{After removed 12L}} \cr & \,\,\,\,\,\,\,{\text{Acid}}\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,{\text{Water}} \cr & \Rightarrow 12 \times \frac{5}{8},\,12\, \times \frac{3}{8} \cr & \Rightarrow - 7.5,\, - 4.5 \cr & {\text{Acid}}\,\,\,\,\,\,\,\,\,\,{\text{Water}} \cr & \,\,\,{\text{20L }}\,\,\,\,\,\,\,\,\,\,{\text{ 12L}} \cr & - 7.5{\text{ }}\,\,\,\,{\text{ }} - 4.5 \cr & {\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ }} + 7.5 \cr & \overline {\,12.5\,\,\,\,\,:\,\,\,\,\,\,15\,\,} \cr & \,\,\,\,\,\,5\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,6 \cr} $$
76
The price of a variety of a commodity is Rs. 7 per kg and that of another is Rs. 12 per kg. Find the ratio in which two varieties should be mixed so that the price of the mixture is Rs. 10 per kg.
Discuss
Answer & Solution
Answer: Option B
Solution:
Alligation mcq question image
77
A milkman has a 15 litres drum with 9 litres milk. He filled the drum with water and now takes out $$\frac{1}{2}$$ litre of the mixture. What is the quantity of milk (in millilitres) in the mixture taken out?
Discuss
Answer & Solution
Answer: Option C
Solution:
\[\begin{array}{*{20}{c}} {{\text{Total capacity}}}&{{\text{Milk}}}&{}&{{\text{Water}}} \\ {15}&9&:&6 \\ {}&3&:&2 \end{array}\]
Half liter is taken out
Milk = 500 ml × $$\frac{3}{5}$$ = 300 ml
78
In what ratio must 45% grape juice be mixed with that of 25% grape juice so as to get a mixture of 40% grape juice?
Discuss
Answer & Solution
Answer: Option B
Solution:
Alligation mcq question image
3 : 1 Answer
79
A drum contains 80 litres of ethanol. 20 litres of this liquid is removed and replaced with water. 20 liters of this mixture is again removed and replaced with water. How much water (in litres) is present in this drum now?
Discuss
Answer & Solution
Answer: Option C
Solution:
After replacing 20 litre of liquid for first time, ratio of
\[\begin{array}{*{20}{c}} {{\text{Ethanol}}}&:&{{\text{Water}}} \\ {60}&:&{20} \end{array}\]
Now, again removing 20 litre of the mixture in the given proportion.
\[\begin{array}{*{20}{c}} {{\text{Ethanol}}}&:&{{\text{Water}}} \\ {45}&:&{15} \end{array}\]
Now, an adding 20 litre water
Total water in the drum = 15 + 20 = 35 litre

$$\eqalign{ & {\bf{Alternate}}\,{\bf{Solution:}} \cr & {\text{Final quantity}} = {\text{Initial quantity}} \times {\left( {1 - \frac{x}{c}} \right)^n} \cr & x \to {\text{ quantity taken out}} \cr & c \to {\text{ total capacity}} \cr & n \to {\text{ number of process}} \cr & = 80 \times {\left( {1 - \frac{{20}}{{80}}} \right)^2} \cr & = 80 \times {\left( {\frac{3}{4}} \right)^2} \cr & = 80 \times \frac{3}{4} \times \frac{3}{4} \cr & {\text{Final quantity of external}} = 45{\text{ litre}} \cr & {\text{and Water}} = 80 - 45 = 35{\text{ litre}} \cr} $$
80
40 litres of 60% concentration of acid solution is added to 35 litres of 80% concentration of acid solution. What is the concentration of acid in the new solution?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{I}} \to \frac{{40 \times 60}}{{100}} = 24 \cr & {\text{II}} \to \frac{{35 \times 80}}{{100}} = 28 \cr & {\text{Acid }}\% = \frac{{52}}{{75}} \times 100 = 69\frac{1}{3}\% \cr} $$