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Ratio - Aptitude MCQ Questions and Solutions with Explanations

Ratio MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Ratio with easy and logical explanations. Ratio MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-9 section-3

Each section contains 100 MCQs on Ratio .
81
In a regiment the ratio between the number of officers to soldiers was 3 : 31 before. In a battle 6 officers and 22 soldiers were killed and the ratio become 1 : 13, the number of officers in the regiment before battle was = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the number of officers and soldiers = 3x, 31x
According to the question,
$$\eqalign{ & \Rightarrow \frac{{3x - 6}}{{31x - 22}} = \frac{1}{{13}} \cr & \Rightarrow 39x - 78 = 31x - 22 \cr & \Rightarrow 8x = 56 \cr & \Rightarrow x = 7 \cr} $$
So, number of officer before
= 3x = 3 × 7 = 21
82
Three containers have their volumes in the ratio 3 : 4 : 5. They are full of mixtures of milk and water. The mixtures contain milk and water in the ratio of (4 : 1), (3 : 1) and (5 : 2) respectively. The mixture of all these three containers are poured into a fourth container. The ratio of milk and water in the fourth container is = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
Milk   :   Water  
4×28×3   :   1×28×3   = 5×28×3
3×35×4   :   1×35×4   = 4×35×4
5×20×5   :   2×20×5   = 7×20×5
1256   :   424  
157   :   53  
83
In a college union, there are 48 students. The ratio of the number of boys to the number of girls is 5 : 3. The number of girls to be added in the union, so that the number of boys to girls in 6 : 5 is = ?
Discuss
Answer & Solution
Answer: Option B
Solution:
And ratio of numbers of boys to the number of girls is 5x : 3x
According to the question,
$$\eqalign{ & \Rightarrow 5x + 3x = 48 \cr & \Rightarrow 8x = 48 \cr & \Rightarrow \boxed{x = 6} \cr & {\text{Boys}} = 30 \cr & {\text{Girl}} = 18 \cr & {\text{Now,}}\frac{{30}}{{18 + y}} = \frac{6}{5} \cr & \Rightarrow 25 = 18 + y \cr & \Rightarrow y = 7 \cr} $$
84
The ratio of milk to water in three containers of equal capacity is 3 : 2, 7 : 3 and 11 : 4 respectively. The contents of the three containers are mixed together. What is the ratio of milk to water after mixing ?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Milk in the mixture}} \cr & = \left( {\frac{3}{5} + \frac{7}{{10}} + \frac{{11}}{{15}}} \right){\text{units}} \cr & = \frac{{61}}{{30}}{\text{units}}{\text{.}} \cr & {\text{Water in the mixture}} \cr & = \left( {\frac{2}{5} + \frac{3}{{10}} + \frac{4}{{15}}} \right){\text{units}} \cr & = \frac{{29}}{{30}}{\text{units}}{\text{.}} \cr & \therefore {\text{Required ratio}} \cr & = \frac{{61}}{{30}}:\frac{{29}}{{30}} \cr & = 61:29 \cr} $$
85
Six coins of gold and silver of equal weights are melted and new coins are cast. The ratio of gold and silver in one of the coins is 2 : 1, in another two coins 3 : 5 and 7 : 5 in the remaining coins. What will be the ratio between gold and silver respectively in the new coins ?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Gold in new coins}} \cr & = \left( {\frac{2}{3} + 2 \times \frac{3}{8} + 3 \times \frac{7}{{12}}} \right){\text{units}} \cr & = \left( {\frac{2}{3} + \frac{3}{4} + \frac{7}{4}} \right){\text{units}} \cr & = \frac{{19}}{6}{\text{units}}{\text{}} \cr & {\text{Silver in new coins}} \cr & = \left( {\frac{1}{3} + 2 \times \frac{5}{8} + 3 \times \frac{5}{{12}}} \right){\text{units}} \cr & = \left( {\frac{1}{3} + \frac{5}{4} + \frac{5}{4}} \right){\text{units}} \cr & = \frac{{17}}{6}{\text{units}}{\text{.}} \cr & \therefore {\text{Required ratio}} \cr & = \frac{{19}}{6}:\frac{{17}}{6} \cr & = 19:17 \cr} $$
86
A dealer buys dry fruit at the rate of Rs. 100, Rs. 80 Rs. 60 per kg. He bought them in the ratio 12 : 15 : 20 by weight. He in total gets 20% profit by selling the first two and at last he finds he has no gain no loss in selling the whole quantity which he had. What was the percentage loss he suffered for the third quantity ?
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the weights of the three varieties be 12x, 15x and 20x kg respectively
Then,

Total C.P. = Rs. (100 × 12x + 80 × 15x + 60 × 20x)
= Rs. (3600x)
S.P. of the first two verities
$$\eqalign{ & = {\text{Rs}}.\left( {\frac{{120}}{{100}} \times 2400x} \right) \cr & = {\text{Rs}}.\left( {2880x} \right) \cr & {\text{S}}{\text{.P}}{\text{. of the third variety}} \cr & = {\text{Rs}}.\left( {3600x - 2880x} \right) \cr & = {\text{Rs}}{\text{. }}720x \cr & {\text{Loss on third variety}} \cr & = {\text{Rs}}{\text{.}}\left( {1200x - 720x} \right) \cr & = {\text{Rs}}{\text{. }}480 \cr & {\text{loss }}\% \cr & = \left( {\frac{{480x}}{{1200x}} \times 100} \right)\% \cr & = 40\% \cr} $$
87
A and B start an enterprise together, with A as active partner. A invests Rs. 4000 and Rs. 2000 more after 8 months. B invests Rs. 5000 and withdraws Rs. 2000 after 9 months. Being the active partner, A takes Rs. 100 per month as allowance,from the profit. What is the share of B if the profit for for the year is 6700 ?
Discuss
Answer & Solution
Answer: Option C
Solution:
A : B
8 × 4000 + 4 × 6000   :  8 × 4000 + 4 × 6000
56 : 54

 

Total profit is Rs. 6700 but Rs. 100 per month is withdrawn by A,
So Rs. 1200 is taken by A in one year. Remaining profit will be distributed between A and B.
According to the question,
56x + 54x = 6700 - 1200
110x = 5500
x = 50
B's share = 54 × 50 = Rs. 2700
88
A's income is Rs. 140 more than B's income ans C's income is Rs. 80 more than D's income. If the ratio of A's and C's income is 2 : 3 and the ratio of B's and D's income is 1 : 2, then the incomes of A, B, C and D are respectively = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
A : C = 2 : 3 or 2x : 3x
B : D = 1 : 2 or y : 2y
According to the question,
2x - 140 = y......(i)
3x- 80 = 2y.......(ii)
multiply equation (i) by 2 and solved
$$\eqalign{ & 4x - 280 = 2y \cr & \underline { \pm 3x \mp 80 = \pm 2y} \cr & x - 200 = 0 \cr & x = 200 \cr} $$
Now put the value of x in equation .... (i)
2 × 200 - 140 = y
$$\boxed{y = 260}$$
A's salary = 2x = 200 × 2 = Rs. 400
B's salary = y = 260 × 1 = Rs. 260
C's salary = 2x = 3 × 200 = Rs. 600
D's salary = 2y = 2 × 260 = Rs. 520
89
Find the fraction which bears the same ratio to $$\frac{1}{{27}}$$ that $$\frac{3}{7}$$ does to $$\frac{5}{9}$$ = ?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Let the fraction is }}x \cr & \Rightarrow x:\frac{1}{{27}}::\frac{3}{7}:\frac{5}{9} \cr & \Rightarrow \frac{x}{{\frac{1}{{27}}}} = \frac{{\frac{3}{7}}}{{\frac{5}{9}}} \cr & \Rightarrow 27x = \frac{{3 \times 9}}{{7 \times 5}} \cr & \Rightarrow x = \frac{1}{{35}} \cr} $$
90
The respective ratio between the monthly salaries of Rene and Som is 5 : 3. Out of her salary Rene gives $$\frac{{\text{1}}}{{\text{6}}}$$ th as rent, $$\frac{{\text{1}}}{{\text{5}}}$$ th to her mother, 30% as her education loan and keeps 25% aside for miscellaneous expenditure. Remaining Rs. 5000 she keeps as savings. What is Som's monthly salary ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the monthly salaries of Rene and Som be 5a and 3a respectively.
Money spent by Rene
$$ = \left( {\frac{1}{6}{\text{of }}5a + \frac{1}{5}{\text{of }}5a + 30\% {\text{ of }}5a + 25\% {\text{ of }}5a} \right)$$
According to given information
$$ \Rightarrow 5a - $$ $$\left[ {5a \times \frac{1}{6} + 5a \times \frac{1}{5} + 5a \times \frac{3}{{10}} + \frac{{5a}}{4}} \right] = $$       $$ 5000$$
$$\eqalign{ & \Rightarrow 5a - \left[ {\frac{{5a}}{6} + a + \frac{{3a}}{2} + \frac{{5a}}{4}} \right] = 5000 \cr & \Rightarrow 5a - \left( {\frac{{55a}}{{12}}} \right) = 5000 \cr & \Rightarrow 60a - 55a = 60000 \cr & \Rightarrow a = 12000 \cr & \therefore {\text{Som's salary}} \cr & = 3 \times 12000 \cr & = {\text{Rs}}{\text{. }}36000 \cr} $$