Ratio - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \frac{A}{D} = {\frac{A}{B}} \times {\frac{B}{C}} \times {\frac{C}{D}} \cr & \,\,\,\,\,\,\,\,\, = {\frac{2}{3}} \times {\frac{4}{5}} \times {\frac{5}{9}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{ {2 \times 4 \times 5} }}{{ {3 \times 5 \times 9} }} \cr & \,\,\,\,\,\,\,\,\, = \frac{8}{{27}} \cr & \,\,\,\,\,\,\,\,\, = 8:27 \cr} $$

Girls : boys = 6 : 5
Hence, girls = $$6 \times \frac{{66}}{{11}}$$  = 36
Boys = 30
New ratio, 30 : (36 + 4) = 3 : 4

$$\eqalign{ & {\text{Let }}x{\text{ be added to each term}}. \cr & {\text{According to question}}, \cr & \frac{{ {7 + x} }}{{11 + x}} = \frac{3}{4} \cr & Or,\,33 + 3x = 28 + 4x \cr & Or,\,x = 5 \cr} $$

Let the numbers be 7x and 11x.
$$\eqalign{ & {\text{According to question}}, \cr & \frac{{ {7x + 7} }}{{ {11x + 7} }} = \frac{2}{3} \cr & Or,\,22x + 14 = 21x + 21 \cr & Or,\,x = 7 \cr & {\text{Smaller}}\,{\text{number}} = 49 \cr} $$

If we go through normal method, It will be quite cumbersome,so,
We will solve this question through options

Taking Option A:
It has P + Q = 3.
The possible value of $$\frac{{\text{P}}}{{\text{Q}}}$$ is $$\frac{1}{2}$$ or $$\frac{2}{1}$$
Using $$\frac{1}{2}$$, we see that on adding 2 in both the numerator and denominator we get $$\frac{3}{4}$$ (not required value)
Similarly we check for $$\frac{2}{1}$$, this will also not give the required value

Option B:
We have $$\frac{1}{3}$$ possible ratio
Then, we get the final value as $$\frac{3}{5}$$ (not = to $$\frac{1}{2}$$)
Hence, rejected

Option C:
Here we have $$\frac{1}{4}$$ or $$\frac{2}{3}$$
Checking for $$\frac{1}{4}$$ we get $$\frac{3}{6}$$ = $$\frac{1}{2}$$
Hence, the option c is correct

Here,Let water = 7x and milk = 3x
Now,
7x + 3x = 30
x = 3
So, water = 7x = 7 × 3 = 21 liter
Milk = 3x = 3 × 3 = 9 liter

Now, we keep milk constant and add water to mixture to get ratio 6 : 1
Let water in this mixture = 6y and milk = y
We have, milk = 9 liter, so y = 9 liter
Water = 6y = 6 × 9 = 54 liter
Then extra water to be added is 33 liter

Let income of A and B be 3x and 2x respectively. Also, their expenditure is 5y and 3y.

Now, according to question,
3x - 5y = 1000 ------- (i) × 3
2x - 3y = 1000 ---------- (ii) × 5
9x - 15y - 10x + 15y = 3000 - 5000
Or, -x = -2000
Or, x = 2000
Then, income of A = 3x = 3 × 2000 = Rs. 6000

Let the two positive numbers be 5x and 3x respectively
According to question,
5x - 3x = 10
Or, x = 5
Then numbers are 25 and 15
Thus, their product = 25 × 15 = 375

We will use work equivalence method,
$$\eqalign{ & \frac{{30}}{{18}} = \frac{{ {\frac{1}{7}} }}{{ {\frac{6}{7}} }} \times \frac{x}{2} \cr & \frac{5}{3} = {\frac{1}{6}} \times \frac{x}{2} \cr & Or,\,x = \frac{{60}}{3} = 20\,{\text{days}} \cr} $$

Given;
3dog = 4 cat
Or, $$\frac{{{\text{dog}}}}{{{\text{cat}}}} = \frac{4}{3}$$
Let cat's 1 leap = 3 meter and dogs 1 leap = 4 meter
Then, ratio of speed of cat and dog = $$\frac{{3 \times 5}}{{4 \times 4}}$$  = 15 : 16