Ratio - Aptitude MCQ Questions and Solutions with Explanations

Let,
⇒ Sonali's age = 5x
⇒ Monali's age = 3x
According to the question,
$$\eqalign{ & \Rightarrow \frac{{5x + 5}}{{3x + 5}} = \frac{{10}}{7} \cr & \Rightarrow \frac{{x + 1}}{{3x + 5}} = \frac{2}{7} \cr & \Rightarrow 7x + 7 = 6x + 10 \cr & \Rightarrow x = 3 \cr} $$
⇒ So, Monali's present age = 3x
= 3 × 3 = 9 years

$$\eqalign{ & {\text{Let,}} \cr & \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{1}{2} \cr} $$
Now check from option to save your valuable time
$$\eqalign{ & {\text{Option D}} \cr & \frac{{a + 3c - 5e}}{{b + 3d - 5f}} \Rightarrow \frac{c}{d}\left[ {\frac{{\frac{a}{c} + 3 - \frac{{5e}}{c}}}{{\frac{b}{d} + 3 - \frac{{5f}}{d}}}} \right] \cr & \Rightarrow \frac{c}{d} = \frac{1}{2}\left( {{\text{Satisfy}}} \right) \cr} $$

Let the age of A & B = 5x and 6x years
According to question,
$$\eqalign{ & \frac{{5x + 7}}{{6x + 7}} = \frac{6}{7} \cr & 35x + 49 = 36x + 42 \cr & x = 7 \cr & {\text{A's Present age}} = 5x \cr & = 5 \times 7 = {\text{35 years}} \cr} $$

Let the number of boys and girls be 4x and 5x respectively.
Then,
$$\eqalign{ & {\text{ = }}\frac{{4x}}{{5x - 100}} = \frac{6}{7} \cr & \Rightarrow 28x = 30x - 600 \cr & \Rightarrow 2x = 600 \cr & \Rightarrow x = 300 \cr & \therefore {\text{Number of boys}} \cr & = 4 \times 300 \cr & = 1200 \cr} $$

Let Sarika's and Gauri's ages one year ago be 3x and 4x years respectively
Sarika's age 1 year hence = (3x + 2) years
Gauri's age 1 year hence = (4x + 2)
$$\eqalign{ & \therefore \frac{{3x + 2}}{{4x + 2}} = \frac{{10}}{{13}} \cr & \Rightarrow 13\left( {3x + 2} \right) = 10\left( {4x + 2} \right) \cr & \Rightarrow 39x + 26 = 40x + 20 \cr & \Rightarrow x = 6 \cr} $$
Hence, Sarika's present age = 3x + 1
= (3 × 6 + 1) years
= 19 years

$$\eqalign{ & {\text{A }}\alpha {\text{ C and B }}\alpha {\text{ C}} \cr & \Rightarrow {\text{A}} = {\text{kC and B}} = {\text{mC}} \cr & {\text{for some constants k and m}}{\text{.}} \cr & \therefore A + B = kC + mC = \left( {{\text{k}} + {\text{m}}} \right){\text{C}} \cr & \Rightarrow \left( {A + B} \right)\alpha {\text{ C}}. \cr & {\text{A}} - {\text{B}} = {\text{kC}} - {\text{mC}} = \left( {{\text{k}} - {\text{m}}} \right){\text{C}} \cr & \Rightarrow \left( {A - B} \right)\alpha {\text{ C}}. \cr & \sqrt {AB} = \sqrt {kC \times mC} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{\text{km}}{{\text{C}}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{\text{km}}.} {\text{C}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{\text{AB}}} {\text{ }}\alpha {\text{ C}} \cr & = \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{kC}}}}{{{\text{mC}}}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{\text{k}}}{{\text{m}}} = {\text{Constant}}{\text{.}} \cr} $$

$$\eqalign{ & a = \frac{{7 \times 24}}{{10 \times 24}} = \frac{{168}}{{240}} \cr & b = \frac{{5 \times 30}}{{8 \times 30}} = \frac{{150}}{{240}} \cr & c = \frac{{2 \times 80}}{{3 \times 80}} = \frac{{160}}{{240}} \cr & {\text{The descending order}} \cr & {\text{ = }}a > c > b \cr} $$

If he spends two months income in three months. It means he saves the third month income in three months.
He saves Rs. 6000 in every 3 months.
So, in one year he saves Rs. 4 × 6000 = Rs. 24000

Ratio of coin's Rs. 1 : 50p : 25P = 3 : 2 : 5

Ratio of coin's value
Rs. 1 : 50p : 25p (value) = 3 : 1 : 1.25
(Multiply by 4 to remove the fraction value in above ratio)
i.e Rs. 1 : 50p : 25p (value) = 12 : 4 : 5

Value of 25p coin in Rs. 252 = $$\frac{5}{{21}} \times 252 = 60 $$
Number of 25p coin is = 60 × 4 = 240

h α t²
⇒ h = kt² for some constant k
⇒ 64 = k × 2² = 4k
⇒ k = 16.
Let the required height be x cm.
Then,
= x = 16 × 6² = (16 × 36) cm = 576 cm.