Ratio - Aptitude MCQ Questions and Solutions with Explanations

Original ratio of the mixture 3:1
Taking out the mixture actually means just taking out the wine because the water anyways is going to be added back.
if we remove 1 part of wine, it makes the ratio 2 : 2 ( 1 part water is added to keep the volume constant )
so, what we have actually done is remove 1 part wine from 3 part Wine.
i.e. $$\frac{1}{3}$$

So, $$\frac{1}{3}$$ mixture drawn.

$$\eqalign{ & = {\text{25}}\% {\text{ of A}} = {\text{35}}\% {\text{ of B}} \cr & \Rightarrow \frac{{25}}{{100}}{\text{A}} = \frac{{35}}{{100}}{\text{B}} \cr & \Rightarrow \frac{{\text{A}}}{4} = \frac{{7{\text{B}}}}{{20}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{7}{{20}} \times 4 = \frac{7}{5} \cr & \Rightarrow {\text{A}}:{\text{B}} = 7:5 \cr} $$

$$\eqalign{ & x = \frac{1}{3}y = \frac{1}{3} \times \frac{1}{2}z = \frac{1}{6}z \cr & Let\,x\, = \frac{1}{3}y = \frac{1}{6}z = k \cr & {\text{Then,}} \cr & x = k,\,y = 3k,\,z = 6k \cr & \therefore {\text{x}}:{\text{y}}:{\text{z}} = k:3k:6k \cr & = 1:3:6 \cr} $$

Rs. 1 : 50 paise : 25 paise
Number of coins 8x : 5x : 3x
$$\eqalign{ & {\text{Value of coins}} \cr & {\text{ = }}\frac{{8x}}{1}:\frac{{5x}}{2}:\frac{{3x}}{4} \cr & \therefore 8x + \frac{{5x}}{2} + \frac{{3x}}{4} = {\text{Rs}}.\,225, \cr & {\text{Given}}\frac{{32x + 10x + 3x}}{4} = 225 \cr & \Rightarrow 45x = 225 \times 4 \cr & \Rightarrow x = \frac{{225 \times 4}}{{45}} \cr & \Rightarrow x = 5 \times 4 = 20 \cr} $$
∴ Number of one rupees coins
= 20 × 8 = 160

P : Q : R
P + Q = R (given)
2x + 7x = 9x
Hence, we don't have sufficient data to insure the values of A, B and C

$$\eqalign{ & {\text{a}}:{\text{b}}\,\,\,\,\,\,\,\,\,\,\,\,{\text{c}}:{\text{d}} \cr & 5:7\,\,\,\,\,\,\,\,\,\,\,\,\,2{\text{a}}:3{\text{b}} \cr & \frac{a}{b} = \frac{5}{7},\,\frac{c}{d} = \frac{{2a}}{{3b}} \cr & = \frac{2}{3} \times \frac{5}{7} \cr & = \frac{{10}}{{21}} \cr & \therefore ac:bd \cr & = \frac{{{\text{ac}}}}{{{\text{bd}}}} \cr & = \frac{5}{7} \times \frac{{10}}{{21}} \cr & = \frac{{50}}{{147}} \cr & = 50:147 \cr} $$

Let 2A = 3B = 4C = k
$$\eqalign{ & {\text{Then,}} \cr & A = \frac{k}{2}, \cr & B = \frac{k}{3}, \cr & C = \frac{k}{{4}} \cr & \therefore {\text{A}}:{\text{B}}:{\text{C}} = \frac{k}{2}:\frac{k}{3}:\frac{k}{4} \cr & = \frac{1}{2}:\frac{1}{3}:\frac{1}{4} {\text{ (Multiply by 12)}} \cr & = 6:4:3 \cr} $$

$$\eqalign{ & = {x^2} + 4{y^2} = 4xy \cr & \Rightarrow {x^2} - 4xy + 4{y^2} = 0 \cr & \Rightarrow \left( {x - 2{y^2}} \right) = 0 \cr & \Rightarrow x = 2y \cr & \Rightarrow \frac{x}{y} = 2 \cr & \Rightarrow x:y = 2:1 \cr} $$

$$\eqalign{ & = \frac{{{{\text{W}}_{\text{2}}}}}{{{{\text{W}}_{\text{1}}}}} = \frac{3}{2}{\text{and}}\frac{{{{\text{W}}_{\text{1}}}}}{{{{\text{W}}_{\text{3}}}}} = \frac{1}{2} \cr & \Rightarrow \frac{{{W_2}}}{{{W_3}}} = \left( {\frac{{{{\text{W}}_{\text{2}}}}}{{{{\text{W}}_{\text{1}}}}} \times \frac{{{{\text{W}}_{\text{1}}}}}{{{{\text{W}}_{\text{3}}}}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} \cr & \Rightarrow {W_2}:{W_3} = 3:4 \cr} $$

$$\eqalign{ & a:b = b:c \cr & \frac{a}{b} = \frac{b}{c} \cr & {b^2} = ac \cr & {b^4} = {a^2}{c^2} \cr & \therefore {a^4}:{b^4} = \frac{{{a^4}}}{{{b^4}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{a^4}}}{{{a^2}{c^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{a^2}}}{{{c^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {a^2}:{c^2} \cr} $$