Ratio - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {4^{3.5}}:{2^5} \cr & = {\left( {{2^2}} \right)^{3.5}}:{2^5} \cr & = {2^{2 \times 3.5}}:{2^5} \cr & = {2^7}:{2^5} \cr & = {2^2}{.2^5}:{2^5} \cr & = {2^2}:1 \cr & = 4:1 \cr} $$

Given,
A : B = 1 : 2 ------(i)
B : C = 3 : 4 ------(ii)
C : D = 6 : 9 ------(iii)
D : E = 12 : 16 ------(iv)

In equation (i) multiply with 3 and In equation (ii) multiply with 2 we get
A : B = 3 : 6 and B : C = 6 : 8
∴ A : B : C = 3 : 6 : 8 -----(v)

In equation (v) multiply with 3 and In equation (iii) multiply with 4 we get
A : B : C = 9 : 18 : 24 and C : D = 24 : 36
∴ A : B : C : D = 9 : 18 : 24 : 36
i.e. A : B : C : D = 3 : 6 : 8 : 12 -----(vi)

Now A : B : C : D = 3 : 6 : 8 : 12 and
D : E = 12 : 16
∴ A : B : C : D = 3 : 6 : 8 : 12 : 16

	A	:	B
	3	:	5
Let	3x	:	5x

L.C.M. of A & B is = 3 × 5 × x = 15x
15x = 225
Given
x = 15
∴ A = 15 × 3 = 45
B = 15 × 3 = 75
Smaller number is = 45

$$\eqalign{ & {\text{Let }}\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \cr & {\text{Then,}} \cr & a = bk, \cr & b = ck, \cr & c = dk \cr & {\text{Also,}} \cr & \Rightarrow \frac{a}{b} \times \frac{b}{c} \times \frac{c}{d} = {k^3} \cr & \Rightarrow {k^3} = \frac{a}{{d}} \cr & \therefore \frac{{{b^3} + {c^3} + {d^3}}}{{{a^3} + {b^3} + {c^3}}} \cr & = \frac{{{b^3} + {c^3} + {d^3}}}{{{{\left( {bk} \right)}^3} + {{\left( {ck} \right)}^3} + {{\left( {dk} \right)}^3}}} \cr & = \frac{{{b^3} + {c^3} + {d^3}}}{{{k^3}\left( {{b^3} + {c^3} + {d^3}} \right)}} \cr & = \frac{1}{{{k^3}}} \cr & = \frac{d}{a} \cr} $$

$$\eqalign{ & = \frac{p}{q} = \frac{3}{4},\frac{r}{s} = \frac{8}{5},\frac{x}{y} = \frac{{10}}{6} \cr & \Rightarrow p = \frac{{3q}}{4},r = \frac{{8s}}{5},x = \frac{{5y}}{3} \cr & \therefore \frac{{psx}}{{qry}} = \frac{{\frac{{3q}}{4} \times s \times \frac{{5y}}{3}}}{{q \times \frac{{8s}}{5} \times y}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\frac{5}{4}qsy}}{{\frac{8}{5}qsy}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{4} \times \frac{5}{8} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{25}}{{32}} \cr} $$

$$\eqalign{ & {\text{Let }}\frac{x}{2} = \frac{y}{3} = \frac{z}{4} = l \cr & {\text{Then,}} \cr & = x = 2l,y = 3l,z = 4l \cr & \therefore \frac{x}{2} = \frac{{2x - 3y + 5z}}{k} \cr & \Rightarrow \frac{{2l}}{2} = \frac{{2 \times 2l - 3 \times 3l + 5 \times 4l}}{k} \cr & \Rightarrow k = 4 - 9 + 20 = 15 \cr} $$

The first mixture contains wine and water in the ratio of 3 : 2
Wine in 3 liters of mixture = $$\frac{3}{5} \times 3 = \frac{9}{5}$$
Water in 3 liters of mixture = $$\frac{2}{5} \times 3 = \frac{6}{5}$$

Let us consider that the first mixture is mixed with the second mixture that has quantity as 9x liters (4x liters of wine and 5x liters of water).

After mixing,
The total quantity of wine = Total quantity of water
$$\eqalign{ & \Rightarrow \frac{9}{5} + 4x = \frac{6}{5} + 5x \cr & \Rightarrow x = \frac{9}{5} - \frac{6}{5} \cr & \Rightarrow x = \frac{3}{5} \cr} $$

Second Mixture required = 9x = $$9 \times \frac{3}{5}$$ = $$5\frac{2}{5}{\text{ litres}}$$

	A	:	B	:	C
	2	:	3	:	5
Let	2x	:	3x	:	5x

C - A = 12000
5x - 2x = 12000
3x = 12000
x = 4000
Monthly salary of B is = 3x = Rs. 12000
∴ Annual salary of B is
= 12 × 1200
= Rs. 144000

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,{\text{A}}:{\text{B}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,5:3 \cr & {\text{Let }}5x:3x = 8x \cr & \Rightarrow {\text{New comers}} \cr & 5y:7y = 12y \cr & \therefore 8x + 12 = 1200 \cr & \Rightarrow 2x + 3y = 300......(i) \cr & {\text{Again,}}\frac{{5x + 5y}}{{3x + 7y}} = \frac{7}{5} \cr & 25x + 25y = 21x + 49y \cr & \Rightarrow 4x - 24y = 0 \cr & \Rightarrow 4x = 24y \cr & \Rightarrow x = 6y...........(ii) \cr & \therefore {\text{From equation }}.......(i) \cr & 12y + 3y = 300 \cr & y = 20 \cr & \therefore x = 6 \times 20 = 120 \cr} $$
The number of students initially
8x = 8 × 120 = 960

$$\eqalign{ & {\text{Let }}\frac{a}{b} = \frac{c}{d} = k \cr & {\text{Then,}} \cr & \Rightarrow a = bk,\,c = dk \cr & \therefore \frac{{ma + nc}}{{mb + nd}} \cr & = \frac{{mbk + ndk}}{{mb + nd}} \cr & = \frac{{k\left( {mb + nd} \right)}}{{\left( {mb + nd} \right)}} \cr & \Rightarrow k = \frac{a}{b} \cr & \Rightarrow a:b \cr} $$