## Ratio Questions and Answers Part-7

1. If $$\frac{x}{2} = \frac{y}{3} = \frac{z}{4} =$$   $$\frac{{2x - 3y + 5z}}{k}{\text{,}}$$    then the value of k is -
a) 12
b) 15
c) 16
d) 18

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

2. A mixture contains wine and water in the ratio 3 : 2 and another mixture contains them in the ratio 4 : 5. How many litres of the latter must be mixed with 3 litres of the former so that the resulting mixture may contain equal quantities of wine and water = ?
a) $$5\frac{2}{5}{\text{ litres}}$$
b) $$5\frac{2}{3}{\text{ litres}}$$
c) $$4\frac{1}{2}{\text{ litres}}$$
d) $$3\frac{3}{4}{\text{ litres}}$$

Explanation: 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{ & \frac{9}{5} + 4x = \frac{6}{5} + 5x \cr & x = \frac{9}{5} - \frac{6}{5} \cr & x = \frac{3}{5} \cr}
Second Mixture required = 9x = $$9 \times \frac{3}{5}$$ = $$5\frac{2}{5}{\text{ litres}}$$

3. The monthly salaries of A, B and C are in the ratio 2 : 3 : 5. If C's monthly salary is Rs. 12000 more than that of A, then B's annual salary is = ?
a) Rs. 120000
b) Rs. 144000
c) Rs. 180000
d) Rs. 240000

Explanation:
 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

4.The ratio of the numbers of boys and girls in a school was 5 : 3. Some new boys and girls were admitted to the school, in the ratio 5 : 7. At this, the total number of students in the school became 1200 and the ratio of boys to girls changed to 7 : 5. The number of students in the school before new admissions was = ?
a) 700
b) 720
c) 900
d) 960

Explanation:
\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,{\text{A}}:{\text{B}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,5:3 \cr & {\text{Let }}5x:3x = 8x \cr & \Rightarrow {\text{New comers}} \cr & 5y:7y = 12y \cr & 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 & {\text{From equation }}.......(i) \cr & 12y + 3y = 300 \cr & y = 20 \cr & x = 6 \times 20 = 120 \cr}
The number of students initially
8x = 8 × 120 = 960

5. If a : b = c : d, then $$\frac{{{\text{ma}} + {\text{nc}}}}{{{\text{mb}} + {\text{nd}}}}$$   is equal to -
a) m : n
b) dm : cn
c) an : mb
d) a : b

Explanation:
\eqalign{ & {\text{Let }}\frac{a}{b} = \frac{c}{d} = k \cr & \Rightarrow a = bk,\,c = dk \cr & \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}

6. If a : b = 5 : 7 and c : d = 2a : 3b then ac : bd is = ?
a)20 : 38
b) 50 : 147
c) 10 : 21
d) 50 : 151

Explanation:
\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 & ac:bd \cr & = \frac{{{\text{ac}}}}{{{\text{bd}}}} \cr & = \frac{5}{7} \times \frac{{10}}{{21}} \cr & = \frac{{50}}{{147}} \cr & = 50:147 \cr}

7.If 2A = 3B = 4C, then A : B : C is equal to -
a) 2 : 3 : 4
b) 3 : 4 : 6
c) 4 : 3 : 2
d) 6 : 4 : 3

Explanation: Let 2A = 3B = 4C = k
\eqalign{ & A = \frac{k}{2}, \cr & B = \frac{k}{3}, \cr & C = \frac{k}{{4}} \cr & {\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}

8. If x2 + 4y2 = 4xy, then x : y is -
a) 1 : 1
b) 1 : 2
c) 2 : 1
d) 1 : 4

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

9. If W1 : W2 = 2 : 3 and W1 : W3 = 1 : 2, then W2 : W3 is -
a) 3 : 4
b) 4 : 3
c) 2 : 3
d) 4 : 5

\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 & \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 & {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 & {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}