## Ratio Questions and Answers Part-9

1. The ratio of the first and second class fares between two railway stations is 4 : 1 and that of the number of passengers travelling by first and second classes is 1 : 40. If on a day Rs. 1100 are collected as total fare, the amount collected from the first class passengers is = ?
a) Rs. 315
b) Rs. 275
c) Rs. 137.50
d) Rs. 100

Explanation:
 1st : 2nd Fare 4x : x Passengers 1 : 40

Total fare = 4x : 40x = 44x
44x = 1100
x = $$\frac{1100}{44}$$ = 25
Fare = 1st class amount received per day
4x = 4 × 25
4x = 100

2. If a : (b + c) = 1 : 3 and c : (a + b) = 5 : 7, then b : (a + c) is equal to.
a) 1 : 2
b) 2 : 3
c) 1 : 3
d) 2 : 1

Explanation:
\eqalign{ & = \frac{a}{{b + c}} = \frac{1}{3} \cr & a = \frac{{b + c}}{3} \cr & \frac{c}{{a + b}} = \frac{5}{7} \cr & 7c = 5a + 5b \cr & 7c = \frac{{5\left( {b + c} \right)}}{3} + 5b \cr & 7c - \frac{5}{3}c = 5b + \frac{5}{3}b \cr & \frac{{16c}}{3} = \frac{{20b}}{3} \cr & 16c = 20b \cr & b = \frac{4}{5}c. \cr & a = \frac{{b + c}}{3} = \frac{{\frac{4}{5}c + c}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{9c}}{5} \times \frac{1}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{5}c. \cr & \frac{b}{{a + c}} = \frac{{\left( {\frac{4}{5}c} \right)}}{{\left( {\frac{3}{5}c + c} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4c}}{5} \times \frac{5}{{8c}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1:2 \cr}

3.If a : b = 3 : 4, b : c = 4 : 7, then $$\frac{{a + b + c}}{c}$$   is equal to -
a) 1
b) 2
c) 3
d) 7

Explanation:
\eqalign{ & = a:b = 3:4 \cr & = b:c = 4:7 \cr & \Rightarrow a:b:c = 3:4:7 \cr & {\text{Let }}a = 3k, \cr & \,\,\,\,\,\,\,\,\,\,b = 4k, \cr & \,\,\,\,\,\,\,\,\,\,c = 7k. \cr & \frac{{a + b + c}}{c} \cr & = \frac{{3k + 4k + 7k}}{{7k}} \cr & = \frac{{14k}}{{7k}} \cr & = 2 \cr}

4.If A : B : C = 2 : 3 : 5 and A = x% of (B + C), then x is equal to -
a) 20
b) 24
c) 25
d) 28

Explanation: Let A = 2k, B = 3k, C = 5k
A = x% of (B + C)
⇒ 2k = x% of (3k + 5k) = x% of 8k
\eqalign{ & \Rightarrow \frac{x}{{100}} = \frac{{{\text{2k}}}}{{{\text{8k}}}} = \frac{1}{4} \cr & x = \frac{{100}}{4} = 25 \cr}

5. If a and b are rational numbers and $$a + b\sqrt 3$$   $$=$$ $$\frac{1}{{2 - \sqrt 3 }}{\text{,}}$$   then a : b is equal to = ?
a) 2 : 1
b) 2 : 3
c) $$\sqrt 3$$ : 1
d) - $$\sqrt 3$$ : 1

Explanation:
\eqalign{ & a + b\sqrt 3 = \frac{1}{{2 - \sqrt 3 }} \cr & \Rightarrow \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow \frac{{2 + \sqrt 3 }}{{4 - 3}} \cr & \Rightarrow 2 + \sqrt 3 \cr}
By rationalisation of denominator
⇒ a + b$$\sqrt 3$$ = 2 + $$\sqrt 3$$
⇒ Now compare the rational & irrational parts
a = 2
b = 1
So, a : b
2 : 1

6. If $${\text{A}}$$ : $${\text{B}}$$ = $$\frac{1}{2}$$ : $$\frac{3}{8}{\text{,}}$$  $${\text{B}}$$ : $${\text{C}}$$ = $$\frac{1}{3}$$ : $$\frac{5}{9}$$ and $${\text{C}}$$ : $${\text{D}}$$ = $$\frac{5}{6}$$ : $$\frac{3}{4}{\text{,}}$$  Then the ratio of A : B : C : D is ?
a) 6 : 4 : 8 : 10
b) 6 : 8 : 9 : 10
c) 8 : 6 : 10 : 9
d) 4 : 6 : 8 : 10

Explanation:
\eqalign{ & {\text{A}}:{\text{B}} = \frac{1}{2}:\frac{3}{8}, \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{4}{3} \cr & {\text{B}}:{\text{C}} = \frac{1}{3}:\frac{5}{9}, \cr & \frac{{\text{B}}}{{\text{C}}} = \frac{1}{3} \times \frac{9}{5} = \frac{3}{5} \cr & {\text{C}}:{\text{D}} = \frac{5}{6}:\frac{3}{4}, \cr & \frac{{\text{C}}}{{\text{D}}} = \frac{{5 \times 4}}{{6 \times 3}} = \frac{{10}}{9} \cr & {\text{A}}:{\text{B}}:{\text{C}}:{\text{D}} \cr & {\text{4}}:{\text{3}} \cr & \,\,\,\,\,\,\,{\text{3}}:{\text{5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ 10}}:{\text{9}} \cr & \Rightarrow {\text{A}}:{\text{B}}:{\text{C}}:{\text{D}} \cr & \,\,\,\,\,\,\,\,\,8:6:10:9 \cr}

7.If A : B : C = 2 : 3 : 4, then ratio $$\frac{{\text{A}}}{{\text{B}}}$$ : $$\frac{{\text{B}}}{{\text{C}}}$$ : $$\frac{{\text{C}}}{{\text{A}}}$$ is equal to?
a) 8 : 9 : 16
b) 8 : 9 : 12
c) 8 : 9 : 24
d) 4 : 9 : 12

Explanation:
\eqalign{ & {\text{A}}:{\text{B}}:{\text{C}} = 2:3:4 \cr & {\text{Let }} \cr & {\text{A}} = 2x,{\text{B}} = 3x,{\text{C}} = 4x \cr & \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} = \frac{{2x}}{{3x}}:\frac{{3x}}{{4x}}:\frac{{4x}}{{2x}} \cr & \Rightarrow \frac{2}{3}:\frac{3}{4}:\frac{2}{1} \cr}
Multiply by the L.C.M of denominator to remove fraction
\eqalign{ & {\text{L}}{\text{.C}}{\text{.M of }}\left( {3,4,1} \right) = 12 \cr & \frac{{\text{A}}}{{\text{B}}}\,\,\,\,\,:\,\,\,\,\,\frac{{\text{B}}}{{\text{C}}}\,\,\,\,\,\,:\,\,\,\,\,\frac{{\text{C}}}{{\text{A}}} \cr & \frac{2}{3} \times 12:\frac{3}{4} \times 12:\frac{2}{1} \times 12 \cr & \,\,\,\,\,\,\,8\,\,\,\,\,\,\,:\,\,\,\,\,\,\,9\,\,\,\,\,\,\,:\,\,\,\,\,\,\,24 \cr}

8. If A : B = 2 : 3, B : C = 2 : 4 and C : D = 2 : 5, then A : D is equal to -
a) 1 : 5
b) 2 : 5
c) 3 : 5
d) 2 : 15

Explanation:
\eqalign{ & = \frac{{\text{A}}}{{\text{B}}} = \frac{2}{3},\,\frac{{\text{B}}}{{\text{C}}} = \frac{2}{4},\,\frac{{\text{C}}}{{\text{D}}} = \frac{2}{5} \cr & \frac{{\text{A}}}{{\text{D}}} = \left( {\frac{{\text{A}}}{{\text{B}}} \times \frac{{\text{B}}}{{\text{C}}} \times \frac{{\text{C}}}{{\text{D}}}} \right) \cr & \frac{2}{3} \times \frac{2}{4} \times \frac{2}{5} = \frac{2}{{15}} \cr & {\text{A}}:{\text{D}} = 2:15. \cr}

9.If 3A = 5B and 4B = 6C, then A : C equal to -
a) 2 : 5
b) 3 : 5
c) 4 : 5
d) 5 : 2

\eqalign{ & = {\text{3A}} = {\text{5B and 4B}} = {\text{6C}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{5}{3}{\text{ and }}\frac{{\text{B}}}{{\text{C}}} = \frac{6}{4} = \frac{3}{2} \cr & \frac{{\text{A}}}{{\text{C}}} = \left( {\frac{{\text{A}}}{{\text{B}}} \times \frac{{\text{B}}}{{\text{C}}}} \right) = \frac{5}{3} \times \frac{3}{2} = \frac{5}{2} \cr & {\text{A}}:{\text{C}} = {\text{5}}:{\text{2}} \cr}
\eqalign{ & = \frac{{\text{A}}}{{\text{B}}} = \frac{2}{1},\,\,\frac{{\text{B}}}{{\text{C}}} = \frac{4}{3},\,\,\frac{{\text{C}}}{{\text{D}}} = \frac{5}{6} \cr & \Rightarrow \frac{{\text{A}}}{{\text{D}}} = \left( {\frac{{\text{A}}}{{\text{B}}} \times \frac{{\text{B}}}{{\text{C}}} \times \frac{{\text{C}}}{{\text{D}}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{2}{1} \times \frac{4}{3} \times \frac{5}{6}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{9} \cr & {\text{A}}:{\text{D}} = {\text{20}}:{\text{9}} \cr}