Ratio - Aptitude MCQ Questions and Solutions with Explanations

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

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

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

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

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

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

Given, A : B = 3 : 4
B : C = 8 : 9
A : B = 3 : 4 (multiply with 2)
i.e. A : B = 6 : 8 and B : C = 8 : 9
∴ A : B : C = 6 : 8 : 9

$$\eqalign{ & \,\,\,{\text{A}}\,\,\,\,\,\,\,{\text{:}}\,\,\,\,\,\,{\text{B}}\,\,\,\,\,\,{\text{:}}\,\,\,\,\,{\text{C}} \cr & 3 \times 4:2 \times 4:2 \times 3 \cr & \,\,\,12\,\,\,\,:\,\,\,\,\,{\text{8}}\,\,\,\,\,:\,\,\,\,{\text{6}} \cr & \,\,\,\boxed{\,\,6\,\,\,\,:\,\,\,\,4\,\,\,\,\,\,:\,\,\,\,3\,\,} \cr} $$

$$\eqalign{ & {\text{Let}}\,\frac{x}{y} = \frac{y}{z} = \frac{z}{u} = k. \cr & {\text{Now, }}\frac{x}{u} = \frac{{64}}{{27}} \cr & \Rightarrow \frac{x}{y} \times \frac{y}{z} \times \frac{z}{u} = \frac{{64}}{{27}} \cr & \Rightarrow {k^3} = {\left( {\frac{4}{3}} \right)^3} \cr & \Rightarrow k = \frac{4}{3}. \cr & {\text{So}},\,x:y = y:z = z:u = 4:3. \cr & \therefore \frac{x}{z} = \frac{x}{y} \times \frac{y}{z} = \frac{4}{3} \times \frac{4}{3} = \frac{{16}}{{9.}} \cr} $$

$$\eqalign{ & p = k,\,q = 2k,\,r = 4k \cr & {\text{Then,}} \cr & \sqrt {5{p^2} + {q^2} + {r^2}} \cr & = \sqrt {5{k^2} + {{\left( {2k} \right)}^2} + {{\left( {4k} \right)}^2}} \cr & = \sqrt {5{k^2} + 4{k^2} + 16{k^2}} \cr & = \sqrt {25{k^2}} \cr & = 5k \cr & = 5p. \cr} $$

$$\eqalign{ & {\text{let A}} = 2{\text{k}},\,{\text{B}} = 3{\text{k,}}\,{\text{C}} = 4{\text{k}} \cr & {\text{Then,}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{2{\text{k}}}}{{3{\text{k}}}} = \frac{2}{3}, \cr & \Rightarrow \frac{{\text{B}}}{{\text{C}}} = \frac{{3{\text{k}}}}{{4{\text{k}}}} = \frac{3}{4}, \cr & \Rightarrow \frac{{\text{C}}}{{\text{A}}} = \frac{{4{\text{k}}}}{{2{\text{k}}}} = 2 \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} = \frac{2}{3}:\frac{3}{4}:2 \cr & = 8:9:24 \cr} $$