Ratio — Part 2 MCQs with Explanations

Ratio of students in 2012 in institutes A and B = 7 : 15 Let number of students in institute A in 2012 = 700 And Number of students in institutes B in 2012 = 1500 25% increase in the number of students in 2013 Now, number of students in Institute A = 700 + 25% of 700 = 875 Number of students in B in 2013 as 26% students increased in B = 1500 + 26% of 1500 = 1890 Current Ratio of the students, $$ = \frac{{875}}{{1890}} = 25:54$$

$$\eqalign{ & \frac{4}{{15}}A = \frac{2}{5}B \cr & \Rightarrow A = \left( {\frac{2}{5} \times \frac{{15}}{4}} \right)B \cr & \Rightarrow A = \frac{3}{2}B \cr & \Rightarrow \frac{A}{B} = \frac{3}{2} \cr & \Rightarrow A:B = 3:2. \cr & \therefore {\text{B's}}\,{\text{share}} = Rs.\left( {1210 \times \frac{2}{5}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,484 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{third}}\,{\text{number}}\,{\text{be}}\,x \cr & {\text{Then,}}\,{\text{first}}\,{\text{number}} \cr & = 120\% \,{\text{of}}\,x = \frac{{120x}}{{100}} = \frac{{6x}}{5} \cr & {\text{Second}}\,{\text{number}} \cr & = 150\% \,{\text{of}}\,x = \frac{{150x}}{{100}} = \frac{{3x}}{2} \cr & \therefore {\text{Ratio}}\,{\text{of}}\,{\text{first}}\,{\text{two}}\,{\text{numbers}} \cr & = {\frac{{6x}}{5}:\frac{{3x}}{2}} = 12x:15x = 4:5 \cr} $$

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively Then, 4x - 3x = 1000 ⇒ x = 1000 ∴ B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x) $$ = \left( {\frac{{140}}{{100}} \times 5x} \right),$$ $$\left( {\frac{{150}}{{100}} \times 7x} \right)$$ and $$\left( {\frac{{175}}{{100}} \times 8x} \right)$$ $$\eqalign{ & = 7x,\,\,\frac{{21x}}{2}{\text{ and }}14x \cr & \therefore {\text{The required ratio}} \cr & = 7x:\frac{{21x}}{2}:14x \cr & = 14x:21x:28x \cr & = 2:3:4 \cr} $$

Quantity of milk $$ = \left( {60 \times \frac{2}{3}} \right)\,{\text{litres}} = 40\,{\text{litres}}$$ Quantity of water in it = (60 - 40) litres = 20 litres New ratio = 1 : 2 Let quantity of water to be added further be x litres $$\eqalign{ & {\text{Then,}}\,{\text{milk : water}} = {\frac{{40}}{{20 + x}}} \cr & {\text{Now}},\, {\frac{{40}}{{20 + x}}} = \frac{1}{2} \cr & \Rightarrow 20 + x = 80 \cr & \Rightarrow x = 60 \cr & \therefore {\text{Quantity}}\,{\text{of}}\,{\text{water}}\,{\text{to}}\,{\text{be}}\,{\text{added = 60}}\,{\text{litres}}{\text{.}} \cr} $$

Originally, let the number of boys and girls in the college be 7x and 8x respectively Their increased number is (120% of 7x) and (110% of 8x) $$\eqalign{ & \Rightarrow \left( {\frac{{120}}{{100}} \times 7x} \right)\,{\text{and}}\,\left( {\frac{{110}}{{100}} \times 8x} \right) \cr & \Rightarrow \frac{{42x}}{5}\,{\text{and}}\,\frac{{44x}}{5} \cr & \therefore {\text{The}}\,{\text{required}}\,{\text{ration}} \cr & = {\frac{{42x}}{5}:\frac{{44x}}{5}} \cr & = 21:22 \cr} $$

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively $${\text{Then}},\,\frac{{2x + 4000}}{{3x + 4000}} = \frac{{40}}{{57}}$$ ⇒ 57(2x + 4000) = 40(3x + 4000) ⇒ 6x = 68,000 ⇒ 3x = 34,000 Sumit's present salary = (3x + 4000) = Rs. (34000 + 4000) = Rs. 38,000

$$\eqalign{ & {x \times 5} = {0.75 \times 8} \cr & \Rightarrow x = \frac{6}{5} \cr & \Rightarrow x = 1.2 \cr} $$

Let the three parts be A, B, C. Then, A : B = 2 : 3 and B : C = 5 : 8 $$\eqalign{ & = \left( {5 \times \frac{3}{5}} \right):\left( {8 \times \frac{3}{5}} \right) = 3:\frac{{24}}{5} \cr & \Rightarrow A:B:C = 2:3:\frac{{24}}{5} = 10:15:24 \cr & \Rightarrow B = {98 \times \frac{{15}}{{49}}} = 30 \cr} $$

$$\eqalign{ & \left( {\frac{1}{2}:\frac{2}{3}:\frac{3}{4}} \right) \times 12 \cr & \Rightarrow 6:8:9 \cr & \therefore {1^{{\text{st}}}}{\text{part's}}\,{\text{share}} \cr & = {\text{Rs}}{\text{.}}\,\left( {\frac{6}{{23}} \times 782} \right) \cr & = {\text{Rs}}{\text{.}}\,6 \times 34 \cr & = {\text{Rs}}{\text{.}}\,204 \cr} $$

$$\eqalign{ & {\text{Let}}, \cr & A = 2k \cr & B = 3k\,{\text{and}} \cr & C\, = 5k \cr & A's\,{\text{new}}\,{\text{salary}} \cr & = \frac{{115}}{{100}}\,of\,2k = {\frac{{115}}{{100}} \times 2k} = \frac{{23k}}{{10}} \cr & B's\,{\text{new}}\,{\text{salary}} \cr & = \frac{{110}}{{100}}\,of\,3k = {\frac{{110}}{{100}} \times 3k} = \frac{{33k}}{{10}} \cr & C's\,{\text{new}}\,{\text{salary}} \cr & = \frac{{120}}{{100}}\,of\,5k = {\frac{{120}}{{100}} \times 5k} = 6k \cr & \therefore {\text{New}}\,{\text{ratio}} \cr & = {\frac{{23k}}{{10}}:\frac{{33k}}{{10}}:6k} \cr & = 23:33:60 \cr} $$

$$\eqalign{ & {\text{Let}}\,40\% \,{\text{of}}\,A = \frac{2}{3}B \cr & {\text{Then}},\,\frac{{40A}}{{100}} = \frac{{2B}}{3} \cr & \Rightarrow \frac{{2A}}{5} = \frac{{2B}}{3} \cr & \Rightarrow \frac{A}{B} = {\frac{2}{3} \times \frac{5}{2}} = \frac{5}{3} \cr & \therefore A:B = 5:3 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fourth}}\,{\text{proportional}}\,{\text{to}}\,5,\,8,\,15\,{\text{be}}\,x \cr & {\text{Then}},\,5:8:15:x \cr & \Rightarrow 5x = {8 \times 15} \cr & x = \frac{{ {8 \times 15} }}{5} = 24 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{number}}\,{\text{be}}\,3x\,{\text{and}}\,5x \cr & {\text{Then}},\,\frac{{3x - 9}}{{5x - 9}} = \frac{{12}}{{23}} \cr & \Rightarrow 23\left( {3x - 9} \right) = 12\left( {5x - 9} \right) \cr & \Rightarrow 9x = 99 \cr & \Rightarrow x = 11 \cr & \therefore {\text{The}}\,{\text{smaller}}\,{\text{number}} \cr & = {3 \times 11} \cr & = 33 \cr} $$

Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively Then, sum of their values $$\eqalign{ & = Rs.\,\left( {\frac{{25x}}{{100}} + \frac{{10 \times 2x}}{{100}} + \frac{{5 \times 3x}}{{100}}} \right) \cr & = Rs.\,\frac{{60x}}{{100}} \cr & \therefore \frac{{60x}}{{100}} = 30 \Leftrightarrow x = \frac{{30 \times 100}}{{60}} = 50 \cr & {\text{Hence,}}\,{\text{the}}\,{\text{number}}\,{\text{of}}\,{\text{5p}}\,{\text{coins}} \cr & = \left( {3 \times 50} \right) \cr & = 150 \cr} $$

Rs. 2.80 = 280 paise ∴ Required ration = 280 : 40 = 7 : 1

Maximum possible number of tables = 6 [∵ 1200 × 6 = 7200] Number of chairs purchased $$\eqalign{ & {\text{ = }}\frac{{{\text{8100}} - {\text{7200}}}}{{{\text{300}}}}{\text{ = }}\frac{{{\text{900}}}}{{{\text{300}}}}{\text{ = 3}}{\text{}} \cr & {\text{Hence,}} \cr & {\text{Required ratio}} = {\text{3}}:{\text{6}} \cr & {\text{ = 1}}:{\text{2}} \cr} $$

$$\eqalign{ & x = \frac{1}{3}y\,\,{\text{and }}y = \frac{1}{2}z \cr & \frac{x}{y} = \frac{1}{3}\,{\text{and }}\frac{y}{z} = \frac{1}{2} \cr & x:y = 1:3 \cr & y:z = 1:2 \times 3\,({\text{multiply}}) \cr & i.e.\,y:z = 3:6 \cr & \therefore x:y:z = 1:3:6 \cr} $$

$$\eqalign{ & x:y = 3:1 \cr & \therefore \frac{x}{y} = \frac{3}{1} \cr & \therefore \frac{{{x^3} - {y^3}}}{{{x^3} + {y^3}}} \cr & \Rightarrow \frac{{{y^3}\left( {\frac{{{x^3}}}{{{y^3}}} - 1} \right)}}{{{y^3}\left( {\frac{{{x^3}}}{{{y^3}}} + 1} \right)}} \cr & {\text{taking }}{{\text{y}}^3}{\text{ common}} \cr & {\text{ = }}\frac{{\frac{{{x^3}}}{{{y^3}}} - 1}}{{\frac{{{x^3}}}{{{y^3}}} + 1}} \cr & \Rightarrow \frac{{27 - 1}}{{27 + 1}} \cr & \Rightarrow \frac{{26}}{{28}} \cr & \Rightarrow \frac{{13}}{{14}} \cr} $$

$$\eqalign{ & {\text{60}}\% \,{\text{of}}\,{\text{A}} = \frac{{\text{3}}}{{\text{4}}}\,{\text{of}}\,{\text{ B}} \cr & \Rightarrow \frac{{60}}{{100}}{\text{A}} = \frac{3}{4}\,{\text{B}} \cr & \Rightarrow \frac{{\text{3}}}{{\text{5}}}{\text{A = }}\frac{{\text{3}}}{{\text{4}}}\,{\text{B}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}}{\text{ = }}\frac{{\text{3}}}{{\text{4}}} \times \frac{{\text{5}}}{{\text{3}}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{5}{4} \cr} $$ ⇒ A : B = 5 : 4

A = 8 : a = 8 : b ⇒ 8a = 8b ⇒ a = b. B = a : 16 = b : 4 ⇒ 4a = 16b ⇒ a = 4b C = a : 8 = b : 8 ⇒ 8a = 8b ⇒ b = a D = 32 : a = b : 2 ⇒ ab =64

The total number of members must be a multiple of the sum ratio terms. 3 + 2 = 5 and 25 is a multiple of 5

$$\eqalign{ & m:n = 3:2 \cr & \frac{m}{n} = \frac{3}{2} \cr & \Rightarrow \frac{{4m + 5n}}{{4m - 5n}} \cr & \Rightarrow \frac{{n\left( {4\frac{m}{n} + 5} \right)}}{{n\left( {4\frac{m}{n} - 5} \right)}} \cr & \Rightarrow \frac{{4 \times \frac{3}{2} + 5}}{{4 \times \frac{3}{2} - 5}} \cr & \Rightarrow \frac{{6 + 5}}{{6 - 5}} \cr & \Rightarrow \frac{{11}}{1} \cr & \Rightarrow 11:1 \cr} $$

A + B = 40 A - B = 4 ∴ A = 22 B = 18 A : B = 22 : 18 = 11 : 9

Each exterior angle of n sided polygon is $${\text{ = }}\left( {\frac{{360}}{n}} \right)$$ And each interior angle of n sided polygon $$\eqalign{ & {\text{ = }}\frac{{\left( {n - 2} \right) \times 180}}{n} \cr & \therefore \frac{{\frac{{\left( {n - 2} \right) \times 180}}{n}}}{{\frac{{360}}{n}}} = \frac{2}{1} \cr & \Rightarrow \frac{{\left( {n - 2} \right)}}{2} = 2 \cr & \Rightarrow n - 2 = 4 \cr & \Rightarrow n = 6 \cr} $$

$$\eqalign{ & \Rightarrow 7:15 = \frac{7}{{15}} = 0.466 \cr & \Rightarrow 15:23 = \frac{{15}}{{23}} = 0.652 \cr & \Rightarrow 17:25 = \frac{{17}}{{25}} = 0.68 \cr & \Rightarrow 21:39 = \frac{{21}}{{39}} = 0.538 \cr} $$ Clearly, 7 : 15 is the lowest

Let the missing number be x. Then, $$\eqalign{ & = \frac{1}{3}:\frac{1}{2}::27y:x \cr & \Rightarrow \frac{3}{4}x = \frac{1}{2} \times {\text{27}}y \cr & \Rightarrow x{\text{ = }}\frac{{27y}}{2} \times \frac{4}{3} \cr & = 18y \cr} $$

$$\eqalign{ & {\text{Harsha = 40}} \cr & {\text{Ritu = 60}} \cr & \frac{{{\text{Harsha}}}}{{{\text{Ritu}}}} = \frac{{40}}{{60}} \cr} $$ ∴ Let x years ago their ages ratio was 3 : 5 $$\eqalign{ & \Rightarrow \frac{{40 - x}}{{60 - x}} = \frac{3}{5} \cr & \Rightarrow 200 - 5x = 180 - 3x \cr & \Rightarrow 2x = 20 \cr & \Rightarrow x = 10 \cr} $$

$$\eqalign{ & {\text{A : B}} \cr & \frac{3}{2}:\frac{8}{3} \cr} $$ take L.C.M. of denominator and multiply) $$ \Rightarrow \frac{3}{2} \times 6:\frac{8}{3} \times 6 = 9x:16x$$ After adding 15 in each we get $$\eqalign{ & \therefore \frac{{9x + 15}}{{16x + 15}} = \frac{5}{3} \times \frac{2}{5} \cr & \Rightarrow \frac{{9x + 15}}{{16x + 15}} = \frac{2}{3}\left( {{\text{cross multiply}}} \right) \cr & \Rightarrow 27x + 45 = 32x + 30 \cr & \Rightarrow 5x = 15 \cr & \Rightarrow x = 3 \cr & \therefore {\text{Smaller number}} \cr & = {\text{9}} \times {\text{3 = 27}} \cr & {\text{Greater number}} \cr & {\text{ = 16}} \times {\text{3 = 48}} \cr} $$

Given, A : B = 6 : 5 B : C = 10 : 9 A : B = 6 : 5 (multiply with 2) i.e. A : B = 12 : 10 ∴ A : B : C = 12 : 10 : 9 ⇒ 12x + 10x + 9x = 1280 ⇒ x = 40 ∴ Share of C = 9 × 40 = 360

$$\eqalign{ & = x:7.5 = 7:17.5 \cr & \Rightarrow 17.5x = 7.5 \times 7 \cr & \Rightarrow x = \frac{{7.5 \times 7}}{{17.5}} \cr & = 3 \cr} $$

$$\eqalign{ & = x:2\frac{1}{3}::21:50 \cr & \Rightarrow 50x = \frac{7}{3} \times 21 \cr & \Rightarrow 50x = 49 \cr & \Rightarrow x = \frac{{49}}{{50}} \cr} $$

$$\eqalign{ & = \sqrt 2 :\left( {1 + \sqrt 3 } \right)::\sqrt 6 :x \cr & \Rightarrow \sqrt {2}x = \sqrt 6 \left( {1 + \sqrt 3 } \right) \cr & \Rightarrow x = \frac{{\sqrt 6 \left( {1 + \sqrt 3 } \right)}}{{\sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \left( {1 + \sqrt 3 } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 + 3 \cr} $$

The ratio of copper and zinc is 5 : 2 i.e 5x : 2x Quantity of initial mixture = 17.5 kg ∴ 5x + 2x = $$\frac{35}{2}$$ ⇒ x = $$\frac{5}{2}$$ Quantity of Copper in initial mixture $$\eqalign{ & = \frac{5}{2} \times 5 \cr & = \frac{{25}}{2}kg \cr} $$ Quantity of Zinc in initial mixture $$\eqalign{ & = \frac{5}{2} \times 2 \cr & = 5\,kg \cr} $$ Now after adding 1.250 kg of zinc = 5 + 1.250 = 6.250 kg = $$\frac{{25}}{4}kg$$ ∴ New Ratio of copper : zinc $$\eqalign{ & = \frac{{25}}{2}:\frac{{25}}{4} \cr & = 2:1 \cr} $$

Let income of A, B and C are 7x, 9x and 12x respectively and expenditure of A, B and C are 8y, 9y and 15y respectively $$\eqalign{ & \Rightarrow {\text{Income}}\,{\text{of,}} \cr & A \times \frac{1}{4} = {\text{Saving}}\,{\text{of}}\,{\text{A}}\,\left( {{\text{given}}} \right) \cr & \Rightarrow 7x - 8y = 7x \times \frac{1}{4} \cr & \Rightarrow 28x - 32y = 7x \cr & \Rightarrow 21x = 32y \cr & \Rightarrow x:y = 32:21 \cr} $$ ∴ The ratio of savings of A, B and C ⇒ (7x - 8y) : (9x - 9y) : (12x - 15y) ⇒ (7 × 32 - 8 × 21) : (9 × 32 - 9 × 21) : (12 × 32 - 15 × 21) ⇒ (224 - 168) : (288 - 189) : (384 - 315) ⇒ 56 : 99 : 69

1 day = 24 hours. ∴ Given ration = 3 : 24 = 1 : 8

Product of 1 st and 4 th terms (extremes) = product of 2 nd and 3 rd terms (means) $$\eqalign{ & \Rightarrow {\text{2}}{\text{.5}}x = {\text{40}} \cr & \Rightarrow x = \frac{{40}}{{2.5}} = 16 \cr} $$

$$\eqalign{ & 20\% \,{\text{of A}} = 30\% \,{\text{of B}} = \frac{1}{6}{\text{of C}} \cr & \Rightarrow \frac{{20{\text{A}}}}{{100}} = \frac{{30{\text{B}}}}{{100}} = \frac{{\text{C}}}{{\text{6}}} \cr & \Rightarrow \frac{{\text{A}}}{5} = \frac{{3{\text{B}}}}{{10}} = \frac{C}{6} = k(say) \cr & \Rightarrow A = 5K,\,B = \frac{{10k}}{3},\,C = 6k \cr & \therefore {\text{A}}:{\text{B}}:{\text{C}} = 5k:\frac{{10k}}{3}:6k \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5:\frac{{10}}{3}:6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 15:10:18 \cr} $$

$$\eqalign{ & {\text{A}}:{\text{B}} \cr & 17:45 \cr & 17x:45x \cr & {\text{Given}} \cr & \Rightarrow 17x \times \frac{1}{3} + 15 = 45x \times \frac{1}{5} \cr & \Rightarrow \frac{{17x + 45}}{3} = 9x \cr & \Rightarrow 17x + 45 = 27x \cr & \Rightarrow 10x = 45 \cr & \Rightarrow x = \frac{{45}}{{10}} \cr & \therefore {\text{Smaller number is }} \cr & = \frac{{45}}{{10}} \times 17 \cr & = \frac{{765}}{{10}} \cr & = 76\frac{1}{2} \cr} $$