Practice MCQ Questions and Answer on Ratio

Solution: Let the numbers be 17x and 45x Then, $$\eqalign{ & = \frac{1}{5}{\text{of }}45x - \frac{1}{3}{\text{of }}17x = 15 \cr & \Leftrightarrow 9x - \frac{{17}}{3}x = 15 \cr & \Leftrightarrow \frac{{10x}}{3} = 15 \cr & \Leftrightarrow x = 15 \times \frac{3}{{10}} = \frac{9}{2} \cr & \therefore {\text{Smaller number}} \cr & = 17x = \left( {17x \times \frac{9}{2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{153}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{76}}\frac{1}{2} \cr} $$

Solution: (a2 + b2) : (b2 + c2) : (c2 + a2) = 34 : 61 : 45 Add all 2(a2 + b2 + c2) = 34 + 61 + 45 2(a2 + b2 + c2) = 140 (a2 + b2 + c2) = 70 c2 = 70 - 34 c2 = 36 c = 6 b2 = 70 - 45 b2 = 25 b = 5 a2 = 70 - 61 a2 = 9 a = 3 (b - a) : (c - b) : (c - a) = (5 - 3) : (6 - 5) : (6 - 3) = 2 : 1 : 3

Solution: Let the weights of the three varieties be 12x, 15x and 20x kg respectively Then, Total C.P. = Rs. (100 × 12x + 80 × 15x + 60 × 20x) = Rs. (3600x) S.P. of the first two verities $$\eqalign{ & = {\text{Rs}}.\left( {\frac{{120}}{{100}} \times 2400x} \right) \cr & = {\text{Rs}}.\left( {2880x} \right) \cr & {\text{S}}{\text{.P}}{\text{. of the third variety}} \cr & = {\text{Rs}}.\left( {3600x - 2880x} \right) \cr & = {\text{Rs}}{\text{. }}720x \cr & {\text{Loss on third variety}} \cr & = {\text{Rs}}{\text{.}}\left( {1200x - 720x} \right) \cr & = {\text{Rs}}{\text{. }}480 \cr & {\text{loss }}\% \cr & = \left( {\frac{{480x}}{{1200x}} \times 100} \right)\% \cr & = 40\% \cr} $$

Solution: $$\eqalign{ & = {\text{25}}\% {\text{ of A}} = {\text{35}}\% {\text{ of B}} \cr & \Rightarrow \frac{{25}}{{100}}{\text{A}} = \frac{{35}}{{100}}{\text{B}} \cr & \Rightarrow \frac{{\text{A}}}{4} = \frac{{7{\text{B}}}}{{20}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{7}{{20}} \times 4 = \frac{7}{5} \cr & \Rightarrow {\text{A}}:{\text{B}} = 7:5 \cr} $$

Solution: Let the 2 number be (2x, 3x) According to the question, $$\eqalign{ & \Rightarrow \frac{{2x + 8}}{{3x + 8}} = \frac{3}{4} \cr & \Rightarrow 8x + 32 = 9x + 24 \cr & \Rightarrow x = 8 \cr} $$ Therefore, the sum of the two numbers is = 2x + 3x = 5x = 5 × 8 = 40

Solution: $$\eqalign{ & {\text{Let C gets }}x{\text{ rupees}} \cr & \therefore {\text{B gets }}x + 8{\text{ rupees}} \cr & \therefore {\text{A gets }}x + 8 + 7{\text{ rupees}} \cr & {\text{Total A}} + {\text{B}} + {\text{C}} \cr & \Rightarrow x + 15 + x + 8 + x = 3x + 23 \cr & \Rightarrow 3x + 23 = {\text{Rs}}{\text{. }}53({\text{given)}} \cr & \Rightarrow 3x = 30 \cr & \Rightarrow x = 10 \cr & \Rightarrow x = {\text{Rs}}{\text{.10}} \cr & \therefore {\text{Ratio of their shares is }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{A}}\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,{\text{B}}\,\,\,\,\,\,\,\,:\,\,\,{\text{C}} \cr & {\text{ = }}\left( {x + 15} \right):\left( {x + 8} \right):\left( x \right) \cr & = \,\,\,\,\,\,\,\,\,25\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,18\,\,\,\,\,:\,\,10 \cr} $$

Solution: 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 = $$9 \times \frac{3}{5}$$ = $$5\frac{2}{5}{\text{ litres}}$$

Solution: \[\begin{array}{*{20}{c}} {}&{{\text{Rs}}{\text{. 2}}}&{}&{{\text{50 Paise }}}&{}&{{\text{25 Paise}}} \\ {{\text{Coins}}}&3&:&5&:&7 \end{array}\] Amount = Number of coins × Face value Amount = $$6:\frac{5}{2}:\frac{7}{4}$$ According to question, $$\eqalign{ & \Rightarrow 6x + \frac{{5x}}{2} + \frac{{7x}}{4} = 51.25 \cr & \Rightarrow 24x + 10x + 7x = 51.25 \times 4 \cr & \Rightarrow 41x = 205 \cr & \Rightarrow x = \frac{{205}}{{41}} \cr & \boxed{x = 5} \cr} $$ ∴ Total number of 50 paise coins = 5 × 5 = 25

Solution: Given, blue and yellow color ratio = 4 : 3 and half part blue and yello ratio = 2 : 3 Let 700 be the painting part in total ∴ Blue part will be: $$\frac{{\text{4}}}{7} \times 700$$   = 400 And yellow will be: $$\frac{{\text{3}}}{7} \times 700$$   = 300 Now, upper half part i.e $$\frac{{700}}{2}$$ = 350 Blue in upper half part → $$\frac{{\text{2}}}{5} \times 350$$   = 140 And, Yellow in upper half part → $$\frac{{\text{3}}}{5} \times 350$$   = 210 Remaining half Blue in lower part → 400 - 140 = 260 Remaining half yellow in lower part → 300 - 90 = 210 Ratio of blue and yellow → 260 : 90 ⇒ 26 : 9

Solution: $$\eqalign{ & x:y = 3:4 \cr & \therefore \frac{x}{y} = \frac{3}{4} \cr & \Rightarrow \frac{{4x - y}}{{2x + 3y}} \cr & \Rightarrow \frac{{y\left( {4\frac{x}{y} - 1} \right)}}{{y\left( {2\frac{x}{y} + 3} \right)}} \cr & \Rightarrow \frac{{4 \times \frac{3}{4} - 1}}{{2 \times \frac{3}{4} + 3}} \cr & \Rightarrow \frac{{\left( {3 - 1} \right) \times 2}}{{3 + 6}} \cr & \Rightarrow \frac{4}{9} \cr & \Rightarrow 4:9 \cr} $$