Ratio - Aptitude MCQ Questions and Solutions with Explanations

Let the 2^nd amount be Rs. x
$$\eqalign{ & {\text{Then,}} \cr & {\text{A's share from }}{{\text{1}}^{{\text{st}}}}{\text{ amount}} \cr & = {\text{Rs}}{\text{.}}\left( {150000 \times \frac{2}{{10}}} \right) \cr & = {\text{Rs}}{\text{. }}30000. \cr & {\text{A's share from }}{{\text{2}}^{{\text{nd}}}}{\text{ amount}} \cr & = {\text{Rs}}{\text{.}}\left( {x \times \frac{5}{{10}}} \right) \cr & = {\text{Rs}}{\text{. }}\frac{x}{2}. \cr & \therefore \frac{x}{2} = 30000\,{\text{or}}\,x = 60000 \cr & {\text{Required ratio}} \cr & = 150000:60000 \cr & = 5:2 \cr} $$

Quantity of copper in 17.5kg of alloy
$$\eqalign{ & = \left( {17.5 \times \frac{5}{7}} \right){\text{kg}} \cr & = 12.5{\text{kg}}. \cr} $$
Quantity of zinc in 17.5kg of alloy
$$\eqalign{ & = \left( {17.5 \times \frac{2}{7}} \right){\text{kg}} \cr & = 5{\text{kg}} \cr & \therefore {\text{Required ratio}} \cr & = 12.5:\left( {5 + 1.25} \right) \cr & = 12.5:6.25 \cr & = 2:1 \cr} $$

Let marks in English, Math and Science are E, M, S respectively
$$\eqalign{ & {\text{Given,}} \cr & {\text{2S}} = {\text{E}} \cr & {\text{S}}:{\text{E}} = 1:2 \cr & {\text{E}}:{\text{M}} = 2:3 \cr & {\text{S}}:{\text{E}}:{\text{M}} \cr & 1:2 \cr & \,\,\,\,\,\,\,\,\,2:3 \cr & \overline {1:2:3} \cr & {\text{Let }}x:2x:3x \cr & \therefore x + 2x + 3x = 180 \cr & \Rightarrow 6x = 180 \cr & \Rightarrow x = 30 \cr & {\text{Marks in Science}} \cr & = 1 \times 30 = 30 \cr & {\text{Marks in English}} \cr & = 2 \times 30 = 60 \cr & {\text{Marks in Maths}} \cr & = 3 \times 30 = 90 \cr} $$

Ratios are 4:3 (4+3=7) and 2:3 (2+3=5). So LCM of 7, 5 is 35.
Now, for a 35-liter mixture.
Let M=milk, W=water

Vessel A,
M : W = 4 : 3 (4+3=7) means in 35-liter mixture. Milk is 4 × $$\frac{{{\text{35}}}}{7}$$   =   20 liter and water is 3 × $$\frac{{{\text{35}}}}{7}$$   =   15 liter.

Vessel B,
M : W = 2 : 3 (2+3=5) means in 35-liter mixture. Milk is 2 × $$\frac{{{\text{35}}}}{5}$$   =  14 liter and water is $$\frac{{{\text{35}}}}{5}$$   =   21 liter.

For Vessel C,
M : W = 1 : 1 (required)
Let the ratio of Vessel A = x and Vessel B = y.
In-Vessel C, both milk and water will be in equal quantity. So,
Net quantity of milk= Net quantity of water
⇒ 20x + 14y = 15x + 21y
⇒ 5x = 7y
⇒ $$\frac{x}{y}$$ = $$\frac{7}{5}$$
So, x : y = 7 : 5
Vessel A : Vessel B = 7 : 5

$$\eqalign{ & {\text{A}}:{\text{B}} = \frac{a}{b}:\frac{c}{d} \cr & = \left( {\frac{a}{b} \times bd} \right):\left( {\frac{c}{d} \times bd} \right) \cr & = ad:bc \cr & \therefore {\text{A's share}} \cr & = Rs.\left( {\frac{{adx}}{{ad + bc}}} \right) \cr} $$

$$\eqalign{ & {\text{A}}:{\text{B}} = 3:7, \cr & {\text{B}}:{\text{C}} = 6:5, \cr & = \left( {6 \times \frac{7}{6}} \right):\left( {5 \times \frac{7}{6}} \right) \cr & = 7:\frac{{35}}{6} \cr & {\text{A}}:{\text{B}}:{\text{C}} = 3:7:\frac{{35}}{6} \cr & = 18:42:35. \cr & {\text{Sum of ratio terms}} \cr & = \left( {18 + 42 + 35} \right) \cr & = 95 \cr & \therefore {\text{B's share}} = \cr & {\text{Rs}}{\text{.}}\left( {33630 \times \frac{{42}}{{95}}} \right) \cr & = {\text{Rs}}{\text{. }}14868 \cr} $$

Given:
⇒ Let the alloys be mixed in the ratio of $$x$$ : $$y$$ (Assumption)
⇒ In 1st alloy, Copper = $$\frac{x}{4}$$
⇒ In 1st alloy, Tin = $$\frac{3x}{4}$$
⇒ In 2nd alloy, Copper = $$\frac{2y}{7}$$
⇒ In 2nd alloy, Tin = $$\frac{5y}{7}$$

To find:
⇒ The ratio in which 2 alloys must be mixed to get a new alloy with a ratio of copper and tin be 3 : 8 =?

Now we have,
$$\eqalign{ & \left( {\frac{x}{4} + \frac{{2y}}{7}} \right) : \left( {\frac{{3x}}{4} + \frac{{5y}}{7}} \right) = 3 : 8 \cr & \Rightarrow \frac{{\left( {\frac{x}{4} + \frac{{2y}}{7}} \right)}}{{ \left( {\frac{{3x}}{4} + \frac{{5y}}{7}} \right) }} = \frac{3}{8} \cr & \Rightarrow \frac{{\frac{{7x + 14y}}{{28}}}}{{ \frac{{21x + 20y}}{{28}} }} = \frac{3}{8} \cr & \Rightarrow \frac{{7x + 14y}}{{21x + 20y}} = \frac{3}{8} \cr & \Rightarrow 56x + 64y = 63x + 60y \cr & \Rightarrow 64y - 60y = 63x - 56x \cr & \Rightarrow 4y = 7x \cr & \therefore \frac{x}{y} = \frac{4}{7} \cr} $$

⇒ Ratio in which 2 alloys must be mixed to get a new alloy with a ratio of copper and tin be 3 : 8 = 4 : 7

	Alcohol	:	Water
	4	:	3
Let	4x	:	3x

∴ 5 litres of water is added to the mixture
$$\eqalign{ & \Rightarrow \frac{{4x}}{{3x + 5}} = \frac{4}{5} \cr & \Rightarrow 20x = 12x + 20 \cr & \Rightarrow 8x = 20 \cr & \Rightarrow x = \frac{{20}}{8} \cr & \Rightarrow x = \frac{5}{2} \cr & {\text{Quantity of alcohol}} \cr & \Rightarrow {\text{4}} \times \frac{5}{2} = 10\,{\text{litres}} \cr} $$

	Zinc	:	Tin
A	5x	:	2x	=	7x
B	3y	:	4y	=	7y

⇒ A ⇒ 7x = 7 kg
x = 1 kg
∴ Zinc in alloy A ⇒ 5kg
Tin in alloy A ⇒ 2 kg
⇒ B ⇒ 7y = 21 kg
y = 3 kg
Zinc in alloy B ⇒ 3 × 3 = 9 kg
Tin in alloy B ⇒ 3 × 4 = 12 kg
∴ After mix - up the ratio of Zinc and Tin in new alloy
$$\eqalign{ & {\text{ Zinc}}:{\text{Tin}} \cr & \,\,\,{\text{A}}\,5\,\,:\,\,\,\,\,2 \cr & \,\,\,{\text{B}}\,9\,\,\,:\,\,\,12 \cr & \overline {{\text{A}} + {\text{B}}:{\text{14}}\,\,\,\,\,{\text{14}}} \cr & \,\,\,\,\boxed{1\,\,\,\,\,:\,\,\,\,\,1} \cr} $$

$$\eqalign{ & {\text{A}}:{\text{B}} = 3:2 \cr & {\text{B}}:{\text{C = }}3:2 \cr & = \left( {3 \times \frac{2}{3}} \right):\left( {2 \times \frac{2}{3}} \right) \cr & = 2:\frac{4}{3} \cr & {\text{A}}:{\text{B}}:{\text{C}} = 3:2:\frac{4}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9:6:4 \cr & \therefore {\text{A's score}} = \left( {361 \times \frac{9}{{19}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 171 \cr} $$