Ratio - Aptitude MCQ Questions and Solutions with Explanations

Ratio of number of men, women and children,
$$\eqalign{ & = \frac{{18}}{6}:\frac{{10}}{5}:\frac{{12}}{3} \cr & = 3:2:4 \cr} $$
Total (Men + Women + Children) = 18
3X + 2X + 4X = 18
9X = 18
X = 2
Hence, number of women = 2X = 2 × 2 = 4
Share of all women = $$\frac{{10 \times 4000}}{{40}}$$  = Rs. 1000 [18 + 10 + 12 = 40]
Thus, share of each woman = $$\frac{{1000}}{4}$$ = Rs. 250

Ratio of Gold and Copper in Alloy A = 5 : 3
Ratio of Gold and Copper in Alloy B = 5 : 11
Amount of Gold in Alloy A = $$\frac{5}{8}$$
Amount of Gold In Alloy B = $$\frac{5}{{16}}$$
Amount of Copper in A = $$\frac{3}{8}$$
Amount of Copper in B = $$\frac{{11}}{{16}}$$
Amount of Gold In C,
= (Amount of gold in A + Amount of gold in B) = $$\frac{5}{8}$$ + $$\frac{5}{{16}}$$ = $$\frac{{10 + 5}}{{16}}$$ = $$\frac{{15}}{{16}}$$
Amount of Copper in C,
= Amount of Copper in A + Amount of Copper in B = $$\frac{3}{8}$$ + $$\frac{{11}}{{16}}$$ = $$\frac{{17}}{{16}}$$
So,
Ratio of Gold and Copper in C,
$$ = \frac{{15}}{{16}}:\frac{{17}}{{16}} = 15:17$$

Let X coins of each type of there
Total Value = Rs. 35
Now,
X + $$\frac{{\text{X}}}{2}$$ + $$\frac{{\text{X}}}{4}$$ = 35
4X + 2X + X = 140
7X = 140
X = 20

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively. When 9 litres of mixture are drawn off, quantity of A in mixture left:
$$\eqalign{ & = \left[ {7x - {\frac{7}{{12}}} \times 9} \right]\, \text{litres} \cr & = \left[ {7x - {\frac{{21}}{4}} } \right]\,{\text{litres}} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & = \left[ {5x - {\frac{5}{{12}}} \times 9} \right]\, \text{litres} \cr & = \left[ {5x - {\frac{{15}}{4}} } \right]\,{\text{litres}} \cr & \therefore \,{\text{ratio becomes}}, \cr & \frac{{ {7x - {\frac{{21}}{4}} } }}{{ {\left(5x - {\frac{{15}}{4}}\right)+9 } }} = \frac{7}{9} \cr & \Rightarrow \frac{{ {28x - 21} }}{{ {20x + 21} }} = \frac{7}{9} \cr & \Rightarrow {252x - 189} = 140x + 147 \cr & \Rightarrow 112x = 336 \cr & \Rightarrow x = 3 \cr & {\text{So the can contained}}, \cr & = 7 \times x \cr & = 7 \times 3 \cr & = 21\,{\text{litres of A initially}}{\text{.}} \cr} $$

Let bucket contains 5x and 3x of liquids A and B respectively.
When 16 litres of mixture are drawn off, quantity of A in mixture left:
$$\eqalign{ & {5x - {\frac{5}{8}} \times 16} = {5x - 10} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & {3x - {\frac{3}{8}} \times 16} = {3x - 6} \cr & {\text{Now the ratio becomes,}} \cr & \frac{{ {5x - 10} }}{{ {3x - 6} }} = \frac{3}{5} \cr & \Rightarrow 25x - 50 = 9x - 18 \cr & \Rightarrow 16x = 32 \cr & \Rightarrow x = 2 \cr & {\text{So, quantity of liquid B initially}}, \cr & = 3 \times 2 = 6\,{\text{litres}} \cr} $$

Let the quantity of the wine in the cask originally be x litres.
Using formula:
Final Amount of solute that is not replaced = Initial Amount × $${\left( {\frac{{{\text{Vol}}{\text{. after removal}}}}{{{\text{Vol}}{\text{. after replacing}}}}} \right)^{\text{N}}}$$
Where N = No. of operation done.
Then ratio of wine to total solution in cask after 4 operations,
$$\eqalign{ & 1 \times { {\left( {\frac{{x - 8}}{x}} \right)} ^4} = \frac{{16}}{{81}} \cr & \Rightarrow 1 \times {\left\{ {\frac{{ {x - 8} }}{x}} \right\}^4} = {\left( {\frac{2}{3}} \right)^4} \cr & \Rightarrow \frac{{ {x - 8} }}{x} = \frac{2}{3} \cr & \Rightarrow 3x - 24 = 2x \cr & \Rightarrow x = 24\,{\text{litres}} \cr} $$

Milk	:	Water
7	:	5
7	:	8
		3 unit

∴ Remember water is added and not milk, so make milk equal but here milk is already equal
3 units = 15 litres
1 units = 5 litres
8 units = 40 litres
Total quantity of water in the new mixture = 40 litres

$$\frac{{\text{x}}}{{\text{y}}} = \frac{5}{2}$$
Means x = 5, y = 2
Putting value of x and y in expression
8 × 5 + 9 × 2 = 58
8 × 5 + 2 × 2 = 44
58 : 44 = 29 : 22

Given;
7 jumps of Tom = 5 jumps of Jerry
Or, $$\frac{{{\text{Tom}}}}{{{\text{Jerry}}}} = \frac{5}{7}$$
Let Jerry's 1 leap = 7 meter and Tom's 1 leap = 5 meter
Then, ratio of speed of Tom and Jerry
= $$\frac{{8 \times 5}}{{6 \times 7}}$$
= $$\frac{{40}}{{42}}$$
= 20 : 21

Ratio of Ducks and Frogs in Pond = 37 : 39
Average of Ducks and Frogs in Pond = 152
So, total number of Ducks and Frogs in the Pond = 2 × 152 = 304
∴ Number of Frogs = $$\frac{{304 \times 39}}{{76}}$$  = 156