A vessel contains 125 litres of wine. 25 litres of wine was taken out of the vessel and replaced by water. Then. 25 litres of mixture was withdrawn and again replaced by water. The operation was repeated for the third time. How much wine is now left in the vessel?
Amount of wine left in the vessel
\[=\left(1-\frac{25}{125}\right)^3\times 125\]
\[=\frac{100\times 100\times 100\times 125}{125\times 125\times 125}\]
\[=64\] litres
Two barrels a contain a mixture of ethanol and gasoline. The content of ethanol is 60% in the first barrel and 30% in the second barrel. In what ratio must the mixtures from the first and the second barrels be taken to form a mixture containing 50% ethanol?
Applying alligation
Hence, the required ratio is 2 : 1
In a glass of milk, the proportion of pure milk and water is 3 : 1, how much of the mixture must be withdrawn and substituted by the water so that the resulting mixture may become half pure and half milk?
Milk and water ratio = 3 : 1
\[\therefore\] Total = 4
\[\therefore\] We take half of pure milk and half of water, then
total remainder
\[= \frac{2}{4}\]
\[= \frac{1}{2}\] unit
The ratio of the quantities of an acid and water in a mixture is 1 : 3. If 5 litres of acid is further added to the mixture, the new ratio becomes 1 : 2. Find the quantity of acid in original mixture.
Let the quantity of acid in original mixture be x litre
and the quantity of water in the original mixture be 3x litre.
Therefore,
\[\frac{x+5}{3x} =\frac{1}{2}\]
\[\implies 2(x+5) = 3x\]
\[\implies 2x + 10 = 3x\]
\[\implies 3x - 2x =10\]
\[\implies x =10\] litres
The ratio of the quantities of an acid and water in a mixture is 1 : 3. If 5 litres of acid is further added to the mixture, the new ratio becomes 1 : 2. The quantity of new mixture in litres is
Let the quantity of acid in original mixture be x litre
and the quantity of water in the original mixture be 3x litre.
Therefore,
\[\frac{x+5}{3x} =\frac{1}{2}\]
\[\implies 2(x+5) = 3x\]
\[\implies 2x + 10 = 3x\]
\[\implies 3x - 2x =10\]
\[\implies x =10\]
Therefore, quantity of a new mixture
\[= 4x +5 = 4(10) +5\]
= 45 litres
Three equal glasses are filled with mixtures of spirit and water. The ratio of the spirit to water is as given below:
| Glass | Ratio |
| 1st | 3 : 4 |
| 2nd | 4 : 5 |
| 3rd | 5 : 6 |
The contents of the three glasses are emptied into a single vessel. What is the ratio of the spirit to water in the mixture now?
Spirit : Water
\[=\left(\frac{3}{7}+\frac{4}{9}+\frac{5}{11}\right):\left(\frac{4}{7}+\frac{5}{9}+\frac{6}{11}\right)\]
= 920 : 1159
Two vessels A and B contain milk and water mixed in the ratio 5 : 3 and 2 : 3. When these mixtures are mixed to form a new mixture containing half milk and half water, they must be taken in the ratio
\[\frac{5}{8}x+\frac{2}{5}y=\frac{3}{8}x+\frac{3}{5}y\]
\[\Rightarrow \frac{x}{y}=\frac{4}{5}\]
\[\Rightarrow \text{Ratio}=4:5\]
A mixture contains milk and water in the ratio 5 : 1. On adding 5 litres of water, the ratio of milk and water becomes 5 : 2. The quantity of milk in the original mixture is
\[\frac{5x}{x+5}=\frac{5}{2}\]
\[\Rightarrow 10x=5x+25\]
\[\Rightarrow x=5\]
\[\therefore\] The quantity of milk in the original mixture is
\[5x=5 \times 5 = 25\] litres
An alloy of gold and silver weighs 50 g. It contains 80% gold. How much gold should be added to the alloy so that the percentage of gold is increased to 90?
Content of gold in alloy
\[= 50 \times \frac{80}{100} = 40 g\]
Let x g of gold should be added to alloy and solve the equation for x.
\[\frac{40+x}{50+x} \times 100 = 90\]
A shink contains exactly 12 litres of water. If water is drained from the shink until it holds exactly 6 litres water less than the quantity drained away, how many litres of water were drained away?
\[Water \ (W) + Drained \ (D) = 12\]
\[\implies W + D = 12\]
Now, \[W=D-6\]
\[\implies D-6+D=12\]
\[\implies D=9\] litres
In a mixture of 60 litres, the ratio of milk and water is 2 : 1. If the ratio of milk and water is to be 1 : 2, then the amount of water to be further added must be
Quantity of milk
\[=\frac{2}{3} \times 60 = 40\] litres
Quantity of water
\[=\frac{1}{3} \times 60 = 20\] litres
Quantity of water added = x litres (suppose)
Now, \[\frac{40}{20+x}=\frac{1}{2}\]
\[\implies x=60\] litres
The wheat sold by grocer contained 10% low quality wheat. What quantity of good quality wheat should be added to 150 kg of wheat so that the percentage of low quality wheat becomes 5%?
Here 10% of 150 kg = 15 kg
\[\therefore\] Good quality of wheat is
(150 - 15) kg = 135 kg
To become the 5% low quality of wheat, we add 150 kg of more wheat.
One litre of water is evaporated from 6 litres of a solution containing 5% salt. Find the percentage of salt in the remaining solution.
Quantity of salt in 6 litres of solution
\[=\frac{2}{100} \times 6 = 0.3\] litres
Percentage of salt in 5 litres of solution
\[=\frac{0.3}{2} \times 100 = 6\%\]
In a mixture of 60 litres, the ratio of milk to water is 2 : 1. If this ratio is to be 1 : 2, then the quantity of water (in litres) to be further added is
Milk = 40 litres
Water = 20 litres
To make the ratio M : W = 1 : 2,
60 litres of water should be added.
Milk and water are in the ratio of 3 : 2 in a mixture of 80 litres. How much water should be added so that the ratio of the milk and water becomes 2 : 3?
- Quantity of milk in the mixture = 48 litres
- Quantity of water in the mixture = 32 litres
Let x litres of water be added in the mixture so that
\[\frac{48}{32+x}=\frac{2}{3}\]
\[\implies 2x + 64 = 144\]
\[\implies x = 40\] litres
How much water must be added to 100 cc of 80% solution of boric acid to reduce it to a 50% solution?
In 100 cc solution, quantity of boric acid = 80 cc
and quantity of water = 20 cc.
Suppose x litres of water be added to make the solution 50%.
\[\therefore 50%(100 + x) = 80\]
\[\implies x = 60\] litres
One type of liquid contains 25% milk, the other contains 30% of milk. A can is filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of milk in the new mixture.
\[\frac{30-x}{x-25} =\frac{6}{4} =\frac{3}{2}\]
\[\implies 5x = 135\]
\[\implies x = 27\]
So, required percentage of milk = 27%.