Alligation or Mixture Questions and Answers

Answer: D

Total solution quantity 57+43 = 100

Out of 100 we are removing 10 liter (i.e. 10%)

so proportionately from honey we will remove = remove 5.7

& proportionately from water we will remove 10% remove = 4.3

Also, we add 10 liter pure honey

So now honey 57-5.7+10 = 61.3

& Water 43-4.3 = 38.7

water: Honey = 38.7 : 61.3 = 387:613

Answer: C

So amount of acid = 2/ 2+1 × 90 = 60L

Amount of water = 90 - 60 = 30L

To make acid to water ratio = 1:2, we simply need

To make water double of acid.

By direct observation, we can say that we should

have 60L × 2 = 120L make the required ratio

We have 30L, we need (120-30) = 90L more water

Answer: B

The alligation rule gives the ratio in which two ingredients must be mixed:

\[ \frac{\text{cheaper quantity}}{\text{dearer quantity}} = \frac{\text{dearer} - \text{mean}}{\text{mean} - \text{cheaper}} \]

\[ \frac{\text{Rs. 7.20 rice}}{\text{Rs. 5.70 rice}} = \frac{6.30 - 5.70}{7.20 - 6.30} = \frac{0.60}{0.90} = \frac{2}{3} \]

Note the numerator and denominator: the quantity of the dearer rice is proportional to the gap on the cheaper side, and vice versa.

Answer: D

B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.

Quantity of water in B = 8k - 5k = 3k.

Quantity of water in container C = 8k - 3k = 5k

Container: A   B    C

Quantity of water: 8k  3k   5k

It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.

5k - 148 = 3k + 148 => 2k = 296 => k = 148

The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.

Answer: B

The operation is performed four times in all - once, then three more.

Wine : water \( = 16 : 65 \) means wine is \( \tfrac{16}{81} \) of the cask:

\[ \left(1 - \frac{8}{T}\right)^{4} = \frac{16}{81} \]

\[ \left(1 - \frac{8}{T}\right)^{2} = \frac{4}{9} \quad\Longrightarrow\quad 1 - \frac{8}{T} = \frac{2}{3} \]

\[ \frac{8}{T} = \frac{1}{3} \quad\Longrightarrow\quad T = 24 \text{ litres} \]

The ratio 16 : 65 has to be converted to the fraction \( \tfrac{16}{81} \) of the whole before the formula applies.

Answer: A

Correct answer: 1 : 4

The gain arises entirely from the water, which costs nothing but is sold at the price of milk. Let the cost of one litre of milk be 1. If the milkman sells \( 1 + x \) litres of mixture containing 1 litre of milk and \( x \) of water, his cost is 1 and his receipt is \( 1 + x \), so \[ \text{gain} = \frac{x}{1} = 25\% = \frac{1}{4} \;\Rightarrow\; x = \frac{1}{4}. \] The ratio of water to milk is therefore \( \frac{1}{4} : 1 = \mathbf{1 : 4} \). The useful shortcut is that the ratio of water to milk equals the gain expressed as a fraction: a 25% gain means 1 : 4, a 20% gain means 1 : 5, and a 50% gain means 1 : 2. The step that trips people is reading the ratio the right way round - the question asks for water to milk, and reversing it gives 4 : 1, which is also offered in many versions of this question.

Answer: B

Let the resulting mixture be 1 kg, and x kg be the amount of wheat of inferior quality.
Therefore, (1-x)kg is the amount of wheat of superior quality.
As the shopkeeper gains Rs.2, the cost of the mixture is Rs.18
14*x + 28*(1-x) = 18
14x - 28x + 28 = 18
14x = 10
x = 5/7
(1 - x) = 2/7
x : (1-x) = 5/7 : 2/7
= 5 : 2

Answer: B

As Ram replaces 10% of kerosene with water, the amount of kerosene left is 90%.

As the replacement is done four times,

After 1st replacement, kerosene left is : 0.9 × 100 = 90 liters

After 2nd replacement, kerosene left is : 0.9 × 90 = 81 liters

After 3rd replacement, kerosene left is : 0.9 × 81 = 72.9 liters

After 4th replacement, kerosene left is : 0.9 × 72.9 = 65.61 liters

Answer: C

\[ \text{acid} = \frac{2}{3} \times 90 = 60 \text{ L}, \qquad \text{water} = 30 \text{ L} \]

Adding water leaves the acid unchanged, so for acid : water \( = 1 : 2 \) the water must reach

\[ 2 \times 60 = 120 \text{ L} \]

\[ \text{water to add} = 120 - 30 = 90 \text{ L} \]

Anchoring on the quantity that does not change - the acid - is what makes this a one-line calculation.

Answer: C

Let the weight of full bottle be 6kg.
Therefore, weight of empty bottle is 1kg and that of water is 5kg.
If x% of water is removed, the weight of the bottle becomes 2kg.
Therefore, the amount of water removed is 4kg.
% of water removed = 4/5 *100 = 80%