Alligation or Mixture Questions and Answers

Answer: C

The water contributes nothing to the value, so all the worth comes from the oil.

\[ 325x = 275(x + 240) \]

\[ 325x = 275x + 66{,}000 \]

\[ 50x = 66{,}000 \quad\Longrightarrow\quad x = 1320 \text{ litres} \]

By alligation, water is priced at 0 and \( \tfrac{\text{oil}}{\text{water}} = \tfrac{275-0}{325-275} = \tfrac{275}{50} = 5.5 \), and \( 5.5 \times 240 = 1320 \).

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: B

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

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

\[ \frac{\text{Rs. 420 rice}}{\text{Rs. 520 rice}} = \frac{520 - 480}{480 - 420} = \frac{40}{60} = \frac{2}{3} \]

So the ratio is \( 2 : 3 \). The mean of 480 lies closer to 520 than to 420, so the dearer variety must be the larger share - which is the check on the direction of the ratio.

Answer: B

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

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

\[ \frac{\text{Rs. 15 wheat}}{\text{Rs. 18 wheat}} = \frac{18 - 17}{17 - 15} = \frac{1}{2} \]

The mean of 17 is nearer the dearer price, so more of the Rs. 18 wheat is needed.

Answer: A

Amount of water in solution = 100 % - 30 % = 70 % of total
Amount of water = 70 % x 500 = 350 gm
Amount of water remains same for both the solutions. Therefore, let amount of new solution be x gm.
x = 350 x (100/ 50)
x = 700
Salt added = (700 - 500) = 200 gm

Answer: A

The water is free, so "no loss" means the revenue from the whole adulterated quantity just covers the cost of the pure milk.

\[ 28(m + 1.5) = 36m \]

\[ 28m + 42 = 36m \quad\Longrightarrow\quad 8m = 42 \]

\[ m = 5.25 \text{ litres} \]

Selling at a price below cost is profitable only because the 1.5 L of water is sold as though it were milk.

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 litres
After 2nd replacement, kerosene left is : 0.9*90 = 81 litres
After 3rd replacement, kerosene left is : 0.9*81 = 72.9 litres
After 4th replacement, kerosene left is : 0.9*72.9 = 65.61 litres

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%

Answer: C

When a fraction of a solution is repeatedly removed and replaced, the original liquid left after \( n \) operations is

\[ A\left(1 - \frac{x}{A}\right)^{n} \]

With \( A = 30 \), \( x = 5 \) and \( n = 3 \):

\[ 30\left(1 - \frac{5}{30}\right)^{3} = 30\left(\frac{5}{6}\right)^{3} \]

\[ = 30 \times \frac{125}{216} = 17.36 \text{ litres} \]

Subtracting \( 3 \times 5 = 15 \) litres would give 15 L, but the later withdrawals remove water as well as milk, so more milk survives than that.

Answer: C

When a fraction of a solution is repeatedly removed and replaced, the original liquid left after \( n \) operations is

\[ A\left(1 - \frac{x}{A}\right)^{n} \]

With \( A = 30 \), \( x = 5 \) and \( n = 3 \):

\[ 30\left(1 - \frac{5}{30}\right)^{3} = 30\left(\frac{5}{6}\right)^{3} \]

\[ = 30 \times \frac{125}{216} = 17.36 \text{ litres} \]

Subtracting \( 3 \times 5 = 15 \) litres would give 15 L, but the later withdrawals remove water as well as milk, so more milk survives than that.