Alligation or Mixture Questions and Answers

Answer: A

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

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

\[ \frac{\text{Rs. 15 sugar}}{\text{Rs. 18 sugar}} = \frac{18 - 16}{16 - 15} = \frac{2}{1} \]

So the ratio is \( 2 : 1 \). The mean price of 16 sits closer to 15 than to 18, so the cheaper sugar must dominate - a check on which way round the ratio goes.

Answer: D

Take each can as 1 unit.

\[ \text{Can 1: water } 1, \text{ wine } 0 \]

\[ \text{Can 2: water } 0.5, \text{ wine } 0.5 \]

\[ \text{total water} = 1.5, \qquad \text{total wine} = 0.5 \]

\[ \text{water} : \text{wine} = 1.5 : 0.5 = 3 : 1 \]

The cans are the same size, which is what lets them be counted as equal units.

Answer: D

Convert each solution to litres of honey and water.

\[ 28 \text{ L at } 4:3 \rightarrow 16 \text{ honey}, \ 12 \text{ water} \]

\[ 21 \text{ L at } 2:1 \rightarrow 14 \text{ honey}, \ 7 \text{ water} \]

\[ 51 \text{ L at } 9:8 \rightarrow 27 \text{ honey}, \ 24 \text{ water} \]

\[ \text{total: } 100 \text{ L}, \ 57 \text{ honey}, \ 43 \text{ water} \]

Removing 10 L takes away 10% of each, then 10 L of pure honey is added:

\[ \text{honey} = 0.9(57) + 10 = 61.3, \qquad \text{water} = 0.9(43) = 38.7 \]

\[ \text{water} : \text{honey} = 387 : 613 \]

The convenient total of exactly 100 L makes the 10% removal easy to apply.

Answer: A

No profit and no loss means the total cost equals the total revenue.

\[ 32(60) + 23x = 28(60 + x) \]

\[ 1920 + 23x = 1680 + 28x \]

\[ 240 = 5x \quad\Longrightarrow\quad x = 48 \text{ litres} \]

By alligation the same result follows from \( \tfrac{60}{x} = \tfrac{28-23}{32-28} = \tfrac{5}{4} \).

Answer: D

Take one unit from each glass and add the parts.

\[ \text{water} = \frac{2}{7} + \frac{4}{11} = \frac{22 + 28}{77} = \frac{50}{77} \]

\[ \text{juice} = \frac{5}{7} + \frac{7}{11} = \frac{55 + 49}{77} = \frac{104}{77} \]

\[ \text{water} : \text{juice} = 50 : 104 = 25 : 52 \]

The question asks for water first, so the ratio is not the reciprocal 52 : 25 - which is offered as a distractor.

Answer: B

Let rice \( = 5k \) and wheat \( = 3k \). Adding 16 kg of wheat changes only the wheat.

\[ \frac{5k}{3k + 16} = \frac{5}{7} \quad\Longrightarrow\quad 35k = 15k + 80 \]

\[ 20k = 80 \quad\Longrightarrow\quad k = 4 \]

\[ \text{new wheat} = 3(4) + 16 = 28 \text{ kg} \]

The question asks for the wheat in the new mixture, not the original 12 kg.

Answer: D

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

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

Here \( A = 40 \), \( x = 4 \) and \( n = 3 \) (once, then twice more):

\[ 40\left(1 - \frac{4}{40}\right)^{3} = 40(0.9)^{3} \]

\[ = 40 \times 0.729 = 29.16 \text{ litres} \]

Simply subtracting \( 3 \times 4 = 12 \) litres would give 28, but each later removal takes out some water as well as juice.

Answer: B

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

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

Work with the fraction of salt in each item.

\[ \text{Item 1} = \frac{2}{13}, \qquad \text{Item 2} = \frac{5}{26}, \qquad \text{mixture} = \frac{7}{39} \]

Over the common denominator 78 these are \( \tfrac{12}{78} \), \( \tfrac{15}{78} \) and \( \tfrac{14}{78} \):

\[ \frac{\text{Item 1}}{\text{Item 2}} = \frac{15 - 14}{14 - 12} = \frac{1}{2} \]

Converting each ratio into a single fraction first is what makes the alligation rule applicable.

Answer: A

The 80 L contains \( 45\% \times 80 = 36 \) L of milk. Adding \( x \) L of pure milk raises both the milk and the total.

\[ \frac{36 + x}{80 + x} = 0.75 \]

\[ 36 + x = 60 + 0.75x \quad\Longrightarrow\quad 0.25x = 24 \]

\[ x = 96 \text{ litres} \]

The non-milk part stays fixed at 44 L, and it must end up as 25% of the total - so the total becomes 176 L, confirming the 96 L added.

Answer: A

Take the quantities as 2, 1 and 3 kg, six kilograms in all.

\[ 2(145) + 1(165) + 3p = 6(180) \]

\[ 290 + 165 + 3p = 1080 \]

\[ 3p = 625 \quad\Longrightarrow\quad p = \text{Rs. } 208.33 \]

The third type is the most expensive, which it has to be for the average of 180 to sit above both known prices.