CAT 2025 Lesson : Mixtures & Alligations - Examples 7 to 13

A vessel had $50$ litres of milk solution (milk and water) that contained $20 \%$ water. Kumaran then added equal quantities of pure milk and water to this vessel. If the resultant solution in the vessel had $60 \%$ milk, then what was the quantity of pure milk (in litres) added to the vessel?

(1) $25$            (2) $50$            (3) $75$            (4) $100$           

Solution

To the initial solution containing $20 \%$ water (or $80 \%$ milk), equal quantities of pure milk and water are added. The final solution contains $60 \%$ milk.

Let the quantity of pure milk and water being added be $x$ litres each. Total quantity of these two is $2x$ and this mixture would contain $50 \%$ water.



$\therefore \dfrac{50}{2x} = \dfrac{10}{20}$ ⇒ $x = 50$

Answer: (2) $50$

$24$ litres of a solution that contains milk and water in the ratio of $5 : 3$ is mixed with $27$ litres of a solution that has milk and water in the ratio $7 : 2$. This solution is then mixed with $26$ litres of a solution that has milk and water in the ratio of $8 : 5$, to form the final solution. What is the ratio of milk and water in the final solution?

Solution

The portion of milk in the three solutions are $\dfrac{5}{8}, \dfrac{7}{9}$ and $\dfrac{8}{13}$.

The alligation method will be difficult as the LCM of denominators $(8, 9, 13) = 936$, is a very big number.

In this question, however, the litres of milk and water in each of the solutions have integral values. Therefore, we do not use alligation and directly find the portion of milk and water in each solution.

Qty (in Litres)	Total	Ratio	Milk	Water
Solution 1	$24$	$5 : 3$	$15$	$9$
Solution $2$	$27$	$7 : 2$	$21$	$6$
Solution $3$	$26$	$8 : 5$	$16$	$10$
Total			$\bm{52}$	$\bm{25}$

When the three solutions are mixed, we get a final solution containing $52$ litres of milk and $25$ litres of water.

∴ Ratio of milk to water in final solution $= 52 : 25$

Answer: $52 : 25$

If $2$ kg of stones are mixed with $18$ kg of rice and sold at the cost price of rice, then what is the profit percentage? [Assume the cost of stones is zero]

(1) $10 \%$            (2) $11.1 \%$            (3) $12.5 \%$            (4) Cannot be determined                                 

Solution

Let the CP of rice be Rs. $1$ per kg.

CP of $18$ kg of rice = Rs. $18$

When $2$ kg of stone is added, we get a total of $20$ kg of adulterated rice, which is then sold at Rs. $x$ per kg.

SP of $20$ kg of rice = Rs. $20$

Profit $\% = \dfrac{20 - 18}{18} \times 100 \% = 11.1 \%$

Alternatively

For every $18$ kg of rice purchased, extra $2$ kg is sold. This additional $2$ kg is the profit.

Profit $\% =\dfrac{2}{18} \times 100 \% = 11.1 \%$

Answer: $(2) 11.1 \%$

A merchant buys and sells lemon concentrate. He wants to sell it at the cost price and achieve a profit of $10$%. He decides to achieve this by adding freely available water. How much water (in ml) must he add to each litre of lemon concentrate?

(1) $90$ ml            (2) $90.9$ ml            (3) $100$ ml            (4) $110$ ml           

Solution

$1$ litre $= 1000$ ml

As the lemon concentrate's SP = CP, the added water added will be the profit.

Let $x$ ml of water be added to $1000$ ml of pure lemon concentrate.

$\dfrac{x}{1000} \times 100 \% = 10 \%$

⇒ $x = 100$ ml

Answer: $(3) 100$ ml

The average marks of students of Classes A and B were $60$ and $80$ respectively. If there are $40$ students in Class A and $60$ students in Class B, then what is their combined average?

Solution

Ratio of number of students $= 40 : 60 = 2 : 3$

The difference in the marks of Classes A and B is $80 - 60 =$ $\bm{20}$ marks.

This difference is to be divided in the ratio of weights. As the ratio is $2 : 3$, we have a total of $2 + 3 =$ $\bm{5}$ parts.

The difference of $20$ marks can be divided into $5$ parts with each part being $\bm{4}$ marks. This is shown in the diagram below.



As the ratio of A to B is $2 :3$,
- A has the strength to pull $2$ parts from B
- B has the strength to pull $3$ parts from A

Therefore, the average will be between $\bm{60}$ and $\bm{80}$. So, to arrive at the weighted average, we either add to A's marks or subtract from B's marks, as shown below.

A pulls the average, from B, with it's strength of $2$ parts ($= 8$ marks).
∴ Weighted Average of Class = B's mark - Value of A's Parts $= 80 - 8 =$ $\bm{72}$ marks

B pulls the average, from A, with it's strength of $3$ parts (or $12$ marks).
∴ Weighted Average of Class = A's mark + Value of B's Parts $= 60 + 12 =$ $\bm{72}$ marks

This is shown in the diagram below.



Answer: $72$ marks

Ram purchased $600$ Oranges at Rs. $27$ each and $360$ Apples at Rs. $39$ each. What was the average price of a fruit?

Solution

Ratio of Oranges to Apples $= 600 : 360 = 5 : 3$

Difference $= 39 - 27 =$ Rs. $12$
Total Parts $= 5 + 3 = 8$ parts
$1$ Part $= \dfrac{12}{8} =$ Rs. $1.5$



Apples pull the average by $3$ parts from the Price of Oranges.

Weighted average Price per fruit $= 27 + (3 \times 1.5) = 31.5$

Answer: Rs. $31.5$

Note: Weighted average computed from an Apple's Price will be $39 - (5 \times 1.5) = 31.5$

A class contains girls and boys in the ratio of $5 : 2$. In a recently conducted math test, the girls scored an average of $84 \%$ and the class as a whole scored an average of $78 \%$. What was the average of the boys in the test?

(1) $63 \%$            (2) $68 \%$            (3) $70 \%$            (4) $73 \%$           

Solution

Boys have pulled the weighted average by $2$ parts from the girls.
∴ $2$ parts $= 84 \% - 78 \% = 6 \%$
⇒ $1$ part $= 3 \%$

Boys average will be $5$ parts away from the weighted average ⇒ $78 \% - (5 \times 3 \%) = 63 \%$

Answer: $(1) 63 \%$