CAT 2025 Lesson : Mixtures & Alligations - Examples 3 to 6

If a shopkeeper mixes $10$ kg of wheat costing Rs. $8$ per kg and $14$ kg of wheat costing Rs. $14$ per kg, and sells the resultant mixture at $10 \%$ profit, what is the price per kg of the wheat sold by the shopkeeper?

Solution

$x_{1}$ and $x_{2}$ are the cost price per kg of the $2$ different wheats – Rs. $\bm{8}$ per kg and Rs. $\bm{14}$ per kg respectively.

$w_{1}$ and $w_{2}$ are the weights of these wheat – $\bm{10}$ kg and $\bm{14}$kg respectively.

We shall now apply the alligation method to find the weighted average cost price per kg of wheat (say $x$).



$\therefore \dfrac{14 - x}{x - 8} = \dfrac{10}{14} = \dfrac{5}{7}$

⇒ $98 - 7x = 5x - 40$
$x = 11.5$

As CP is Rs. $11.5$ per kg and profit is $10 \%$,
SP $= 11.5 + 1.15 =$ Rs. $12.65$ per kg

Answer: Rs. $12.65$ per kg

A vessel contains $80$ litres of a solution that contains $70 \%$ milk and the rest water. If $16$ litres of this solution is replaced with pure milk, what percentage of the resultant solution is milk?

Solution

If $16$ litres are removed from the vessel, we are left with $64$ litres of the $70 \%$ milk solution. To this, $16$ litres of $100 \%$ milk is added.

Therefore, the resultant solution contains $64$ litres of $70 \%$ milk solution and $16$ litres of $100 \%$ milk.

Let $x \%$ be the percentage of milk in the final mixture.



$\therefore \dfrac{100 - x}{x - 70} = \dfrac{4}{1}$

⇒ $100 - x = 4x - 280$
⇒ $x = 76 \%$

Answer: $76 \%$

Alloys A and B comprise of two metals – copper and zinc. Alloy A has copper and zinc in the ratio of $3 : 5$. When alloys A and B are mixed in the ratio of $3 : 1$, the resultant alloy has copper and zinc in the ratio of $37 : 59$. What is the ratio of copper to zinc in alloy B?

(1) $5 : 12$            (2) $5 : 7$            (3) $7 : 12$            (4) $7 : 5$           

Solution

As explained in Example 3, when mixture of mixtures are involved, we find the portion of $1$ of the items in the base mixtures and then mix these in the given ratio.

Portion of Copper in Alloy A $= \dfrac{3}{3 + 5} = \dfrac{3}{8}$

Portion of Copper in the final alloy $= \dfrac{37}{37 + 59} = \dfrac{37}{96}$

(We can now scale up $\dfrac{3}{8}$ to $\dfrac{3 \times 12}{8 \times 12} = \dfrac{36}{96}$ for ease in calculations with a common denominator)

Let the portion of copper in Alloy B be $x$.



$\therefore x - \dfrac{37}{96} = \dfrac{3}{96}$

⇒ $x = \dfrac{40}{96} = \dfrac{5}{12}$

Ratio of copper to zinc in Alloy B $= 5 : 7$

Answer: (2) $5 : 7$