Ratio & Proportion — Case-Study Based Question

Case study:
Amit runs a grocery store in a busy neighbourhood market. He stocks two varieties of lentils.

The premium quality sells at ₹ 280 per kg.

The standard quality sells at ₹ 240 per kg.

Customers in his area are price-conscious, so Amit wants to create a mixture of the two varieties that appears high quality but is also affordable. He decides to sell the mixture at ₹ 265 per kg.

In what ratio should the ₹ 240/kg and ₹ 280/kg varieties be mixed to obtain the mixture costing ₹ 265 per kg?

Answer

Let x be the quantity of ₹ 240/kg variety and y be the quantity of ₹ 280/kg variety.

Amit sells the mixture at ₹ 265 per kg.

∴ 240x + 280y = 265(x + y)

⇒ 240x + 280y = 265x + 265y

⇒ 240x - 265x + 280y - 265y = 0

⇒ -25x + 15y = 0

⇒ 15y = 25x

⇒ $\dfrac{y}{x} = \dfrac{25}{15}$

⇒ $\dfrac{y}{x} = \dfrac{5}{3}$

⇒ y : x = 5 : 3.

Hence, the ratio of ₹ 280/kg to ₹ 240/kg varieties is 5 : 3.

Case study:
In a school chemistry lab, the lab assistant has two bottles of acid solution on his shelf:

Solution A: 55% acid concentration

Solution B: 80% acid concentration

The science teacher wants a 60% acid solution for tomorrow’s experiment that is to be used by the students.

In what ratio should the 55% solution and the 80% solution be mixed to obtain a 60% acid solution?

Answer

Let x be the quantity of 55% solution and y be the quantity of 80% solution.

Required concentration of solution by mixing two solutions = 60%.

∴ 55x + 80y = 60(x + y)

⇒ 55x + 80y = 60x + 60y

⇒ 55x - 60x + 80y - 60y = 0

⇒ -5x + 20y = 0

⇒ 20y = 5x

⇒ $\dfrac{x}{y} = \dfrac{20}{5}$

⇒ $\dfrac{x}{y} = \dfrac{4}{1}$

⇒ x : y = 4 : 1.

Hence, the ratio of 55% solution to 80% solution is 4 : 1.