The owner of a milk store finds that he can sell 980 liters milk each

1.	The owner of a milk store finds that he can sell 980 liters milk each week at Rs. 14 per liter and 1220 liters of milk each week at Rs. 16 per liter. Assuming a linear relationship between selling price and demand, how many liters could he sell weekly at Rs. 17 per liter.
Answer» Assuming: x denotes the price per liter, and y denote the quality of the milk sold at this price. Since there is a linear relationship between the price and the quality, the line representing this Given: Relationship passes through (14, 980) and (16, 1220). To find: How many liters could he sell weekly at Rs. 17 per liter. Explanation: So, the equation of the line passing through these points is y-y₁ = \(\Big(\frac{y_2-y_1}{x_2-x_1}\Big)(x-x_1)\) y - 980 = \(\frac{1220-980}{16-14}\) (x - 14) ⇒ y – 980 = 120(x - 14) ⇒ 120 x - y - 700 = 0 When x = 17 then we have, 120(17) – y – 700 = 0 ⇒ y = 1340 Hence, the owner of the milk store can shell 1340 litres of milk at Rs. 17 per litre.

The owner of a milk store finds that he can sell 980 liters milk each week at Rs. 14 per liter and 1220 liters of milk each week at Rs. 16 per liter. Assuming a linear relationship between selling price and demand, how many liters could he sell weekly at Rs. 17 per liter.

Answer»

Assuming:

x denotes the price per liter, and y denote the quality of the milk sold at this price.

Since there is a linear relationship between the price and the quality, the line representing this

Given: Relationship passes through (14, 980) and (16, 1220).

To find: How many liters could he sell weekly at Rs. 17 per liter.

Explanation:

So, the equation of the line passing through these points is

y-y₁ = \(\Big(\frac{y_2-y_1}{x_2-x_1}\Big)(x-x_1)\)

y - 980 = \(\frac{1220-980}{16-14}\) (x - 14)

⇒ y – 980 = 120(x - 14)

⇒ 120 x - y - 700 = 0

When x = 17 then we have,

120(17) – y – 700 = 0

⇒ y = 1340

Hence, the owner of the milk store can shell 1340 litres of milk at Rs. 17 per litre.