Represent the following situations mathematically:A cottage industry p

1.	Represent the following situations mathematically:A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.
Answer» Let the number of toys produced on that day be ‘x’. ∴ The cost of production (in rupees) of each toy that day = 55 – x ∴ The total cost of production (in rupees) that day = x(55 – x) ∴ x(55 – x) = – 750 55x – x²= 750 -x² + 55x – 750 = 0 x² – 55x + 750 = 0 ∴ The number of toys produced that day satisfies the quadratic equation x² – 55x + 750 = 0. Now, we have to find out the value of ’x’ x² – 55x + 750 = 0 x² – 30x – 25x + 750 = 0 x(x – 30) – 25(x – 30) = 0 (x – 30) (x – 25) = 0 If x – 30 = 0, then x = 30 If x – 25 = 0, then x = 25 ∴ x = 30 OR x = 25.

Represent the following situations mathematically:A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.

Answer»

Let the number of toys produced on that day be ‘x’.

∴ The cost of production (in rupees) of each toy that day = 55 – x

∴ The total cost of production (in rupees) that day = x(55 – x)

∴ x(55 – x) = – 750

55x – x²= 750

-x² + 55x – 750 = 0

x² – 55x + 750 = 0

∴ The number of toys produced that day satisfies the quadratic equation x² – 55x + 750 = 0.

Now, we have to find out the value of ’x’

x² – 55x + 750 = 0

x² – 30x – 25x + 750 = 0

x(x – 30) – 25(x – 30) = 0

(x – 30) (x – 25) = 0

If x – 30 = 0, then x = 30

If x – 25 = 0, then x = 25

∴ x = 30 OR x = 25.