Question

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 articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation to find x.

Answer 1

SOLUTION :

Given : Total cost of production of toy = 750

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

Therefore the cost of production of each toy that day = (55 - x)

So, the total cost production that day = Number of toys × cost of production of each toy that day

Total cost of production of toy =  x × (55 - x)

A.T.Q..

x(55 - x) = 750

55x - x² = 750

-x² + 55x - 750 = 0

x² - 55x + 750 = 0

Hence, the required quadratic equation is x² - 55x + 750 = 0

By factorisation method :  

x² - 55x + 750 = 0

x²- 30x - 25x + 750 = 0

x(x - 30) - 25(x - 30) = 0

(x - 30)(x - 25) = 0

x - 30 = 0

x = 30

x - 25 = 0

x = 25

Hence, the number of toys produced (x) = 25 or 30 .

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Answer 2

Answer is given below:-

Total cost of production of toy = 750

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

Therefore the cost of production of each toy that day = (55 - x)

So, the total cost production that day = Number of toys × cost of production of each toy that day

Total cost of production of toy =  x × (55 - x)

A.T.Q..

x(55 - x) = 750

55x - x² = 750

-x² + 55x - 750 = 0

x² - 55x + 750 = 0

Hence, the required quadratic equation is x² - 55x + 750 = 0
By factorisation method :  

x² - 55x + 750 = 0

x²- 30x - 25x + 750 = 0

x(x - 30) - 25(x - 30) = 0

(x - 30)(x - 25) = 0

x - 30 = 0

x = 30

x - 25 = 0

x = 25

Hence, the number of toys produced (x) = 25 or 30 .