Question

A cottage industry produces a certain no. of wooden toys in a day. The cost of production of each toy in rupees was found to be 55 minus the no. of toys produced in a day. On that particular day the total cost of production was Rs. 750. What is the no. of toys produced on that day?

Answer 1

Hint: Let the no. of toys produced in a day be x then the cost of production of each toy will be $ 55 - x $ . The cost of production of x toys is 750, which means $ x\left( {55 - x} \right) = 750 $ . Solve this equation and find the no. of toys produced on that day.

Complete step-by-step answer:
We are given that a cottage industry produces a certain no. of wooden toys in a day.
Let us assume the total no. of toys produced on that day is x.
The cost of production of each toy in rupees was found to be 55 minus the no. of toys produced in a day.
Therefore, the cost of each toy will be $ 55 - x $ rupees.
On that particular day the total cost of production was Rs.750.
This means the cost of x toys is Rs.750
 $
  x\left( {55 - x} \right) = 750 \\
   \Rightarrow 55x - {x^2} = 750 \\
   \Rightarrow {x^2} - 55x + 750 = 0 \\
 $
Solve the above quadratic equation to find the no. of toys produced.
 $
  {x^2} - 55x + 750 = 0 \\
   \Rightarrow {x^2} - 25x - 30x + \left( {30 \times 25} \right) = 0 \\
   \Rightarrow x\left( {x - 25} \right) - 30\left( {x - 25} \right) = 0 \\
   \Rightarrow \left( {x - 25} \right)\left( {x - 30} \right) = 0 \\
   \Rightarrow x - 25 = 0\& x - 30 = 0 \\
   \Rightarrow x = 25,x = 30 \\
  \therefore x = 25,30 \\
 $
Therefore, the no. of toys produced on that particular day was 25 or 30.
When the no. of toys produced is 25, the cost of production of each toy will be 55-25=30 rupees.
When the no. of toys produced is 30, the cost of production of each toy will be 55-30=25 rupees.

Note: We have got a quadratic equation in one variable in the above solution. Every quadratic equation will definitely have 2 solutions because the highest degree of the variable will be 2. If the equation is linear then the number of solutions is only 1. The no. of solutions of an equation varies with the highest power of its variable.