Question

Mr.Kasam runs a small business of making pots. He makes a certain number of pots on a daily basis. Production cost of each pot is Rs.40 more than 10 times the total number of pots he makes in one day. If the production cost of all pots per day is Rs.600, find the cost of one pot and the number of pots he makes per day.

Answer 1

Hint: In this question, we are given a word problem for production of a certain number of pots on a daily basis. Using the given information, we need to find the cost of one pot and number of pots per day. For this, we will first suppose the number of pots per day to be x and then use the given information to form equation solving which will give us a value of x. Hence, we will calculate the number of pots per day. We are also given the cost of one pot in terms of number of pots per day. So, we will find the cost of one pot. Equation formed will be quadratic, so we will use a quadratic formula to find the value of x. For any equation $a{{x}^{2}}+bx+c$ quadratic formula is given by $\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step by step answer:
For solving this sum, let us first suppose that the number of pots Mr.Kasam makes per day are x. As we are given that, the cost of per pot is of Rs.40 more than 10 times the total number of pots. So, the cost of one pot will become equal to Rs.(40+10x).
Hence, number of pots = x.
Cost of one pot = Rs.(40+10x).
Therefore, total production in one day will be the product of the number of pots and cost of one pot.
Hence, total production in one day $\Rightarrow Rs.\left( x\times \left( 40+10x \right) \right)$.
But we are given total production in one day as Rs.600. Therefore,
\[\begin{align}
  & \Rightarrow x\left( 40+10x \right)=600 \\
 & \Rightarrow 40x+10{{x}^{2}}=600 \\
 & \Rightarrow 10{{x}^{2}}+40x-600=0 \\
\end{align}\]
Taking 10 common from left side and taking it to right side, we get:
\[\Rightarrow {{x}^{2}}+4x-60=0\]
This is a quadratic equation and hence, we will get two values of x.
Let us now use the quadratic formula to calculate the value of x.
As we know, quadratic formula for $a{{x}^{2}}+bx+c$ is given as $\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ so for this equation we get:
\[\begin{align}
  & \Rightarrow x=\dfrac{-4\pm \sqrt{16+240}}{2} \\
 & \Rightarrow x=\dfrac{-4\pm \sqrt{256}}{2} \\
\end{align}\]
As we know, ${{\left( 16 \right)}^{2}}=256$ so $\sqrt{256}=16$. Hence, we get:
\[\begin{align}
  & \Rightarrow x=\dfrac{-4\pm 16}{2} \\
 & \Rightarrow x=\dfrac{-4+16}{2},x=\dfrac{-4-16}{2} \\
 & \Rightarrow x=\dfrac{12}{2},x=\dfrac{-20}{2} \\
 & \Rightarrow x=6,x=-10 \\
\end{align}\]
Since, the number of pots cannot be negative. So rejecting x = -10 we get x = 6.
Now, x was supposed to be the number of pots.
So, the number of pots that Mr.Kasam makes in a day are 6.
Cost of one pot $\Rightarrow Rs.\left( 40+10x \right)=\left( 40+10\left( 6 \right) \right)=\left( 40+60 \right)=100$.

Therefore, cost of one pot is Rs.100

Note: While solving quadratic equations, students should take care of the sign. Do not forget to solve negative values too. Make equations from sentences carefully by taking proper variables. Do not forget to write units for rupees. Students can also solve quadratic equations using splitting the middle term method.