Question

A certain number of tennis balls were purchased for Rs 450. Five more balls could have been purchased in the same amount if each ball was cheaper by Rs 15. The number of balls purchased was __________.
(a) 10
(b) 15
(c) 20
(d) 25

Answer 1

Hint: To solve this question, we will take the number of balls as x. Then we will find the cost of each ball and subtract 15 from it. We have to multiply this modified cost with (x + 5) and this will be equal to 450. Thus, we will get a quadratic equation and find the value of x.

Complete step-by-step answer:
Let us assume that the number of tennis balls that can be bought with Rs 450 is x.
So, the original cost of each tennis ball will be $\dfrac{450}{x}$.
Now, it is given that if each tennis ball was cheaper by Rs 15, 5 more balls could have been bought for the same amount, i.e. Rs 450.
The modified cost of the ball will be:
$\begin{aligned} & \Rightarrow \left( \text{original cost}-15 \right) \\\ & \Rightarrow \left( \dfrac{450}{x}-15 \right) \\\ \end{aligned}$
Now, with the reduced cost of each tennis ball, we can buy 5 more balls, i.e. (x + 5) balls, with the same amount of Rs 450.
Therefore,
$\begin{aligned} & \Rightarrow \left( \text{modified cost of each ball} \right)\times \left( x+5 \right)=450 \\\ & \Rightarrow \left( \dfrac{450}{x}-15 \right)\left( x+5 \right)=450 \\\ \end{aligned}$
We will manipulate this equation to form a quadratic equation.
$\begin{aligned} & \Rightarrow \left( \dfrac{450-15x}{x} \right)\left( x+5 \right)=450 \\\ & \Rightarrow 450x-15{{x}^{2}}+450\left( 5 \right)-15x\left( 5 \right)=450x \\\ \end{aligned}$
450x gets cancelled out and we divide the whole equation by 15
$\begin{aligned} & \Rightarrow 450x-15{{x}^{2}}+450\left( 5 \right)-15x\left( 5 \right)=450x \\\ & \Rightarrow -{{x}^{2}}+150-5x=0 \\\ \end{aligned}$
To solve this quadratic equation, we write this equation in standard form $A{{x}^{2}}+Bx+C=0$, by multiplying the equation by -1.
$\begin{aligned} & \Rightarrow -1\left( -{{x}^{2}}-5x+150 \right)=0 \\\ & \Rightarrow {{x}^{2}}+5x-150=0 \\\ \end{aligned}$
We need to find factors such that the sum of the factors is 5 and product of factors is -150. Thus, 10 and -15 are the two factors
$\begin{aligned} & \Rightarrow {{x}^{2}}+15x-10x-150=0 \\\ & \Rightarrow x\left( x+15 \right)-10\left( x+15 \right)=0 \\\ & \Rightarrow \left( x+15 \right)\left( x-10 \right)=0 \\\ \end{aligned}$
Now, either x + 15 = 0 or x – 10 = 0. But x cannot be negative, so, x = 10.
Thus, the number of balls purchased were 10.
Therefore, option (a) is the correct option.

Note: Students can opt for option verification, by substituting various options in the equation $\left( \dfrac{450}{x}-15 \right)\left( x+5 \right)=450$. This will save the efforts to solve the quadratic equation. Moreover, the quadratic equation can be solved by other methods, like completing the square method or formula method.