Question

# The total cost of $10$ erasers and $5$ sharpeners is at least $Rs.{\rm{ }}65$. The cost of each eraser cannot exceed$Rs.{\rm{ }}4$. Find the minimum possible cost of each sharpenersA. $Rs.\,{\rm{ 6}}$B. $Rs.{\rm{ 5}}{\rm{.50}}$C. $Rs.{\rm{ 5}}$D. $Rs.{\rm{ 6}}{\rm{.50}}$

Hint:Here we have to find the minimum cost of sharpener, for that we will initially form an inequality from the given initial condition, and also form an inequality using the rate of eraser. Let us solve the inequalities by substituting one in the other and finally it will help us to find the required value.

Here, it is given that the total cost of 10 erasers and 5 sharpeners is at least$Rs.{\rm{ }}65$. And. the cost of each eraser cannot exceed $Rs.{\rm{ }}4$.
We have to find the minimum possible cost of sharpener.
Let us cost one eraser to be $e$ and the cost of one sharpener to be $s$ respectively.
According to question,
We can write and mark it by equation (1)
$65\; < {\rm{ }}10e{\rm{ }} + {\rm{ }}5s$………….. (1)
Again, cost of eraser cannot exceed
$e\; \le {\rm{ }}4{\rm{ }}$……………………… (2)
Multiply both side by 10 in equation (2)
$10e{\rm{ }}\; \le {\rm{ }}40\;$…………………… (3)
Now let us add $5s$ on both sides of the inequality (3) we get
$10e + 5s{\rm{ }}\; \le {\rm{ }}40\; + 5s$…………. (4)
By comparing (1) and (4) we get,
$65{\rm{ < }}10e + 5s \le {\rm{ }}40 + {\rm{ }}5s$
From the above equation we get,
$65{\rm{ < }}40 + {\rm{ }}5s$
By solving the above equation we get,
$65 - 40{\rm{ < }}5s$
On subtracting the left hand side of the above equation we get,
$25{\rm{ < }}\;\;5s$
Let us divide the above inequality by 25 both sides we get,
$s{\rm{ > }}\dfrac{{25}}{5}$
That is $s{\rm{ > }}5$
Therefore, the minimum price of a sharpener is $Rs.{\rm{ }}5$

So, the correct answer is “Option C”.

Note:While solving the above problem we have reduced $\le$ to ${\rm{ < }}$because while solving two inequalities the priority is given to the minimum inequality that is $<$ has the minimum strength when compared to $\le$. Similarly if we have $\ge$and $>$we will prefer $>$.