Question

A storekeeper wants to mix two types of flour to get $300\mathrm{pounds}$, so he can sell it by $\$2.50$ per pound.

If he uses flour worth $\$2.40$ a pound with another flour worth $\$3.00$ a pound, how many pounds of each does he use?

Answer 1

Step-1: Find the equations:

Let the quantity of two flour be $x$ and $y$ in pounds.

Quantity of flour when two types of flour is mixed is $300\mathrm{pounds}$.

So, $x+y=300$

The cost of first flour is $\$2.40$ per pound and second flour is $\$3.00$ a pound.

Total cost of selling $x$ and $y$ pounds of flour will be $2.40x+3y$.

Storekeeper wants to sell mix flour of $300\mathrm{pounds}$ in $\$2.50$ per pound.

So, cost of selling mix flour is $300\times \left(2.50\right)$.

Then we can write,

$\begin{array}{rcl}2.40x+3y& =& 300\times \left(2.50\right)\\ \therefore 2.40x+3y& =& 750\end{array}$

Step-2: Solve the equations for $x$.

We have two equations,

$\begin{array}{rcl}x+y& =& 300\dots \left(1\right)\\ 2.40x+3y& =& 750\dots \left(2\right)\end{array}$

Multiplying equation $\left(1\right)$ with $3$ and subtracting equation $\left(2\right)$ from it.

$\begin{array}{rcl}3\left(x+y\right)-\left(2.40x+3y\right)& =& 3\times 300-750\\ & \Rightarrow & 3x+3y-2.40x-3y=900-750\\ & \Rightarrow & 0.60x=150\\ & \Rightarrow & x=\frac{150}{0.60}\left(\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}0.60\right)\\ \therefore x& =& 250\mathrm{pounds}\end{array}$

Step-3: Find $y$:

Substitute $x=250$ in $x+y=300$.

$\begin{array}{rcl}x+y& =& 300\\ & \Rightarrow & 250+y=300\\ & \Rightarrow & y=300-250\left(\mathrm{Subtracting}\mathrm{both}\mathrm{sides}\mathrm{by}250\right)\\ \therefore y& =& 50\mathrm{pounds}\end{array}$

Hence, storekeeper must use $250\mathrm{pounds}$ flour worth $\$2.40$ a pound with flour $50\mathrm{pounds}$ worth $\$3.00$ a pound.