Question

Mr. Abernathy purchased a selection of wrenches for his shop. The total cost was $78. He bought the same number of $1.50 and $2.50 wrenches, and half that many $4 wrenches. The number of $3 wrenches was one more than the number of $4 wrenches. How many of each did he purchase?

Answer 1

$*Let \: number \: of \: \ 1.50 \: wrenches \\purchased = x$

$* number \: of \: \ 2.50 \: wrenches \\purchased = x$

$* number \: of \: \ 4 \: wrenches \\purchased = \frac{x}{2}$

$* number \: of \: \ 3 \: wrenches \\purchased = \frac{x}{2} + 1$

/* According to the problem given */

$Total \:cost \:of \: all \: wrenches = \ 78$

$\implies 1.50x + 2.50x + 4\times \frac{x}{2} + 3 \times \Big( \frac{x}{2} + 1 \Big) = 78$

$\implies 4x+ 2x+ \frac{3x}{2} + 3= 78$

$\implies 6x+ \frac{3x}{2} = 78-3$

$\implies \frac{12x + 3x}{2} = 75$

$\implies \frac{15x}{2} = 75$

$\implies x= 75 \times \frac{2}{15}$

$\implies x= 5 \times 2$

$\implies x= 10$

Therefore.,

$* number \: of \: \ 1.50 \: wrenches \\purchased = x \green {= 10}$

$* number \: of \: \ 2.50 \: wrenches \\purchased = x\green {= 10}$

$* number \: of \: \ 4 \: wrenches \\purchased = \frac{x}{2} \\= \frac{10}{2}\\\green {= 5}$

$* number \: of \: \ 3 \: wrenches \\purchased = \frac{x}{2} + 1\\= 5 + 1 \\\green { = 6 }$

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