Question

1. When a store had 3/5 of the shirts that were on display left to sell, they brought out another30 from the stockroom. The store likes to keep at least 150 shirts on display.A) Write an inequality the store manager could use to describe the scenario.B) Solve the inequality.C) Explain what the solution means in the situation.

Answer 1

The store had 3/5 of the shirts that were on display left to sell.

This can be written algebraically if x is the number of shirts on display that day, what's remaining is

`(3)/(5)x`

They brought out another 30 from the stockroom since the store likes to keep at least 150 shirts on display.

This means that the number of shirts there at that moment was;

`(3)/(5)x+30`

This number must at least be equal to 150.

A. Therefore, the inequality is;

`(3)/(5)x+30\ge150`

B.

`\begin{gathered} (3)/(5)x+30\ge150 \\ (3)/(5)x\ge150-30 \\ (3)/(5)x\ge120 \\ \text{cross multiply} \\ 3x\ge600 \\ \text{divide both sides by 3} \\ x\ge200 \end{gathered}`

C. The solution means that the store had at least 200 shirts at the start of their sale.