SAT Mathematics Practice Test 7

Ken earned $\mathrm{\$}8$ per hour for the first 10 hours he worked, so he earned a total of $\mathrm{\$}80$ f or the first 10 hours he worked. For the rest of the week, Ken was paid at the rate of $\mathrm{\$}10$ per hour. Let x be the number of hours he will work for the rest of the week. The total of Ken’s earnings, in dollars, for the week will be 10x + 80. He saves 90% of his earnings each week, so this week he will save 0.9(10x + 80) dollars. The inequality 0.9(10x + 80) ≥ 270 represents the condition that he will save at least $\mathrm{\$}270$ for the week. Factoring 10 out of the expression $10\mathit{x}+80$ gives $10\left(\mathit{x}+8\right)$. The product of 10 and 0.9 is 9, so the inequality can be rewritten as $9\left(\mathit{x}+8\right)\ge 270$. Dividing both sides of this inequality by 9 yields $\mathit{x}+8\ge 30$, so $\mathit{x}\ge 22$. Therefore, the least number of hours Ken must work the rest of the week to save at least $\mathrm{\$}270$ for the week is 22.