SAT Level 5 Heart of Algebra Problem with Solution

Today I would like to give you a difficult Heart of Algebra problem with a solution. Additional methods of solution will be provided in the next few days.

Level 5 Heart of Algebra

A worker earns $12 per hour for the first 40 hours he works in any given week, and $18 per hour for each hour above 40 that he works each week. If the worker saves 75% of his earnings each week, what is the least number of hours he must work in a week to save at least $441 for the week?

A) 6
B) 8
C) 46
D) 47

* Informal solution: If $441 represents 75% of the worker’s earnings, then the worker’s total earnings is 441/0.75 = $588.

For the first 40 hours, the worker earns 12⋅40=480 dollars. So, the remaining amount that the worker needs to earn is 588-480=108 dollars. Therefore, the number of additional hours above 40 that the worker will work is 108/18 = 6.

The total number of hours that the worker must work is therefore 40 + 6 = 46, choice C.

Notes: (1) We change a percent to a decimal by moving the decimal point to the left 2 places. The number 75 has a “hidden” decimal point at the end of the number (75 = 75. or 75.0). When we move this decimal point to the left two places we get .75 or 0.75.

(2) We can find the worker’s total earnings formally as follows:

We are given that 441 is 75% of the worker’s total earnings. So, we have 441 = 0.75T, where T is the worker’s total earnings. We divide each side of this equation by 0.75 to get T = 441/0.75 = 588.

(3) Be careful that you do not accidentally choose 6 as the answer. 6 is the number of hours above 40 that the worker must work. The question is asking for the total number of hours, which is 40 + 6.

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