Inequality Word Problem for the SAT with Solutions Today I would like to give a solution to the SAT problem from last Wednesday on Inequalities. You can see the original post here: Inequality Word Problem for the SAT Level 5 Heart of Algebra A worker earns $12 per hour for the first 40 hours he works in any given week, $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. So, 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. Algebraic solution 1: Let x be the number of hours that the worker works above 40 hours. We need to solve the following inequality for x: 0.75(12⋅40 + 18x) ≥ 441 480 + 18x ≥ 441/0.75 480 + 18x ≥ 588 18x ≥ 108 x ≥ 6 So, the least number of hours the worker needs to work is 40 + 6 = 46, choice C. Notes: (1) For the first 40 hours, the worker earns 12 ⋅ 40 = 480 dollars. (2) For x hours above 40, the worker earns 18x dollars. (3) using notes (1) and (2), we see that he worker earns a total of 12 ⋅ 40 + 18x = 480 + 18x dollars. (4) We are given that 75% of the total earned must be at least 441. So, 0.75T = 441, where T = 12 ⋅ 40 + 18x (see note (2) in the previous solution). (5) Remember that we let x represent the number of hours the worker works above 40. So, at the end, we need to add 40 to the result. Algebraic solution 2: This time we let x be the total number of hours that the worker works, where x must be at least 40. We need to solve the following inequality for x: 0.75(12⋅40 + 18(x – 40)) ≥ 441 480 + 18x – 720 ≥ 441/0.75 18x – 240 ≥ 588 18x ≥ 828 x ≥ 46 So, the least number of hours the worker needs to work is 46, choice C. Notes: (1) This time we are letting x represent a number greater than or equal to 40. If x = 40, the total earnings is 12 ⋅ 40 + 18 ⋅ 0 If x = 41, the total earnings is 12 ⋅ 40 + 18 ⋅ 1 If x = 42, the total earnings is 12 ⋅ 40 + 18 ⋅ 2 Observe the relationship between x and the last number in the expression (both in bold). We subtract 40 from the x-value to get that number. This shows that the total earnings is 12 ⋅ 40 + 18(x – 40). (2) This time, the value that we get for x is the answer to the question. Share this problem with your Facebook friends that are preparing for the SAT If you’re preparing for the SAT, you may want to check out the Get 800 collection of SAT math books. Comments comments