The integers y and z are positive. A group of y juniors and z seniors took a test. The average (arithmetic mean) score of the juniors was 75, the average score of the seniors was 87, and the average score of all the juniors and seniors combined was 84. In the table below, pick two numbers that are consistent with the information that is given. In the first column, select the row that shows the value of y, and in the second column, select the row that shows the value of z.

Official Explanation We're given two groups of test scores—one averaging 75 and the other 87—that combine to average 84. At first glance, this may seem odd, because the average of 75 and 87 is 81, not 84. The reason for this seeming contradiction is that simply averaging 75 and 87 does not account for that fact that the number of juniors (y) and the number of seniors (z) may be different. The fact that the overall average is closer to the seniors' average tells us that there must be more seniors than juniors—the greater number of higher scores pulls the average up. So, we know that z > y. There are two ways to set up an equation to solve for y and z. One takes advantage of the fact that while the juniors' average and the seniors' average may not straightforwardly combine, both the sums of their scores and the number of students do. We know this: (sum of juniors' scores) + (sum of seniors' scores) = (sum of all scores). We can work backward from an average to a sum: average score = (sum of scores) ÷ (number of students). (average score) × (number of students) = sum of scores. Hence, (sum of juniors' scores) + (sum of seniors' scores) = (overall average score) × (total number of students). 75y + 87z = 84x (total number of students) 75y + 87z = 84y + 2) 75y + 87z = 84y + 84z 3z = YW z=3y We can also use the differences between the averages to deduce this relationship: If we had one score of 75 and one score of 87, the average would be 81. Note that this hypothetical average is exactly between 75 and 87; the differences between the terms and the average balance out: 81 – 75 = 6, and 87 – 81 = 6. This will also be true of our actual average of 84, once we account for the relative sizes of y and z: (84 – 75)y should balance out with (87 – 84)z: (84-75)y =(87- 84)z 9y = 3z 3y=Z Both approaches lead us to the same result, so on Test Day we want to use the approach that seems most straightforward. Having deduced (see stimulus explanation, above) that z = 3y, finding a pair of values that satisfy this condition should be straightforward. We'll test a possible value for y by multiplying it by 3. If the result is among the answer choices, we will have found the right answer. Since y is substantially smaller, we'll start with the smallest possible value, y = 18. If y = 18, then z = 3y = 54. That value matches one of the answer choices, so we have found our answers. Now we just need to be careful to select the proper values in the proper column—not to enter the value for z in the column for y by mistake. We might take a moment to verify that our result makes sense logically; we determined above that there must be more seniors than juniors, since the overall average is closer to the seniors' average than the juniors' average. The correct answers are 18 for the value of y (the number of juniors) and 54 for the value of z (the number of seniors).

Jr........Total.........Sr. 75.......84..........87 Y Z (87-84)/(84-75) = Y/Z 3/9= 1/3 = Y/Z So consistent values for Y and Z are 18 and 54 respectively.

(75y+87z)/y+z = 84 so z=3y hence y =18,z=54 Posted from my mobile device

The global average of seniors and juniors is 84. This can be tranlsated in: (75y+87z)/(y+z) = 84 solving, we get: z=3y Let's take the first possible value for y: 18. This makes z = 3(18) = 54. Possible answer. Let's take the second possible value for y: 27. This makes z = 3(27) = 81. We don't have this value (nor any other greater value) in the table so the only possible answer is 18 and 54.