rishabhmishra wrote:
Q.13 students in a class had a median score 10 and an average score of 15. If all scores were in integers and mike scored higher than all other students in the quiz, What can be the lowest possible score of mike?
A. More than 23
B. 23
C. 22
D. 21
E. 20
You can do this problem without algebra.
Regardless of method, this problem has a wrinkle that may approach cognitive dissonance; we need the lowest possible high score by maximizing the not-highest scores.
In steps, it's not so bad. Start with what we know: how to find the sum of the scores.
1) Total of all scores?
\(A * n = S\) (15 * 13) = 195
We need to "use up" as many of these points as possible. We are trying to keep Mike's high score small.
To use up points, maximize scores of all but Mike, in two different ways (steps 2 and 3).
2) Maximize some of the others' scores with the median. Those values have an easily determined fixed maximum.
13 people. Median = 10: 6 people are to the left of median, and 6 are to the right.
Allot 10 points each to the 6 on the left:
10, 10, 10, 10, 10, 10,
10 __ __ __ __ __ __
Total? 10 * 7 = 70
Remaining points? (195 - 70) = 125 points
3) With 125 points remaining, use answer choices to assign scores to the other 6.
Subtract Mike's score from 125. Then give a score of ONE fewer than Mike to as many people as possible.*
I started with D) 21 = Mike's score
125 - 21 = 104
Assign 20, which is ONE fewer than Mike, to as many as possible:
20, 20, 20, 20 . . .and 24
(104 - 80) = 24
Answer D will not work.
The fifth person has to "take" 24 points. Not allowed. Mike is highest.
Try C) Mike = 22
125 - 22 = 103
Assign 21, which is ONE fewer than Mike, to as many as possible:
21, 21, 21, 21 . . .19
(103 - 84) = 19
That works.
Mike has high score of 22.
Four others took maximum of 21.
Fifth person takes 19.
Answer C
*
Others' scores must be: integers; smaller than Mike's; AND maximized in order to give Mike the fewest points for the "win." _________________