13 students in a class had a medium score 10 and an average score of

Sorry and thanks now i rectify the problem

It must be more than 23 ...so option a...

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Since 10 is the medial score and we have to keep Mike's score the lowest. to Keep Mike's score lowest, we have to keep other score highest possible. First 6 scores can't be more than 10, so just consider them 10; this makes a total of 70 till 7th score.
total is 195 (13x15=150)
We are left with 125 (195-70) to be distributed in 6 spots in which the last spot, which is Mike's, needs to be highest.
The size spot can take the following values :
20,20,20,20,22,23 OR
20,21,21,21,21,21 OR
20,20,21,21,21,22 (this one satisfies both conditions - Lowest possible score of Mike but higher than others.

brother OA is correct i can explain you how?

total of 13 students will be 15*13= 195
so to minimize mike we need to make numbers before median equal i.e. all the no. till 7th digits are 10 so
10*7=70
195-70=125
now if mike score is x then other previous 5 digits will be x-1 to minimize mike
so total will be
5(x-1)+x=125
5x-5+x=125
6x=130
x=21.66
so x must be integer so it can be rounded to greater one
then answer is 22(c)
hope you like my explanation

I agree that the answer must be 22. Corrected my solution!

As per the question, let Mike score be x & let the sample space be -

\({10, 10, 10, 10, 10, 10, 10, x-1, x-1, x-1, x-1, x-1, x}\)

6x = 130 --> x = 22(Because x has to be unique).

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."

13 student, and median is 10, average is 15
so total score is 195 (15*13)

XXXXXX10 XXXXXX
A) 6 students lower than 10
B) 6 students higher than 10

assume that A) students all got 10 so 10*6 =60 [ because you have to find out the lowest possible for mike]

195 - 60 - 10 = 125

125 = 5(x-1) +5
X= 21.8, so 22

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