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

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13 students in a class had a medium score 10 and an average score of  [#permalink]

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Updated on: 17 Mar 2018, 01:13
1
10
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Difficulty:

95% (hard)

Question Stats:

23% (03:07) correct 77% (02:53) wrong based on 112 sessions

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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

Originally posted by rishabhmishra on 17 Mar 2018, 00:45.
Last edited by rishabhmishra on 17 Mar 2018, 01:13, edited 1 time in total.
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Re: 13 students in a class had a medium score 10 and an average score of  [#permalink]

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17 Mar 2018, 00:59
rishabhmishra wrote:
Q.13 students in a class had a medium 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

rishabhmishra - Did you mean median score?
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Re: 13 students in a class had a medium score 10 and an average score of  [#permalink]

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17 Mar 2018, 01:15
pushpitkc wrote:
rishabhmishra wrote:
Q.13 students in a class had a medium 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

rishabhmishra - Did you mean median score?

Sorry and thanks now i rectify the problem
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13 students in a class had a medium score 10 and an average score of  [#permalink]

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17 Mar 2018, 01:51
2
Out of 13 students in the class, the median score is 10(making the 7th score - 10)
We know that the score of the 7 students cannot be greater than 10. Let the score
of these 7 students be 10 and the total be 7*10 = 70
We also know the sum of the scores is 15*13 = 195

If Mike's score is x, $$195 - 70 - x$$ must be an integer such that x is the highest score!

$$125 - x$$ is an integer
When x=21, even if the other 5 scores are 20 each, we will not get 125 as the total.
When x=22, the other 5 scores are 21,21,21,20,20 and Mike's score is the highest.

Therefore, Mike's lowest score could be 22(Option C) such that the other students score an integer score
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Re: 13 students in a class had a medium score 10 and an average score of  [#permalink]

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17 Mar 2018, 01:52
It must be more than 23 ...so option a...

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Re: 13 students in a class had a medium score 10 and an average score of  [#permalink]

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17 Mar 2018, 02:55
1
1
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.
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Re: 13 students in a class had a medium score 10 and an average score of  [#permalink]

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17 Mar 2018, 02:56
pushpitkc wrote:
Out of 13 students in the class, the median score is 10(making the 7th score - 10)
We know that the score of the 7 students cannot be greater than 10. Let the score
of these 7 students be 10 and the total be 7*10 = 70
We also know the sum of the scores is 15*13 = 195

If Mike's score is x, $$195 - 70 - x$$ must be an integer such that x is the highest score!

$$125 - x$$ is an integer when x=20. The other scores are 19,19,19,19,19,10
Therefore, Mike's lowest score could be 20(Option E) such that the other students score an integer score

Please check the OA, rishabhmishra

brother OA is correct i can explain you how?
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13 students in a class had a medium score 10 and an average score of  [#permalink]

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17 Mar 2018, 03:05
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

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
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Re: 13 students in a class had a medium score 10 and an average score of  [#permalink]

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17 Mar 2018, 05:57
rishabhmishra wrote:
pushpitkc wrote:
Out of 13 students in the class, the median score is 10(making the 7th score - 10)
We know that the score of the 7 students cannot be greater than 10. Let the score
of these 7 students be 10 and the total be 7*10 = 70
We also know the sum of the scores is 15*13 = 195

If Mike's score is x, $$195 - 70 - x$$ must be an integer such that x is the highest score!

$$125 - x$$ is an integer when x=20. The other scores are 19,19,19,19,19,10
Therefore, Mike's lowest score could be 20(Option E) such that the other students score an integer score

Please check the OA, rishabhmishra

brother OA is correct i can explain you how?

I agree that the answer must be 22. Corrected my solution!
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Re: 13 students in a class had a medium score 10 and an average score of  [#permalink]

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18 Mar 2018, 05:42
1
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

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).
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13 students in a class had a medium score 10 and an average score of  [#permalink]

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18 Mar 2018, 12:35
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.

*Others' scores must be: integers; smaller than Mike's; AND maximized in order to give Mike the fewest points for the "win."
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