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Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
13 students in a certain class had a median score of 10 and an average
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24 Aug 2021, 04:40
24 Aug 2021, 04:40
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Question Stats:
57% (03:05) correct
42% (02:57) wrong based on 28 sessions
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13 students in a certain class including Mike, have a median score of 10 and an average score of 15.
All the scores are integers and Mike scored higher than all other students in the quiz.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
All the scores are integers and Mike scored higher than all other students in the quiz.
Quantity A |
Quantity B |
Lowest possible score of Mike |
23 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: 13 students in a certain class had a median score of 10 and an average
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03 Sep 2021, 21:46
03 Sep 2021, 21:46
1
Explanation:
\(a_1, a_2, ........., a_7, .........., a_{12}, a_{13}\)
Given:
\(a_1, a_2, ........., 10, .........., a_{12}, a_{13}\), and
\(\frac{a_1 + a_2 + ......... + 10 + .......... + a_{12} + a_{13}}{13} = 15\)
i.e.\(a_1 + a_2 + ......... + 10 + .......... + a_{12} + a_{13} = 195\)
Since, we need to find the minimum value of \(a_{13}\), we must keep all other scores maximum
\(10 + 10 + 10 + 10 + 10 + 10 + 10 + a_8 + .......... + a_{12} + a_{13} = 195\)
\(70 + a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} = 195\)
\(a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} = 125\)
\(a_{13} = 125 - (a_8 + a_9 + a_{10} + a_{11} + a_{12})\)
Now, if \(a_8 = a_9 = a_{10} = a_{11} = a_{12} = 20\)
\(a_{13} = 125 - 5(20) = 25\)
if \(a_8 = a_9 = a_{10} = a_{11} = a_{12} = 21\)
\(a_{13} = 125 - 5(21) = 20\), which cannot be the case as \(a_{13}\) has to be the maximum value of all
What if we take \(a_8 = a_9 = a_{10} = a_{11} = 21\) and \(a_{12} = 20\)
\(a_{13} = 125 - 4(21) - 20 = 21\), which again is not possible
Let us take \(a_8 = a_9 = a_{10} = 21\) and \(a_{11} = a_{12} = 20\)
\(a_{13} = 125 - 3(21) - 2(20) = 22\), POSSIBLE
Let us take \(a_8 = a_9 = 21\) and \(a_{10} = a_{11} = a_{12} = 20\)
\(a_{13} = 125 - 2(21) - 3(20) = 23\), the value starts increasing
Col. A: \(22\)
Col. B: \(23\)
Hence, option B
\(a_1, a_2, ........., a_7, .........., a_{12}, a_{13}\)
Given:
\(a_1, a_2, ........., 10, .........., a_{12}, a_{13}\), and
\(\frac{a_1 + a_2 + ......... + 10 + .......... + a_{12} + a_{13}}{13} = 15\)
i.e.\(a_1 + a_2 + ......... + 10 + .......... + a_{12} + a_{13} = 195\)
Since, we need to find the minimum value of \(a_{13}\), we must keep all other scores maximum
\(10 + 10 + 10 + 10 + 10 + 10 + 10 + a_8 + .......... + a_{12} + a_{13} = 195\)
\(70 + a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} = 195\)
\(a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} = 125\)
\(a_{13} = 125 - (a_8 + a_9 + a_{10} + a_{11} + a_{12})\)
Now, if \(a_8 = a_9 = a_{10} = a_{11} = a_{12} = 20\)
\(a_{13} = 125 - 5(20) = 25\)
if \(a_8 = a_9 = a_{10} = a_{11} = a_{12} = 21\)
\(a_{13} = 125 - 5(21) = 20\), which cannot be the case as \(a_{13}\) has to be the maximum value of all
What if we take \(a_8 = a_9 = a_{10} = a_{11} = 21\) and \(a_{12} = 20\)
\(a_{13} = 125 - 4(21) - 20 = 21\), which again is not possible
Let us take \(a_8 = a_9 = a_{10} = 21\) and \(a_{11} = a_{12} = 20\)
\(a_{13} = 125 - 3(21) - 2(20) = 22\), POSSIBLE
Let us take \(a_8 = a_9 = 21\) and \(a_{10} = a_{11} = a_{12} = 20\)
\(a_{13} = 125 - 2(21) - 3(20) = 23\), the value starts increasing
Col. A: \(22\)
Col. B: \(23\)
Hence, option B
General Discussion
Re: 13 students in a certain class had a median score of 10 and an average
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25 Aug 2021, 02:10
25 Aug 2021, 02:10
solution please.
