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n: 64
S: 66
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
76% (02:40) correct
24%
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wrong
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A university professor teaches an introductory class with n students. On the first exam, the average (arithmetic mean) of the n scores of these students was exactly 72.5. After the exam, a new student transferred into the professor's class. The student had taken the same exam in another professor's class. The new average score on the exam, including the n scores of the original students, and the new student's score of S, was exactly 72.4.
Select for n and for S values that are jointly consistent. Make only 2 selections, one for each.
Select for n and for S values that are jointly consistent. Make only 2 selections, one for each.
| n | S | |
| 62 | ||
| 63 | ||
| 64 | ||
| 65 | ||
| 66 |
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Official Answer
n: 64
S: 66
poojaarora1818
Joined: 30 Jul 2019
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10 Dec 2024, 19:21
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Bismuth83
A university professor teaches an introductory class with n students. On the first exam, the average (arithmetic mean) of the n scores of these students was exactly 72.5. After the exam, a new student transferred into the professor's class. The student had taken the same exam in another professor's class. The new average score on the exam, including the n scores of the original students, and the new student's score of S, was exactly 72.4.
Select for n and for S values that are jointly consistent. Make only 2 selections, one for each.
We know the average mean on the first exam = 72.5Select for n and for S values that are jointly consistent. Make only 2 selections, one for each.
No. of students = n
The sum of the scores on the first exam = 72.5n
As mentioned above after the exam a new student is transferred into the professor's class.
So. now the no.of the students would be n+1
The new student's score would be s The new average score on the exam including the n scores of the original students and the new student's score of s would be
72.5n + s/n+1 = 72.4
By solving the above equation we get 0.1n = 72.4 - s
Now, to equate the above equation we look at the values for n and s that best match up and we get the value for n as 64 and s as 66
Hence, n = 64 and s = 66
General Discussion
11 Dec 2024, 09:00
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1. Let's number all of the students from 1 to n+1 and define \(a_i\) as the score of the i th student.
2. Using the formula for the mean, it is known that \(\frac{a_1 + a_2 + ... + a_n}{n} = 72.5\).
3. When the new student's score is included in the mean, we have: \(\frac{a_1 + a_2 + ... + a_n + a_{n+1}}{n + 1} = 72.4 \rightarrow \frac{a_1 + a_2 + ... + a_n + S}{n + 1} = 72.4\).
4. Notice that the sum of the first n students' scores can substituted in the second equation. \(\frac{a_1 + a_2 + ... + a_n}{n} = 72.5 \rightarrow a_1 + a_2 + ... + a_n = 72.5n\). \(\frac{a_1 + a_2 + ... + a_n + S}{n + 1} = 72.4 \rightarrow \frac{72.5n + S}{n + 1} = 72.4\).
5. Now, let's simplify this equation: \(\frac{72.5n + S}{n + 1} = 72.4 \rightarrow 72.5n + S = 72.4n + 72.4 \rightarrow 0.1n + S = 72.4\).
6. 0.1n can be 6.2, 6.3, 6.4, 6.5, or 6.6 and S can be 62, 63, 64, 65, or 66. See how the 0.4 in 72.4 is dependent on 0.1n. So, 0.1n = 6.4.
7. Our answer will be: n - 64 and S - 66.
2. Using the formula for the mean, it is known that \(\frac{a_1 + a_2 + ... + a_n}{n} = 72.5\).
3. When the new student's score is included in the mean, we have: \(\frac{a_1 + a_2 + ... + a_n + a_{n+1}}{n + 1} = 72.4 \rightarrow \frac{a_1 + a_2 + ... + a_n + S}{n + 1} = 72.4\).
4. Notice that the sum of the first n students' scores can substituted in the second equation. \(\frac{a_1 + a_2 + ... + a_n}{n} = 72.5 \rightarrow a_1 + a_2 + ... + a_n = 72.5n\). \(\frac{a_1 + a_2 + ... + a_n + S}{n + 1} = 72.4 \rightarrow \frac{72.5n + S}{n + 1} = 72.4\).
5. Now, let's simplify this equation: \(\frac{72.5n + S}{n + 1} = 72.4 \rightarrow 72.5n + S = 72.4n + 72.4 \rightarrow 0.1n + S = 72.4\).
6. 0.1n can be 6.2, 6.3, 6.4, 6.5, or 6.6 and S can be 62, 63, 64, 65, or 66. See how the 0.4 in 72.4 is dependent on 0.1n. So, 0.1n = 6.4.
7. Our answer will be: n - 64 and S - 66.
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