A university professor teaches an introductory class with n students.

Question

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.

MartyMurray · Accepted Answer

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.

Find the relationship between n and S

Average = Sum/Number

Average × Number = Sum

(Original Average × Original Number) + S = New Average × New Number

72.5n + S = 72.4(n + 1)

72.5n + S = 72.4n + 72.4

0.1n + S = 72.4

Find two values in the list that fit the relationship

62

63

64

65

66

Since all the values are integers, to get 4 in the tenths place in the sum of 0.1n + S, n must be 64 so that 0.1n has 4 in the tenths place.

Thus, n = 64, and S = 72.4 - 6.4 = 66.

Correct answer:   64 ,  66

poojaarora1818 · Answer

We know the average mean on the first exam = 72.5
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

Bismuth83 · Answer

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  .

GMATinsight · Answer

Answer: Option C | E (64 | 66) Please check the video for a detailed explanation. (It's good... I think..

) https://www.youtube.com/watch?v=tM02p_4LAlc To check the cyclicity and many more such videos of questions and their solution please visit the Data Insight playlist here D on't forget to LIKE and SUBSCRIBE the channel If you like them, then you may opt for a modular 100% VIDEO BASED COURSE course on statistics and many more at very low prices. LINK: https://gmatinsight.com/gmat-2/

egmat · Answer

Step 1: Set Up the Data

Original class:  n  students with average  72.5 
 Therefore , total of original scores =  72.5n

New student scores  S , so:
- New number of students =  n + 1 
- New total =  72.5n + S 
- New average =  72.4

Step 2: Write the Equation 
 Average Formula:  (Total of all scores) / (Number of students) = Average

(72.5n + S) / (n + 1) = 72.4

Step 3: Solve for S in terms of n 
72.5n + S = 72.4(n + 1)
72.5n + S = 72.4n + 72.4
S = 72.4n - 72.5n + 72.4
S = -0.1n + 72.4

This gives us:   S = 72.4 - 0.1n

Step 4: Test the Answer Choices 
We need S to be one of our choices:  62, 63, 64, 65, 66

- If n =  62 : S = 72.4 - 6.2 =  66.2  (not a choice)
- If n =  63 : S = 72.4 - 6.3 =  66.1  (not a choice)
- If n =  64 : S = 72.4 - 6.4 =  66   (This works!) 
- If n =  65 : S = 72.4 - 6.5 =  65.9  (not a choice)
- If n =  66 : S = 72.4 - 6.6 =  65.8  (not a choice)

Step 5: Verify 
With n =  64  and S =  66 :
- Original total = 64 x 72.5 =  4640 
- New total = 4640 + 66 =  4706 
- New average = 4706 / 65 =  72.4   (Confirmed!)

Answer: n = 64, S = 66

A university professor teaches an introductory class with n students.

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