Mrs. Walter gave an exam in a mathematics class of five students. She

Question

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71,76,80,82, and 91. What was the last score Mrs. Walters entered?

(A) 71
(B) 76
(C) 80
(D) 82
(E) 91

Show Answer

Official Answer

C

(A) 71 
(B) 76
(C) 80 
(D) 82
(E) 91

EgmatQuantExpert · Answer

Solution

Given:  
 • Mrs. Walter gave an exam in a mathematics class of five students.
• She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered.
• Mrs. Walter noticed that after each score was entered, the average was always an integer.
• The scores (listed in ascending order) were 71, 76, 80, 82 and 91.

To find:  
 • The value of the last score Mrs. Walters entered.

Approach and Working:  
As in each step the average is integer, we can say:
 • First number is divisible by 1.  (can be any of the 5 numbers) 
• The sum of the first two numbers is even.  (even sum can be obtained in multiple ways, eg. 76 + 80 or 80 + 82 or 71 + 91 etc.) 
• Sum of the first three numbers is divisible by 3.
• Sum of the first four numbers is divisible by 4.
• Sum of the first five numbers is divisible 5, which is equal to 71 + 76 + 80 + 82 + 91 = 400.

Now, if we want to get the sum of the first 3 numbers is divisible by 3, then the numbers must be 76, 82 and 91 only, as these three numbers give remainder 1, when divided by 3  (1 + 1 + 1 = 3, hence, their sum is divisible by 3) .
 • Sum of the first 3 numbers = 76 + 82 + 91 = 249

As the sum of first 3 numbers is 249 and the sum of first 4 numbers is divisible by 4, the 4th number must be 71.
 • And therefore, the last number entered is 80.

Hence, the correct answer is option C.

Answer: C

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-NP-e1533292144892.png

ThatDudeKnows · Answer

The sum is 400, which is divisible by 4, and the total after the fourth score must also be divisible by 4.
That means that the fifth score is divisible by 4. There are two options: 76 or 80.
If the fifth score is 80 and the overall average is also 80 (400/5), that means the other four scores would need to average 80.
71 is under by 9. 76 is under by 4. That's a total of 13 under. 82 is over by 2. 91 is over by 11. That a total of 13 over. 
The under and over offset, so the average of those four is 80. We are all set with 80 as the fifth score.

Answer choice C.

Kinshook · Answer

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71,76,80,82, and 91. What was the last score Mrs. Walters entered?

(A) 71
(B) 76
(C) 80
(D) 82
(E) 91

Show Answer

Official Answer

C

What was the last score Mrs. Walters entered?

Let the last score entered by Mrs Walters be x.

(71+76+80+82+91+x)/6 is an integer
(400+x)/6 is an integer

(A) 71 : 471/6 is NOT an integer
(B) 76: 476/6 is NOT an integer
 (C) 80: 480/6 is an integer 
(D) 82: 482/6 is NOT an integer
(E) 91: 491/6 is NOT an integer

IMO C

HarshavardhanR · Answer

Scores: 71, 76, 80, 82, 91.

After each score entered, the average was an integer.

(1) 1st number added: Whatever the first number is, the average would always be an integer (the same integer)

(2) 2nd number added: for the average to be an integer, the sum of the 2 numbers entered has to be even.

Cases:

a) 71, 91
b) 76,80
c) 80, 82
d) 76, 84

(3) 3rd number added: for the average to be an integer, the sum of the 3 numbers entered has to be a multiple of 3.

Let's explore the possibilities with the 4 cases above.
 
a) 71, 91. Sum 162 = 3k. For the sum after adding the next number to still be a multiple of 3, we need to add a number which is a multiple of 3. But none of the 5 numbers given (71, 76, 80, 82, 91) are multiples of 3. Thus, these two numbers are not the first two numbers.

b) 76, 80. Sum 156 = 3k. For the sum after adding the next number to still be a multiple of 3, we need to add a number which is a multiple of 3. But none of the 5 numbers given (71, 76, 80, 82, 91) are multiples of 3. Thus, these two numbers are not the first two numbers.

c) 80, 82. Sum 162 = 3k. For the sum after adding the next number to still be a multiple of 3, we need to add a number which is a multiple of 3. But none of the 5 numbers given (71, 76, 80, 82, 91) are multiples of 3. Thus, these two numbers are not the first two numbers.

d) 76, 82. Sum 158 = 3k+2. For the sum after adding the next number to be a multiple of 3, we need to add a number of the form 3k+1. Among the remaining 3 numbers, 91 fits.

Hence,

First 3 numbers: (76, 82), and 91.

Now,

(4) 4th number added: for the average to be an integer, the sum of the 4 numbers entered has to be a multiple of 4.

Sum of 1st 3 numbers = 76 + 82 + 91 = 249.

The 4th number has to be 71 (if we take 80 as the 4th number, the sum 249 + 80 is not even an even number, it is not divisible by 4).

So, the first 4 numbers -> (76,82), 91, 71.

Hence, the last number entered = 80. Choice C.

Hope this helps.
Harsha

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