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Avg of 20 students = 20
So, total= 140

Now, Sum of 11 students that get each of possible whole no is 0+1+...+10=55

Now, sum of rest 9 students marks= 140-55= 85

For getting lowest 2 students' marks, We have to take max marks for most of students.
If rest 9 students got 10 marks, then their sum = 90, Not possible
Next will be 8 students got 10 marks and the one left student gets 5 marks.
I.e. 8*10+1*5=85
This means that lowest possible marks scored by two students will be 5 (as 1 student had already score 5 in first 11 students group)

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Bunuel
In a class with 20 students, a test was administered, scored only in whole numbers from 0 to 10. At least one student got every possible score, and the average was 7. What is the lowest score that two students could have received?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8


I could be wrong but does anyone find the language of this question problematic ?

"At least one student got every possible score".

"What is the lowest score that two students could have received?" If the correct answer choice is 5 then wouldn't it mean that 5 is the total of minimum marks that two students could have received.

I understand that we are dealing with whole numbers only in this question so last two students could not have gotten equal marks i.e. 2.5.
Language did throw me off. Does GMAC also frame their question in similar language ?
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Bunuel
In a class with 20 students, a test was administered, scored only in whole numbers from 0 to 10. At least one student got every possible score, and the average was 7. What is the lowest score that two students could have received?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Use the concept of Deviation (Deficit = Excess)

11 students got scores from 0, 1, 2, 3, ..., 7, 8, 9, 10.
Since average is 7, for now, Deficit = 7 + 6 + 5 + ... +1 = 28 and Excess = 1 + 2 + 3 = 6
For now, deficit is higher by 22 so other 9 students must make up for it. Also the excess must be as high as possible so that we can find the minimum possible value for one of the leftover students.

Of the 9 students, 8 must be at 10 to increase the excess by 3*8 = 24.
Deficit = 28, Excess = 6 + 24 = 30. Now excess is 2 more than deficit.
So the last student makes a deficit of 2 and hence can have a minimum value of 7 - 2 = 5

Answer (B)

Here is a post discussing the deviation method: https://anaprep.com/arithmetic-usefulness-of-deviations/
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