A student’s average (arithmetic mean) test score on four tests is 78.

Question

A student’s average (arithmetic mean) test score on four tests is 78. If each test is scored out of 100, which of the following can be the student’s score on the fifth test so that the student’s average score on the five tests increases by an integer value?

(A) 82
(B) 87
(C) 89
(D) 93
(E) 95

(A) 82
(B) 87
(C) 89
(D) 93
(E) 95

chetan2u · Answer

total score in four tests =  4*78 = 312 
now the average after 5 tests should be an integer, let the marks in 5th be x
so total = 312+x and average =  \frac{(312+x)}{5} = an integer
so  312+x  must be divisible by 5, meaning units digit of 312+x must be 5 or 0
 so units digit of x can be  
1) 5-2=3, or
2) 10-2=8

in the given choices, D has a 3 in units digit
so marks are  312+93=405  and average =  \frac{405}{5}=81

D

KHow · Answer

Could someone provide an algebraic approach to this problem? I seem to be getting stuck on this. Thank you!!

-KHow

Staphyk · Answer

Hello! Let’s define some variables let x= student score on the 5th test
y= increased integer value of the average
Now total of 4 students =4*78=312. Total score of 5students=312+x and average =312+x/5
New average as a result=78+y
Therefore 312+x/5=78+y —> 312+x=390+5y
x-78=5y —> x-78/5=y since y has to be an integer then x-78/5 has to be an integer and using the answer choices only (93-78)/5 is an integer so Answer is D 
Hope it helps

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energetics · Answer

The formula is Avg=Sum/n items
Originally, 78=S/4, or Sum = 4*78 = 312 
What the question is asking is to make 312+1 more item divisible by 5 with 0 remainder, i.e. (312+x)/5 
Something is divisible by 5 when there is a 0 or 5 on in the units digit.
Looking at the answers, only D fits.

ScottTargetTestPrep · Answer

We can let n = the 5th test score and create the equation:

average = (78 x 4 + n)/5

average = (312 + n)/5

So we see that 312 + n must be divisible by 5, so n could be 93.

Answer: D

SchruteDwight · Answer

(new average - old average) * 1/n = (x-78)/5 difference must be a multiple of 5

78+5 = 83, hence 93-78=5k, done.

Mohammadmo · Answer

(4*78) +X /5 =?
312+X /5=?
Just numbers which their unit digits are 0 or 5 can dividede by 5 and make an integer number.
So option D would be the answer.
312+93/5= 405/5=81

Posted from my mobile device

gracie · Answer

4*78=312
new score=(312+x)/5
to be multiple of 5, 312+x must have units digit of 3 or 8
only 93 fits
D

ThickGuitar · Answer

To raise the average by an interger, that means the difference between the mark for the 5th test and the current average has to be divisible by 5 for there to be an integer increase. You can find this by subtracting 78 from the answers. 93-78 = 15 which is divisible by 5.

A student’s average (arithmetic mean) test score on four tests is 78.

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A student’s average (arithmetic mean) test score on four tests is 78.

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