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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.  [#permalink]

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New post 25 Sep 2018, 04:57
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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

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Re: A student’s average (arithmetic mean) test score on four tests is 78.  [#permalink]

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New post 26 Sep 2018, 00:06
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Bunuel wrote:
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


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
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Re: A student’s average (arithmetic mean) test score on four tests is 78.  [#permalink]

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New post 04 Feb 2019, 07:40
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Could someone provide an algebraic approach to this problem? I seem to be getting stuck on this. Thank you!!

-KHow

Bunuel wrote:
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
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A student’s average (arithmetic mean) test score on four tests is 78.  [#permalink]

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New post 04 Feb 2019, 09:11
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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
KHow wrote:
Could someone provide an algebraic approach to this problem? I seem to be getting stuck on this. Thank you!!

-KHow

Bunuel wrote:
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


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Re: A student’s average (arithmetic mean) test score on four tests is 78.  [#permalink]

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New post 04 Feb 2019, 09:30
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.
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Re: A student’s average (arithmetic mean) test score on four tests is 78.  [#permalink]

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New post 06 Feb 2019, 20:06
Bunuel wrote:
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


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
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Re: A student’s average (arithmetic mean) test score on four tests is 78.   [#permalink] 06 Feb 2019, 20:06
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