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A student's average (arithmetic mean) test score on 4 tests is 78. What must be the student's score on a 5th test for the student's average score on the 5 tests to be 80?

(A) 80 (B) 82 (C) 84 (D) 86 (E) 88

Practice Questions Question: 30 Page: 156 Difficulty: 600

A student's average (arithmetic mean) test score on 4 tests is 78. What must be the student's score on a 5th test for the student's average score on the 5 tests to be 80?

(A) 80 (B) 82 (C) 84 (D) 86 (E) 88

Since the student's average test score on 4 tests is 78, then the sum of the first 4 test scores is 4*78=312. We need the sum of ALL 5 scores to be 5*80=400, so if \(x\) is the 5th test score, then \(312+x=400\) --> \(x=88\).

Re: A student's average (arithmetic mean) test score on 4 tests [#permalink]

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26 Aug 2012, 07:13

Let the score in the 5th test be x. The sum of the first 4 tests = 78*4=312 Now it implies 312+x/5=80 Therefore x=88
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A student's average (arithmetic mean) test score on 4 tests is 78. What must be the student's score on a 5th test for the student's average score on the 5 tests to be 80?

(A) 80 (B) 82 (C) 84 (D) 86 (E) 88

Since the student's average test score on 4 tests is 78, then the sum of the first 4 test scores is 4*78=312. We need the sum of ALL 5 scores to be 5*80=400, so if \(x\) is the 5th test score, then \(312+x=400\) --> \(x=88\).

Re: A students average (arithmetic mean) test score on 4 tests [#permalink]

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16 Dec 2012, 03:17

Ans: total score after 4 tests=312 and score needed after 5 tests= 5x80=400 the student scores the difference in the 5th test which is 88 , (E).
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Re: A student's average (arithmetic mean) test score on 4 tests [#permalink]

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12 Apr 2013, 09:42

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Examining the known average can simply things. We need to reach 80. We currently have 78, four times. 78 is -2 away from 80. Having this four times weights our average -8 off what is desires. To 'catch up' to the average in one test, we must make up the full 8 point deficit in one test. We need an 88. 88 - 8 = 80.

Re: A student's average (arithmetic mean) test score on 4 tests [#permalink]

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06 Jun 2014, 05:08

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I will suggest a little different solution (not a very great trick) but very useful for finding averages since it can save your "TIME" at the tests... Average of 5 tests to be 80 suggests - in each test the student scored 80 (and this can make average 80). SO - In 5th test the student has to score 80 for sure, plus any deficit to make up for the average. We know in 4 tests his averge was 78, which means he is 2 markes short of average(80 in this case) in previous 4 tests and he has to make up for this deficit in the 5th test. Hence he has to score 2 * 4 (deficit * no of tests) = 8. Remember, he has to score 80 also in the 5th test. Hence in total - 80 + 8 = 88. Choice E

Here is how I would approach - 80 in the 5th test, and 2 *4 = 8 totl 88 n the 5th tests. SO it took few seconds to do the ques... Practise this kind of technique while solving questions n averages has been very handy for m... All the best!!

Re: A student's average (arithmetic mean) test score on 4 tests [#permalink]

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05 Jul 2014, 07:04

zerosleep wrote:

I will suggest a little different solution (not a very great trick) but very useful for finding averages since it can save your "TIME" at the tests... Average of 5 tests to be 80 suggests - in each test the student scored 80 (and this can make average 80). SO - In 5th test the student has to score 80 for sure, plus any deficit to make up for the average. We know in 4 tests his averge was 78, which means he is 2 markes short of average(80 in this case) in previous 4 tests and he has to make up for this deficit in the 5th test. Hence he has to score 2 * 4 (deficit * no of tests) = 8. Remember, he has to score 80 also in the 5th test. Hence in total - 80 + 8 = 88. Choice E

Here is how I would approach - 80 in the 5th test, and 2 *4 = 8 totl 88 n the 5th tests. SO it took few seconds to do the ques... Practise this kind of technique while solving questions n averages has been very handy for m... All the best!!

Thanks zerosleep.. I always followed traditional approach.. now i would try your way when encountered with question on averages

Re: A student's average (arithmetic mean) test score on 4 tests [#permalink]

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10 Sep 2014, 02:39

Bunuel wrote:

A student's average (arithmetic mean) test score on 4 tests is 78. What must be the student's score on a 5th test for the student's average score on the 5 tests to be 80?

(A) 80 (B) 82 (C) 84 (D) 86 (E) 88

Practice Questions Question: 30 Page: 156 Difficulty: 600

total of 4 test = 4*78 = 312 if the mean of 5 test has to be 80 then total has to be 400.

400-312=88.

so in the fifth test the score should be 88 for the mean to become 80

Re: A student's average (arithmetic mean) test score on 4 tests [#permalink]

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25 May 2016, 06:18

Bunuel wrote:

A student's average (arithmetic mean) test score on 4 tests is 78. What must be the student's score on a 5th test for the student's average score on the 5 tests to be 80?

(A) 80 (B) 82 (C) 84 (D) 86 (E) 88

To solve, we use the average equation:

average = sum/quantity

We don't know the individual scores for each of the first 4 tests, but because their average is 78, we can think of the first 4 scores as 4 scores of 78 each. We then multiply 78 by 4 to get the sum of those four tests:

78 x 4 = 312

Let's let n = the score on the 5th test. Thus, the sum of the 5 tests is 312 + 5th test score, or 312 + n. Also, because the new average will be 80, we substitute 80 for the new average. We can now solve for n.

Average = sum/quantity

80 = (312 + n)/5

400 = 312 + n

88 = n

The answer is E.
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Re: A student's average (arithmetic mean) test score on 4 tests [#permalink]

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25 May 2016, 07:46

Bunuel wrote:

A student's average (arithmetic mean) test score on 4 tests is 78. What must be the student's score on a 5th test for the student's average score on the 5 tests to be 80?

(A) 80 (B) 82 (C) 84 (D) 86 (E) 88

Practice Questions Question: 30 Page: 156 Difficulty: 600

To have avg score of 80 implies that student needs 80 marks in all 5 tests.

That means 2 marks increase in all 4 tests= 4*2= 8 total marks increase in 4 tests + 80 marks in 5th tests= 88 marks

E is the answer
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