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A student's average (arithmetic mean) test score on 4 tests is 78. Wha

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A student's average (arithmetic mean) test score on 4 tests is 78. Wha  [#permalink]

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Originally posted by Bunuel on 26 Aug 2012, 03:07.
Last edited by Bunuel on 04 Feb 2019, 05:32, edited 1 time in total.
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Re: A student's average (arithmetic mean) test score on 4 tests is 78. Wha  [#permalink]

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New post 26 Aug 2012, 03:08
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SOLUTION:

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 \(312+x=400\) --> \(x=88\).

Answer: E.
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Re: A student's average (arithmetic mean) test score on 4 tests is 78. Wha  [#permalink]

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New post 12 Apr 2013, 10:42
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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.
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Re: A student's average (arithmetic mean) test score on 4 tests is 78. Wha  [#permalink]

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New post 15 Dec 2012, 08:28
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I follow Manhattan advice on this one and write the average formula as a first thing:

in order to reach an average of 80 over 5 test the students has to reach 400 points; he now has 78*4= 312.

400-312= 88.

Answer E
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Re: A student's average (arithmetic mean) test score on 4 tests is 78. Wha  [#permalink]

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New post 06 Jun 2014, 02:56
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Let students score be

78..... 78..... 78....... 78

Average after 5th test should be 80

So score in 5th exam should at least be 80 & also it should "cover up the loses" in the earlier tests

Loses earlier = 2 per test = 2 * 4 = 8

80 + 8 = 88

Answer = E
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Re: A student's average (arithmetic mean) test score on 4 tests is 78. Wha  [#permalink]

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New post 06 Jun 2014, 06:08
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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!!
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Re: A student's average (arithmetic mean) test score on 4 tests is 78. Wha  [#permalink]

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New post 25 May 2016, 07:18
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 is 78. Wha  [#permalink]

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New post 29 Jan 2018, 16:02
Hi All,

The standard average formula 'approach' is likely the easiest for most Test Takers to use on these types of prompts. However, there is another way to think about this prompt - since the average of the first 4 tests is 78, we can think of those tests as four 78s. For a 5th exam to raise the overall average to 80, that exam has to score 80 points on its own AND raise each of the other 4 tests to 80. Those 4 tests would require an extra 2 points apiece...

80 + 4(2) = 88 points

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A student's average (arithmetic mean) test score on 4 tests is 78. Wha  [#permalink]

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New post Updated on: 01 May 2019, 08:43
Hi,

For questions like these we have a very quick method called deviation method.
Fisrt, Average is nothing but adding up everything and dividing it equally.

Let me take a moment to explain how this method works.
Say we are told average weight of a group of 7 persons is 60 Kg. If new person who weighs 76 Kgs joins the group, what will be the new average?

There are 7 people in the group initially and 1 person joins the group. And this new person's weight is 76Kgs.
Now just for a moment assume that the new persons weight was 60kg, which is the same as the initial average .Then if new persons weight was 60 Kgs then the average would remain unchanged. But the new persons weight is 76 Kgs which is 16 kgs more than 60 Kgs.
Average is dividing the weight equally among all people. So this deviation of 16kgs would be divided equally among 8 people of the group.

New Avg= 60 +16/8
New Avg= 62
In general:
New avg= Assumed value +or- (deviation/n+1)


let's take one more Say we were told that 4 Cell Phones were sold at average price of 1200 $, if 5 cell phone is sold at 1100$ what is the new average?

Here if the 5th Cell phone were to sell at 1200$ the new average would have been the same.

So lets assume the price at which fifth cellphone is sold as 1200. Now since the fifth Cellphone is sold at 1100 which is $ 100 less than the assumed price , So this deviation of - $100 would be divided equally among the 5 Cellphones in the group.
hence
new avge= 1200- (100/5)
new avg= 1200-20
new avg=1180/-





So in general ,
New Avg= Assumed new value +or- (deviation)/n+1

No lets try using this in above question,

Lets say the student scores 78 on the fifth test. Now since the new average is increasing the deviation must have been added to the assumed value.lets say the deviation is x
So
80=78+(deviation/5)
80=78+x/5
2=x/5
so x=10
This means deviation 10 from assumed 78 Hence the score on 5 Test must have been 88.


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Originally posted by Probus on 15 Aug 2018, 16:07.
Last edited by Probus on 01 May 2019, 08:43, edited 1 time in total.
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Re: A student's average (arithmetic mean) test score on 4 tests is 78. Wha  [#permalink]

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New post 28 Aug 2018, 14:54
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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


My first approach would be identical to the ones shown above. However, we can also solve the question using the weighted averages formula:

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

We want the TOTAL average to equal 80
4/5 of the 5 scores have an average of 78
Let x = average of the 1 remaining test, which is worth 1/5 of the total average.

We get: 80 = (4/5)(78) + (1/5)(x)
Multiply both sides by 5 to get: 400 = 312 + x
Solve: x = 88

Answer: E

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Re: A student's average (arithmetic mean) test score on 4 tests is 78. Wha  [#permalink]

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New post 30 Aug 2018, 14:11
While there have already been some efficient approaches to this question posted in this thread, I thought that I would also show how this question can be solved quickly using the weighted average mapping strategy (the link contains a list of questions that can be used to practice using the weighted average mapping strategy, also knows as the tug of war).

In general, the weighted average mapping strategy can be used to solve problems like this, where 1 more item is being added to a group. One weight will be the number in the original group (4 in this question), and the other weight will be 1, the size of the "group" of one being added; these are shown above the line in the diagram below. We know the average of the original group (78) and the final weighted average (80), so we can fill these in below the line in the diagram. We need to solve for the "average" of the group of 1 (which will be the score on the 5th test) by making the ratio of the weights equal to the ratio of the distances. The ratio of the weights is 4:1, and the smaller distance is 2, so the larger distance must be 8. Thus, the score of the 5th test is 80+8 = 88.

Image

When I drew the diagram above by hand, I timed myself and found that it took me just 12 seconds. Please let me know if you have any questions, or if you want me to post a video solution!
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Re: A student s average (arithmetic mean) test score on 4 tests  [#permalink]

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Re: A student s average (arithmetic mean) test score on 4 tests   [#permalink] 18 Feb 2019, 05:17
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