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Because her test turned out to be more diffi-cult than she

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Because her test turned out to be more diffi-cult than she [#permalink] New post 16 May 2017, 08:32
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62% (01:52) correct 37% (01:25) wrong based on 107 sessions


Because her test turned out to be more difficult than she intended it to be, a teacher decided to adjust the grades by deducting only half the number of points a student missed. For example, if a student missed 10 points, she received a 95 instead of a 90. Before the grades were adjusted the class average was A. What was the average after the adjustment?

A) \(50 + \frac{A}{2}\)

B) \(\frac{1}{2}\) * \((100 - A)\)

C) \(100 -\) \(\frac{A}{2}\)

D) \(\frac{50 + A}{2}\)

E) \(A + 25\)
[Reveal] Spoiler: OA

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Re: Because her test turned out to be more diffi-cult than she [#permalink] New post 18 Dec 2017, 17:47
please explain the answer?
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Re: Because her test turned out to be more diffi-cult than she [#permalink] New post 18 Dec 2017, 21:07
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divyavivek wrote:
please explain the answer?


I got it just by plugging in numbers.

Let's say, A=80.

How do I get a spread for an 80 average? Easy, four tests 90,70, 90,70.

Then I do the teacher's changes: 95+85+95+85/4= 90

Now, which of the answers=90?

A
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Re: Because her test turned out to be more diffi-cult than she [#permalink] New post 20 Jan 2018, 03:18
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Alternately, can we not solve this as, 1/2* (100-A)+A
Hence, 1/2*(100-80)+80= 90
Please let me know if this is right?
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Re: Because her test turned out to be more diffi-cult than she [#permalink] New post 20 Jan 2018, 06:56
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The easiest way to do this question is by making up numbers. However, there's no need to make different numbers!

You have variables in the answer choices so you definitely want to find A and the plug it in, and it's an average question so you should immediately put down \(\frac{Sum}{Count=Avg}\)

We can make life easy for ourselves by saying there's only 2 people in the class so Count=2. We can say they're original score was both 90. The new scores would've been both 95.

Therefore, the new avg is simply 95!

Plug in 90 in the answer choices and see which gives you 95, which is A.

Just remember, variables in the answer choices don't do algebra. Plug and Chug!

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Re: Because her test turned out to be more diffi-cult than she [#permalink] New post 02 Apr 2018, 05:30
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Original mark average - A =( X1 + X2 + X3 + .... Xn) / (n), where Xn is mark resicived by the n number of student.

After readjustment

Student mark = (100- (100-Xn)/2) or 50+ 1/2 * Xn and the readjustment average

average = ( 50+1/2*X1 +50+1/2*X2 +50+1/2*X3 .....50+1/2*Xn) / n
average = (50* n+ 1/2 * ( X1 + X2 + X3 + .... Xn)) / n

average = 50 + 1/2 * A
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Re: Because her test turned out to be more diffi-cult than she [#permalink] New post 04 Apr 2018, 02:05
Assuming there is only one student in the class. The class average earlier was A. Now, after the adjustment, the new average would be A + ((100 - A)/2). Hence the answer is 'A'
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Re: Because her test turned out to be more diffi-cult than she [#permalink] New post 02 May 2018, 09:45
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Carcass wrote:


Because her test turned out to be more difficult than she intended it to be, a teacher decided to adjust the grades by deducting only half the number of points a student missed. For example, if a student missed 10 points, she received a 95 instead of a 90. Before the grades were adjusted the class average was A. What was the average after the adjustment?

A) \(50 + \frac{A}{2}\)

B) \(\frac{1}{2}\) * \((100 - A)\)

C) \(100 -\) \(\frac{A}{2}\)

D) \(\frac{50 + A}{2}\)

E) \(A + 25\)


Since the average is A, we see that, on average, each student misses (100 - A) points. Since the teacher decided to deduct only half the number of points a student missed, on average, she deducted (100 - A)/2 points from each student. Therefore, the new average is:

100 - (100 - A)/2 = 100 - 100/2 + A/2 = 50 + A/2

Answer: A
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Re: Because her test turned out to be more diffi-cult than she [#permalink] New post 02 May 2018, 11:48
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The way I did this was make up a scenario. I chose a class of 5 students and gave them random numbers for the test, 80, 60, 40, 30, 90. The average A here was 60. Then I did the adjustment for each grade and that gave the average 80. From there I worked out a formula. It took me a little less than 4mins. Experts, is this too much time for this kind of problem?
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Re: Because her test turned out to be more diffi-cult than she [#permalink] New post 03 Nov 2018, 09:05
JeffTargetTestPrep wrote:
Carcass wrote:


Because her test turned out to be more difficult than she intended it to be, a teacher decided to adjust the grades by deducting only half the number of points a student missed. For example, if a student missed 10 points, she received a 95 instead of a 90. Before the grades were adjusted the class average was A. What was the average after the adjustment?

A) \(50 + \frac{A}{2}\)

B) \(\frac{1}{2}\) * \((100 - A)\)

C) \(100 -\) \(\frac{A}{2}\)

D) \(\frac{50 + A}{2}\)

E) \(A + 25\)


Since the average is A, we see that, on average, each student misses (100 - A) points. Since the teacher decided to deduct only half the number of points a student missed, on average, she deducted (100 - A)/2 points from each student. Therefore, the new average is:

100 - (100 - A)/2 = 100 - 100/2 + A/2 = 50 + A/2

Answer: A


We are assuming total score is 100
Re: Because her test turned out to be more diffi-cult than she   [#permalink] 03 Nov 2018, 09:05
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