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Re: Because her test turned out to be more diffi-cult than she [#permalink]
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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]
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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]
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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]
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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]
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



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]
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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]
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
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Re: Because her test turned out to be more diffi-cult than she [#permalink]
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