Author 
Message 
TAGS:


Moderator
Joined: 18 Apr 2015
Posts: 5917
Followers: 96
Kudos [?]:
1158
[1]
, given: 5488

Because her test turned out to be more difficult than she [#permalink]
16 May 2017, 08:32
1
This post received KUDOS
Question Stats:
56% (02:01) correct
43% (01:18) wrong based on 71 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) \(A + 25\)
_________________
Get the 2 FREE GREPrepclub Tests




Intern
Joined: 22 Nov 2017
Posts: 6
Followers: 0
Kudos [?]:
5
[0], given: 2

Re: Because her test turned out to be more difficult than she [#permalink]
18 Dec 2017, 17:47
please explain the answer?



Intern
Joined: 15 Dec 2017
Posts: 1
Followers: 0
Kudos [?]:
1
[1]
, given: 0

Re: Because her test turned out to be more difficult than she [#permalink]
18 Dec 2017, 21:07
1
This post received KUDOS
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



Intern
Joined: 17 Jul 2017
Posts: 15
Followers: 0
Kudos [?]:
7
[1]
, given: 13

Re: Because her test turned out to be more difficult than she [#permalink]
20 Jan 2018, 03:18
1
This post received KUDOS
Alternately, can we not solve this as, 1/2* (100A)+A Hence, 1/2*(10080)+80= 90 Please let me know if this is right?



Manager
Joined: 15 Jan 2018
Posts: 147
GMAT 1: Q V
Followers: 3
Kudos [?]:
182
[4]
, given: 0

Re: Because her test turned out to be more difficult than she [#permalink]
20 Jan 2018, 06:56
4
This post received KUDOS
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! Best, Tae@SherpaPrep
_________________
    
Need help with GRE math? Check out our groundbreaking books and app.



Intern
Joined: 15 Mar 2018
Posts: 32
Followers: 0
Kudos [?]:
8
[2]
, given: 1

Re: Because her test turned out to be more difficult than she [#permalink]
02 Apr 2018, 05:30
2
This post received KUDOS
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 (100Xn)/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



Intern
Joined: 20 Mar 2018
Posts: 34
Followers: 1
Kudos [?]:
24
[0], given: 2

Re: Because her test turned out to be more difficult than she [#permalink]
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'



Target Test Prep Representative
Status: Head GRE Instructor
Affiliations: Target Test Prep
Joined: 09 May 2016
Posts: 161
Location: United States
Followers: 4
Kudos [?]:
125
[1]
, given: 0

Re: Because her test turned out to be more difficult than she [#permalink]
02 May 2018, 09:45
1
This post received KUDOS
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
_________________
Jeffery Miller
Head of GRE Instruction
GRE Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Manager
Joined: 27 Feb 2017
Posts: 189
Followers: 0
Kudos [?]:
49
[1]
, given: 15

Re: Because her test turned out to be more difficult than she [#permalink]
02 May 2018, 11:48
1
This post received KUDOS
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?



Intern
Joined: 27 Oct 2018
Posts: 49
Followers: 0
Kudos [?]:
12
[0], given: 27

Re: Because her test turned out to be more difficult than she [#permalink]
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 difficult than she
[#permalink]
03 Nov 2018, 09:05





