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A group of students took an exam that was scored from 0 to 1

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Carcass
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A group of students took an exam that was scored from 0 to 1 [#permalink] New post   21 May 2020, 10:37
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A group of students took an exam that was scored from 0 to 100 points. 70 percent of the students in the group passed the exam, and these students received an average score of 86. The average score on the exam for the entire group was 74.


Quantity A
Quantity B
The median score on the exam for the entire group of students
Two times the average score of all the students who did not pass the exam



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   05 Jan 2021, 10:08
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Carcass wrote:
A group of students took an exam that was scored from 0 to 100 points. 70 percent of the students in the group passed the exam, and these students received an average score of 86. The average score on the exam for the entire group was 74.

Quantity A
Quantity B
The median score on the exam for the entire group of students
Two times the average score of all the students who did not pass the exam



We can apply the weighted averages formula here...
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Let x = the average score of the students who FAILED the exam.
We can write: 74 = (0.7)(86) + (0.3)(x)
Expend: 74 = 60.2 + 0.3x
Subtract 60.2 from both sides of the equation to get: 13.8 = 0.3x
Solve: x = 13.8/0.3 = 46
So, 46 is the average score of the students who failed the exam

Consider these two conflicting cases:

case i: There are 100 students in total, 30 students all scored exactly 46, and the remaining 70 students all scored exactly 86
In this case, the median score for the entire group = 86
And the average score of all the students who failed the exam = 46
We get:
QUANTITY A: 86
QUANTITY B: 2(46) = 92
In this case, Quantity B is greater

case ii: There are 100 students in total, and 30 students all scored exactly 46. 18 of the remaining students all scored exactly 60, and the remaining 52 students scored 95 (these values meet the condition that the students who passed the exam have an average score of 86)
In this case, the median score for the entire group = 95
And the average score of all the students who failed the exam = 46
We get:
QUANTITY A: 95
QUANTITY B: 2(46) = 92
In this case, Quantity A is greater

Answer: D
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mikearox
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   27 May 2020, 21:19
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We can find the average score of the students who did not pass:

For the 70% that did pass : 0.7x (x is the total number of students)
86 = sum(passing scores)/0.7x
sum(passing scores) = 60.2x

Then we can find the sum(all scores), using x as 100% of the students
74 = sum(all scores)/x
sum(all scores) = 74x

If we remove the passing scores from all scores we get the sum(failing scores)
sum(all scores) - sum(passing scores) = sum(failing scores)
74x - 60.2x = 13.8x

And the average score
avg = 13.8x/(0.3x) = 46

But since 46 * 2 = 92 , this is in the range of "passing scores", so we don't know whether the median score of all students is greater or less than this. (if it were below the range of passing scores, we could say option A is correct, since we know that 70% will include the median)
Then, the answer is D
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   02 Dec 2020, 13:29
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The problem does not say a certain score is why someone does or does not pass an exam. Therefore, let's use 100 students (instead of % and algebra) but please look at the previous poster's response. The total score of the 30 students who failed is 1,380 points, for an average of 46. This average doubled is quantity B, 92. Now for me, I think it is quite easy to imagine the median lower than this number, but equal or higher is more difficult. Let's look at those 30 students again and say that 15 of them scored a 92 and 15 scored a 0. (maybe some were failed because of not showing their work or something). With the remaining 70 students, can we get them to have an average of 86 while at the same time having at least 35 of them over 92? Not too difficult a situation to create.
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   03 Dec 2020, 01:21
We have not given any idea about the distribution of the number, therefore it is not possible to predict the median and that is why we can not make any comparison here. So, D will be the answer.
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   03 Dec 2020, 03:26
endurance wrote:
We have not given any idea about the distribution of the number, therefore it is not possible to predict the median and that is why we can not make any comparison here. So, D will be the answer.

If the median had to be over 100 (because quantity B said 2.3 times the avg score of the lower group) then could we not make a comparison without the distribution?
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   18 Dec 2020, 21:53
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mikearox wrote:
We can find the average score of the students who did not pass:

For the 70% that did pass : 0.7x (x is the total number of students)
86 = sum(passing scores)/0.7x
sum(passing scores) = 60.2x

Then we can find the sum(all scores), using x as 100% of the students
74 = sum(all scores)/x
sum(all scores) = 74x

If we remove the passing scores from all scores we get the sum(failing scores)
sum(all scores) - sum(passing scores) = sum(failing scores)
74x - 60.2x = 13.8x

And the average score
avg = 13.8x/(0.3x) = 46

But since 46 * 2 = 92 , this is in the range of "passing scores", so we don't know whether the median score of all students is greater or less than this. (if it were below the range of passing scores, we could say option A is correct, since we know that 70% will include the median)
Then, the answer is D



I think this answer is incomplete because it doesn't take into the account the constraint placed on the passing group which is that there average must be 86.

We know that the lowest possible score anyone could have scored is 74.000000001 ==> 74. To get the median the highest possible point we put roughly (n/2-1) people on the lowest score and roughly (n/2+1) on the score that makes the average true. Let's say we have 1001 people. We have the expression 86 = (500*74 + 501x)/1001. Solving for x we get ~98. Therefore our upper bound for the median is about ~98%.

(Shortcut way to think about this is the difference from 86 (the mean) to 74 (the lowest possible score) is 12. Since this is how far down we have to go, to get the means to be equal we need to go up 12 as well. That's 98.

We know that the median can be less than the quantity of be (92). An example, let all students score an 86.

So D.
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   05 Jan 2021, 09:38
Can someone please help with this question?
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   05 Jan 2021, 12:08
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GreenlightTestPrep wrote:
Carcass wrote:
A group of students took an exam that was scored from 0 to 100 points. 70 percent of the students in the group passed the exam, and these students received an average score of 86. The average score on the exam for the entire group was 74.

Quantity A
Quantity B
The median score on the exam for the entire group of students
Two times the average score of all the students who did not pass the exam



We can apply the weighted averages formula here...
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Let x = the average score of the students who FAILED the exam.
We can write: 74 = (0.7)(86) + (0.3)(x)
Expend: 74 = 60.2 + 0.3x
Subtract 60.2 from both sides of the equation to get: 13.8 = 0.3x
Solve: x = 13.8/0.3 = 46
So, 46 is the average score of the students who failed the exam

Consider these two conflicting cases:

case i: There are 100 students in total, 30 students all scored exactly 46, and the remaining 70 students all scored exactly 86
In this case, the median score for the entire group = 86
And the average score of all the students who failed the exam = 46
We get:
QUANTITY A: 86
QUANTITY B: 2(46) = 92
In this case, Quantity B is greater

case ii: There are 100 students in total, and 30 students all scored exactly 46. 18 of the remaining students all scored exactly 60, and the remaining 52 students scored 95 (these values meet the condition that the students who passed the exam have an average score of 86)
In this case, the median score for the entire group = 95
And the average score of all the students who failed the exam = 46
We get:
QUANTITY A: 95
QUANTITY B: 2(46) = 92
In this case, Quantity A is greater

Answer: D


Thanks for detailed explanation..It indeed is very helpful..
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   05 Jan 2021, 12:45
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arorni wrote:
GreenlightTestPrep wrote:
Carcass wrote:
A group of students took an exam that was scored from 0 to 100 points. 70 percent of the students in the group passed the exam, and these students received an average score of 86. The average score on the exam for the entire group was 74.

Quantity A
Quantity B
The median score on the exam for the entire group of students
Two times the average score of all the students who did not pass the exam



We can apply the weighted averages formula here...
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Let x = the average score of the students who FAILED the exam.
We can write: 74 = (0.7)(86) + (0.3)(x)
Expend: 74 = 60.2 + 0.3x
Subtract 60.2 from both sides of the equation to get: 13.8 = 0.3x
Solve: x = 13.8/0.3 = 46
So, 46 is the average score of the students who failed the exam

Consider these two conflicting cases:

case i: There are 100 students in total, 30 students all scored exactly 46, and the remaining 70 students all scored exactly 86
In this case, the median score for the entire group = 86
And the average score of all the students who failed the exam = 46
We get:
QUANTITY A: 86
QUANTITY B: 2(46) = 92
In this case, Quantity B is greater

case ii: There are 100 students in total, and 30 students all scored exactly 46. 18 of the remaining students all scored exactly 60, and the remaining 52 students scored 95 (these values meet the condition that the students who passed the exam have an average score of 86)
In this case, the median score for the entire group = 95
And the average score of all the students who failed the exam = 46
We get:
QUANTITY A: 95
QUANTITY B: 2(46) = 92
In this case, Quantity A is greater

Answer: D


Thanks for detailed explanation..It indeed is very helpful..


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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   08 Aug 2021, 10:26
Thank you all for your elaboration, one thing that I’m still confused about is how to get the highest/lowest possible median of the entire group, I didn’t understand the above examples well ..
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   27 Sep 2021, 03:22
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The explanations here aren't simple. My take: Try to prove that A> B and A=B to get the answer D(after initial analysis). set the no of students as 100. You have B as 92.
A: (50th value + 51st value)/2 >=92. Try to make values from 30-49 so small that nos from 50-100 should be 93. Is it possible ? Yes.
93 is 7 away from 86. That means. If 50 values are 93, This increment should be compensated with lower values from 31-49 which is possible(50/19 = 2.6, 2.6*7=18.2,86-18.2>46<93)
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Re: A group of students took an exam that was scored from 0 to 1 [#permalink] New post   13 Jan 2024, 21:54
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