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Prof Liu gave the same quiz to the students in her morning class and

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Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   Updated on: 23 Jan 2024, 06:52
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Prof Liu gave the same quiz to the students in her morning class and in her afternoon class. The average score for the two classes combined was 84. Which class had more students ?

(1) The average score for the students in the morning class was 80
(2) The average score for the students in the afternoon class was 86
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C

Originally posted by aniketm.87@gmail.com on 08 May 2016, 21:49.
Last edited by Bunuel on 23 Jan 2024, 06:52, edited 2 times in total.
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   14 May 2016, 07:20
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aniketm.87@gmail.com wrote:
Can someone please shed some light on this?

Thanks in advance



Hi aniket,

you are correct that the average is closer to the higher quantity..

this is weighted average method and you can even find the ratio of two items..
Morning class/ evening class = \(\frac{(86-84)}{(84-80)} = \frac{2}{4} = \frac{1}{2}\)....
so evening class will have two times the morning class
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   08 Dec 2017, 13:01
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aniketm.87@gmail.com wrote:
Prof Liu gave the same quiz to the students in her morning class and in her afternoon class. the average score for the two classes combined was 84. which class had more students ?

(1) The average score for the students in the morning class was 80
(2) the average score for the students in the afternoon class was 86


This is a great question testing the foundations of weighted averages. We must remember that the weighted average will always be closer to the average of the group that has a larger quantity. Let’s test this theory with a simple example.

Let’s say there are 5 women in a room with an average age of 40 and 3 men in a room with an average age of 20. The weighted average age of all the people in the room is as follows:

(5 x 40) + (3 x 20)/(5 + 3) = (200 + 60)/8 = 260/8 = 32.5 years old

Notice that the weighted average of the ages of all people in the room is closer to 40 (the average age of the 5 women) than it is to 20 (the average age of the 3 men). The reason for this is that there are a greater number of women than there are men.

Let’s apply this principle to the above problem.

We are given that the combined average score for the morning and afternoon class is 84. We need to determine which class had more students.

Statement One Alone:

The average score for the students in the morning class was 80.

Since we do not have any information regarding the average score of the afternoon class, we do not have enough information to answer the question. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

The average score for the students in the afternoon class was 86.

Since we do not have any information regarding the average score of the morning class, we do not have enough information to answer the question. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using the information from statements one and two, we see that the average score of the morning class was 80 and that the average score from the afternoon class was 86. Using the example provided above, we know that since the average score from the afternoon class is closer to the combined (i.e., weighted) average of 84, there are more students in the afternoon class.

Answer: C
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   14 May 2016, 07:24
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Ram025 wrote:
X- no.in mrng class
Y- no. in aftn class
Equation be- 80X + 86Y =168
now you know which is greater. You need both [ C]


hi
just a point..
the equation will be..
\(80x + 86y = 84(x+y)..\)
and from this you can get the ratio of number of students in two classes..
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   08 May 2016, 21:49
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This was my soln: need to confirm :

Soln :

80 ——— 84 ——86

Since the final avg is more towards the aft class avg, more students should be present in aft class. Hence ans is C)
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   14 May 2016, 05:24
Can someone please shed some light on this?

Thanks in advance
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   14 May 2016, 06:54
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X- no.in mrng class
Y- no. in aftn class
Equation be- 80X + 86Y =168
now you know which is greater. You need both [ C]
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   30 May 2016, 17:27
Hi Chetan, thanks for the explanation is very useful
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   22 Jun 2017, 17:14
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I chose (E), My rationale:
1) clearly insuff. Only info about morning class
2) clearly insuff. Only info about afternoon class
1/2) Insuff to determine the number of students in each class because I assumed the average scores could just be results of a majority of the class doing better/worse and not give any info about the actual number of students in each class.

Can someone please help explain how much logic if flawed?
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   23 Jun 2017, 01:04
bjohnson1392 wrote:
I chose (E), My rationale:
1) clearly insuff. Only info about morning class
2) clearly insuff. Only info about afternoon class
1/2) Insuff to determine the number of students in each class because I assumed the average scores could just be results of a majority of the class doing better/worse and not give any info about the actual number of students in each class.

Can someone please help explain how much logic if flawed?


bjohnson1392
Even if majority of students score better, the average still did not exceed 86 which is closer to the mean average of both class
Let's assume
Class A average = 40 among 5 students so assume \(A={40,40,40,40,40}\)
Class A average remains 40 if 1 student gets 80 and \(A={20,20,30,40,90}\)

In the question the average of both classes didn't deviate much, the majority of students in both classes must have scored within the mean of either of the classes
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   28 Oct 2017, 16:35
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chetan2u wrote:
aniketm.87@gmail.com wrote:
Can someone please shed some light on this?

Thanks in advance



Hi aniket,

you are correct that the average is closer to the higher quantity..

this is weighted average method and you can even find the ratio of two items..
Morning class/ evening class = \(\frac{(86-84)}{(84-80)} = \frac{2}{4} = \frac{1}{2}\)....
so evening class will have two times the morning class



Chetan Based on your explanation. This is what the solution should be. It looks like you have the morning and evening numbers inverted in your original calc
Morning class/ evening class = \(\frac{(84-80)}{(86-84)} = \frac{4}{2} = \frac{2}{1}\)....

Based on the calc above it shows morning class will have two times the evening class. I don't think that is the correct solution here. I am aware that the average moves toward the item with higher qty.
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   28 Oct 2017, 16:55
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KD25 wrote:
chetan2u wrote:
aniketm.87@gmail.com wrote:
Can someone please shed some light on this?

Thanks in advance



Hi aniket,

you are correct that the average is closer to the higher quantity..

this is weighted average method and you can even find the ratio of two items..
Morning class/ evening class = \(\frac{(86-84)}{(84-80)} = \frac{2}{4} = \frac{1}{2}\)....
so evening class will have two times the morning class



Chetan Based on your explanation. This is what the solution should be. It looks like you have the morning and evening numbers inverted in your original calc
Morning class/ evening class = \(\frac{(84-80)}{(86-84)} = \frac{4}{2} = \frac{2}{1}\)....

Based on the calc above it shows morning class will have two times the evening class. I don't think that is the correct solution here. I am aware that the average moves toward the item with higher qty.



Hi....
No it's correct the way I have written...
Average of morning is 80 and average of afternoon is 86
Overall average is 84..

The morning/evening will be (evening average-average)/(average-morning average) =(86-84)/(84-80)=2/4
There have to be more of the item that is closer to average..
Here 84 is closer 86 so items of 86 will be more than those of 80.

KD25 in response to your post below..

say in a test girls average is 80, and boys average is 50... class average is 70..
now if I dont have boys, the class average is 80 itself , same as that of girls..
so what is the average of boys doing - IT is getting the average of class from 80 to 70
and similarly average of girls is bringing the class average up from 50 to 70..
this is the reason the RATIO depends on the opposite element..
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   28 Oct 2017, 17:19
The part that I am not getting is if the ratio says Morning/Evening = (evening average-average)/(average-morning average) ----> why are you subtracting the evening part in the numerator, even though it corresponds to the morning students ?
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   03 Jan 2018, 20:40
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M ---- A
80 ---- 86
--- 84
2 ----- 4

So, Afternoon class had more students. Ans C
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   27 Mar 2018, 20:42
Quick question

If we were instead given that the combined average was 83 instead of 84, would the answer be (E)?
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   27 Mar 2018, 22:46
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geniecc wrote:
Quick question

If we were instead given that the combined average was 83 instead of 84, would the answer be (E)?


Hello

If we were given that combined average was 83, then looking at the two statements, we would have mathematically concluded that both morning and afternoon classes had equal number of students (since 83 is equal distance away from 80 as well as 86). So we would have concluded that neither of the two classes had more students, so the question is still answered. Still answer should have been C i think .

But I dont think that a question in GMAT would confuse us like this. If a GMAT question asks '..which class has more students..' then after combining the statements:

Either data would be insufficient to answer the question, i.e, we wouldn't be able to conclude which class has more students, in this case answer would be E
OR the information would lead to one particular class having more students than the rest, in this case answer would be C

So personally I think if GMAT asks this question exactly (with 80 and 86 in the two statements respectively) then they would not give combined average as 83.
But yes, they can instead ask a question, "Does morning class have more students than the evening class" and then even if the students turn out to be equal we can mark answer as C because now we know that morning class does NOT have more students (they are equal), so the 'asked' question has been answered as NO with surety.

I hope I have been able to clarify myself
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   19 Apr 2023, 09:18
aniketm.87@gmail.com wrote:
Can someone please shed some light on this?

Thanks in advance


see algebraic way
sum of morning scores=SM
sum of afternoon scores=SA
total morning students=TM
total afternoon students=TA

\( \frac{Σ(SM)+ Σ(SA)}{TM+TA}=84\)
Σ(SM)+Σ(SA)=84TM+84TA

(1) \(\frac{Σ(SM)}{TM}=80\)
Σ(SM)=80TM

replace it in the previous formula getting
80TM+Σ(SA)=84TM+84TA
Σ(SA)=4TM+84TA
insufficient, since we can say nothing about the relation between TM and TA

(2) Same procedure as (1). Σ(SA)=86TA. Insufficient

(1)+(2) having Σ(SM)=80TM and Σ(SA)=86TA
replace them in the original formula Σ(SM)+Σ(SA)=84TM+84TA

80TM+86TA=84TM+84TA
2TA=4TM, SIMPLIFY
TA=2TM

both together are sufficient

IMO C

hope it helps ;)
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   23 Jan 2024, 05:19
ScottTargetTestPrep wrote:
aniketm.87@gmail.com wrote:
Prof Liu gave the same quiz to the students in her morning class and in her afternoon class. the average score for the two classes combined was 84. which class had more students ?

(1) The average score for the students in the morning class was 80
(2) the average score for the students in the afternoon class was 86


This is a great question testing the foundations of weighted averages. We must remember that the weighted average will always be closer to the average of the group that has a larger quantity. Let’s test this theory with a simple example.



Answer: C


thanks! just confirming -by larger quantity do you mean larger number of people? because in this case the individual scores could've been higher for a certain class whilst the number of people could've been the same?
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink] New post   23 Jan 2024, 07:00
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tickledpink001 wrote:
ScottTargetTestPrep wrote:
aniketm.87@gmail.com wrote:
Prof Liu gave the same quiz to the students in her morning class and in her afternoon class. the average score for the two classes combined was 84. which class had more students ?

(1) The average score for the students in the morning class was 80
(2) the average score for the students in the afternoon class was 86


This is a great question testing the foundations of weighted averages. We must remember that the weighted average will always be closer to the average of the group that has a larger quantity. Let’s test this theory with a simple example.



Answer: C


thanks! just confirming -by larger quantity do you mean larger number of people? because in this case the individual scores could've been higher for a certain class whilst the number of people could've been the same?


    Class A's average score = 80.
    Class B's average score = 86.
    Combined average score = 84.

Since the combined average score of 84 is closer to Class B's average, then Class B has more students.

For an algebraic proof, refer to the earlier posts.
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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink]
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