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aniketm.87@gmail.com
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Ram025
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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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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Can someone please shed some light on this?

Thanks in advance
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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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Hi Chetan, thanks for the explanation is very useful
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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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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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chetan2u
aniketm.87@gmail.com
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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chetan2u
aniketm.87@gmail.com
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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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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M ---- A
80 ---- 86
--- 84
2 ----- 4

So, Afternoon class had more students. Ans C
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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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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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aniketm.87@gmail.com
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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ScottTargetTestPrep
aniketm.87@gmail.com
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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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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