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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]

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08 May 2016, 20:49
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 08 May 2016, 22:12, edited 1 time in total.
Renamed the topic and edited the question.

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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink]

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08 May 2016, 20:49
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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]

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14 May 2016, 04:24
Can someone please shed some light on this?

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Re: Prof Liu gave the same quiz to the students in her morning class and [#permalink]

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14 May 2016, 05:54
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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]

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14 May 2016, 06:20
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aniketm.87@gmail.com wrote:
Can someone please shed some light on this?

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]

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14 May 2016, 06:24
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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]

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30 May 2016, 16:27
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]

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22 Jun 2017, 16:14
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.

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Prof Liu gave the same quiz to the students in her morning class and [#permalink]

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23 Jun 2017, 00:04
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.

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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Prof Liu gave the same quiz to the students in her morning class and [#permalink]

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28 Oct 2017, 15:35
chetan2u wrote:
aniketm.87@gmail.com wrote:
Can someone please shed some light on this?

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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Prof Liu gave the same quiz to the students in her morning class and [#permalink]

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28 Oct 2017, 15:55
KD25 wrote:
chetan2u wrote:
aniketm.87@gmail.com wrote:
Can someone please shed some light on this?

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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Prof Liu gave the same quiz to the students in her morning class and [#permalink]

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28 Oct 2017, 16:19
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]

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08 Dec 2017, 12:01
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.

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Re: Prof Liu gave the same quiz to the students in her morning class and   [#permalink] 08 Dec 2017, 12:01
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