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A teacher gave the same test to three history classes: A, B,
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Updated on: 21 Dec 2013, 05:28

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A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?

A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?

A. 74 B. 75 C. 76 D. 77 E. 78

ratio is 4:6:5 , numbers are 4x, 6x, 5x

total scores of each class is (65*4x + 6x * 80 + 77*5x) = 260x+480x+385x = 1125x

Re: A teacher gave the same test to three history classes: A, B,
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28 Oct 2014, 18:28

1

is this the fastest way? I am def. error prone when I do these longer multiplications...so wondering if there is a better way to estimate it? I looked at the numbers and thought majority of students got 80 or 77 so the average of all would be skewed that way and just picked 78. is there a way i could have thought about that and got the right answer or just multiply everything?

Re: A teacher gave the same test to three history classes: A, B,
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29 Oct 2014, 01:54

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1

angelfire213 wrote:

is this the fastest way? I am def. error prone when I do these longer multiplications...so wondering if there is a better way to estimate it? I looked at the numbers and thought majority of students got 80 or 77 so the average of all would be skewed that way and just picked 78. is there a way i could have thought about that and got the right answer or just multiply everything?

Re: A teacher gave the same test to three history classes: A, B,
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29 Oct 2014, 03:12

1

so there isn't a way around multiplying the numbers out?

PareshGmat wrote:

angelfire213 wrote:

is this the fastest way? I am def. error prone when I do these longer multiplications...so wondering if there is a better way to estimate it? I looked at the numbers and thought majority of students got 80 or 77 so the average of all would be skewed that way and just picked 78. is there a way i could have thought about that and got the right answer or just multiply everything?

Re: A teacher gave the same test to three history classes: A, B,
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29 Oct 2014, 03:24

angelfire213 wrote:

so there isn't a way around multiplying the numbers out?

PareshGmat wrote:

angelfire213 wrote:

is this the fastest way? I am def. error prone when I do these longer multiplications...so wondering if there is a better way to estimate it? I looked at the numbers and thought majority of students got 80 or 77 so the average of all would be skewed that way and just picked 78. is there a way i could have thought about that and got the right answer or just multiply everything?

However, as you can see, the numbers are "tricked" such that it cannot be simplified at this point (Not divisible by 3); hence the calculation
_________________

Re: A teacher gave the same test to three history classes: A, B,
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02 Nov 2014, 17:45

3

A good way to answer this question and have a good guess is realizing the fact that they are all almost in the same proportion. 4:6:5 is close to 5:5:5.

So you can calculate the average of all three direcly \((65+80+77)/3=74\)

Since you can realize that there is slightly more weight to a higher average (80 and 77 are at slighlty higher proportion) you can guess that the answer will be 75.

This would be a good aproach if faced with a similar question when running out of time.

Re: A teacher gave the same test to three history classes: A, B,
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04 Jan 2015, 10:10

I used a longer way - for some reason the concept of weighted average manage to slip my mind even though I love it...

Anyway, I still did it in about 90 seconds. So, it is not that bad I guess: 4x+6x+5x=15x The number of students. I picked 60 as 15x, because it is the LCM of 4,6,5. So, x=4

Then it goes like this: 65=s/4x | 65=s/16 | s= 1040 80=s/6x | s= 1920 77= s/20 | s= 1540

Re: A teacher gave the same test to three history classes: A, B,
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25 Jul 2016, 20:44

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2

A slightly simpler way to calculate the average would be to choose a number between 65 and 80, say 65, and subtract it from all the numbers, ie \((65-65, 77-65, 80-65)\) -> \((0, 12, 15)\).

calculate the average of the new set of numbers: \((0*4+12*5+15*6)/15=150/15=10\)

Now add this value with the number we chose, \(65+10=75\).

Re: A teacher gave the same test to three history classes: A, B,
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25 Sep 2016, 04:55

Can this question worked in a opposite way? I mean

A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. If the average score for the three classes combined is 75 what is the ratio of the numbers of students in each class who took the test?

Re: A teacher gave the same test to three history classes: A, B,
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25 Sep 2016, 07:04

albany09 wrote:

A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?

A. 74 B. 75 C. 76 D. 77 E. 78

Score ( A : B : C ) = 65 : 80 : 77 Number ( A : B : C ) = 4 : 6 : 5

So, Weighted average score is = \(\frac{(65*4)+( 80*6)+(77*5)}{( 4 + 6 + 5 )}\)

Or, Weighted average score is = \(\frac{(260)+(480)+(385)}{(15)}\)

Or, Weighted average score is = \(\frac{(1125)}{(15)}\)

Or, Weighted average score is = \(75\) So, Answer will be (B) 75 _________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

Re: A teacher gave the same test to three history classes: A, B,
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25 Sep 2016, 09:34

Ndkms wrote:

Can this question worked in a opposite way? I mean

A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. If the average score for the three classes combined is 75 what is the ratio of the numbers of students in each class who took the test?

Let the ratio of students in 3 classes be a : b : c

So, \(75\) = \(\frac{(65*a) + (80*b) + (77*c)}{(a + b + c)}\)

Or, 75a + 75b + 75c = 65a + 80b + 77c

Or, 10a = ( 5b + 2c )

Hope you understand now, that what we need... We need the total number of students and relationship between the number of students in 2 classes.... _________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

A teacher gave the same test to three history classes: A, B,
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25 Sep 2016, 12:33

A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?

A. 74 B. 75 C. 76 D. 77 E. 78

4*65+6*80+5*77=1125 4+6+5=15 1125/15=75=average score for three classes B.

Re: A teacher gave the same test to three history classes: A, B,
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26 Jan 2018, 10:46

albany09 wrote:

A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?

A. 74 B. 75 C. 76 D. 77 E. 78

The ratio of A : B : C = 4x : 6x : 5x, so the total is 15x. We can create a weighted average equation:

Re: A teacher gave the same test to three history classes: A, B,
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30 Jan 2018, 10:08

Top Contributor

albany09 wrote:

A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?

A. 74 B. 75 C. 76 D. 77 E. 78

This is a weighted averages question, so we can use the following formula:

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Ratio of classes A, B, C = 4 : 6 : 5, respectively 4 + 6 + 5 = 15 So, class A represents 4/15 of the TOTAL population Class B represents 6/15 of the TOTAL population Class C represents 5/15 of the TOTAL population

So, weighted average of 3 groups combined = (4/15)(65) + (6/15)(80) + (5/15)(77) = 260/15 + 480/15 + 385/15 = 1125/15 = 75

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Re: A teacher gave the same test to three history classes: A, B,
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27 May 2019, 18:47

albany09 wrote:

A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?

A. 74 B. 75 C. 76 D. 77 E. 78

Total score of Class A = 65*4 = 260

Total score of Class B = 80*6 = 480

Total score of Class A = 77*5 = 385

Average score for the three classes combined = (260+480+385)/15 = 75

Answer B _________________

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