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Manager  Joined: 03 Oct 2008
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A teacher gave the same test to three history classes: A, B,  [#permalink]

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28 00:00

Difficulty:   25% (medium)

Question Stats: 78% (02:19) correct 22% (02:38) wrong based on 617 sessions

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

Originally posted by albany09 on 07 Oct 2008, 10:16.
Last edited by Bunuel on 21 Dec 2013, 05:28, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: math--teacher and students  [#permalink]

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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

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

total number of students = 15x

average = 1125x/15x = 75

B is the answer
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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?
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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11
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?

Combined Average $$= \frac{65 * 4 + 80 * 6 + 77 * 5}{4+6+5} = \frac{1125}{15} = 75$$

Answer = B
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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?

Combined Average $$= \frac{65 * 4 + 80 * 6 + 77 * 5}{4+6+5} = \frac{1125}{15} = 75$$

Answer = B
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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?

Combined Average $$= \frac{65 * 4 + 80 * 6 + 77 * 5}{4+6+5} = \frac{1125}{15} = 75$$

Answer = B

Not that I'm aware of............. Usually I would have done in this way to avoid the multiplication

$$\frac{65*4}{15} + \frac{80*6}{15} + \frac{77*5}{15}$$

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
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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.

Hope it helps.
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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

Total:
M=S/N
M= 4500/60
M=450/6
M=75 ANS B
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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10
1
Hi angelfire123
Yes there is faster way. 4+6+5=15 so look for a number that is multiple of 15. Only B looks good. Hope it is clear
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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Konstantin1983 wrote:
Hi angelfire123
Yes there is faster way. 4+6+5=15 so look for a number that is multiple of 15. Only B looks good. Hope it is clear

This does not work. Consider the case where the ratios are 3:7:5.

65*3 + 80*7 + 77*5 = 1140

1140 / 15 = 76
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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2
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, $$65+10=75$$.
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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since we do not know the mean you can meanwhile assume the mean to be 70

So,

65 80 77
-5 +10 +7
ratio 4 6 5
-20 +60 +35

So, mean = 70 + (-20 +60 +35)/(4+6+5)= 75

this is deviation method,we calculate the total deviation from the assumed mean and then find the actual mean.
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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?
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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

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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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....

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A teacher gave the same test to three history classes: A, B,  [#permalink]

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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. 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.
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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:

[65(4x) + 80(6x) + 77(5x)]/15x

5[13(4x) + 16(6x) + 77x]/15x

[13(4x) + 16(6x) + 77x]/3x

(52x + 96x + 77x)/3x

225x/3x = 75

Answer: B
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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) + ...

For more information on weighted averages, you can watch this free GMAT Prep Now video: http://www.gmatprepnow.com/module/gmat-statistics?id=805

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

Answer: B

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GMAT 1: 600 Q40 V33 Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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Huh ! so basically no one managed to suggested a easier way to solve it in like 30 or so secs.. took me 90 sec using weighted avg.
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Re: A teacher gave the same test to three history classes: A, B,  [#permalink]

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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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