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# The incomplete table above shows a distribution of scores

Author Message
Intern
Joined: 28 Dec 2004
Posts: 35
The incomplete table above shows a distribution of scores [#permalink]

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07 May 2005, 08:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The incomplete table above shows a distribution of scores for a class of 20 students. If the average score is 78, what score is missing from the table?

Score # of students
------- ---------------
83....................5
70....................6
92....................3
X.....................5
64....................1

A 73
B 75
C 77
D 79
E 81

I got this by randomly guessing between the two most reasonable answers. Besides doing all the math out, can anyone suggest a quick method?
CIO
Joined: 09 Mar 2003
Posts: 463

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07 May 2005, 08:56
i can think of two ways to do this, both are pretty time intensive. That said, I think just doing this one straight out requires no more than 2 minutes of just a bunch of multiplication.

Meaning, it's not complicated math, it's just a lot of it. And it can be done quickly. I don't think, if this were a real question, I would take my chances with trying to guess educatedly at the answer when I'd be sure to get it right with just a little math.

I got C, 77, in under two minutes, just by doing this:

83(5) + 70(6) + 92(3) + 5x + 64 = 78(20)
415 + 420 + 276 + 5x + 64 = 1560
1175 + 5x = 1560
5x = 385
x = 77

It took me longer to type it than to do it.
Intern
Joined: 28 Dec 2004
Posts: 35

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07 May 2005, 08:59
Thanks for the tip. I thought I was missing the trick.
Manager
Joined: 05 May 2005
Posts: 92
Location: Kyiv, Ukraine

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08 May 2005, 06:28
do you think balance method will work here?

83-78=5 5*5=25

70-78=-8 -8*6=-48

92-78=14 14*3=42

78-64=14 -14*1=-14

Sum=25-48+42-14=5 5x=5 x=1 and then the average is 79

is there a mistake somewhere?

thanks
Intern
Joined: 06 May 2005
Posts: 17

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09 May 2005, 11:55
I recommend this method if the calculations are labor intensive.
Note that here we are using the property that the sum of deviations from the mean is zero.

ie Sigma(x-mean) = 0

where Sigma stands for the summation and x stands for the individual data points.

-Srinivas

july05 wrote:
do you think balance method will work here?

83-78=5 5*5=25

70-78=-8 -8*6=-48

92-78=14 14*3=42

78-64=14 -14*1=-14

Sum=25-48+42-14=5 5x=5 x=1 and then the average is 79

is there a mistake somewhere?

thanks
Manager
Joined: 05 May 2005
Posts: 92
Location: Kyiv, Ukraine

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10 May 2005, 01:45
Srinivas,

could you please show how to work this formula using this problem?

thanks a lot.
Intern
Joined: 06 May 2005
Posts: 17

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13 May 2005, 10:57
Score # of students
------- ---------------
83....................5
70....................6
92....................3
X.....................5
64....................1

83-78=5 5*5=25

70-78=-8 -8*6=-48

92-78=14 14*3=42

78-64=14 -14*1=-14

Sum=25-48+42-14=5 5x=5 x=1 and then the average is 79

=================================================

I might have to repeat what was posted earlier which is very useful for my explanation below. So, please refer to the above table.

In the formula, x represents individual numbers Or scores in this problem.

Mean is the average which is given as 78 (in this problem).

The average or mean is given as 78.

The scores(numbers) are 83,83,..(5 times because there are 5 students who got 83), 70,70..(6times), 92., 92, 92, X (5 times), and 64.

Keep in mind that deviations are positive and negative and the sign is important.

The deviation of 83 from the mean is 83-78 = 15

because we have 5 students who scored 83, To add up the deviations of scores of these 5 students, we need to multiply 15 by 5.

So, you have 15 * 5 = 75.

Similarly, the deviation of 70 is 70-78 = -8.

Since, there are 6 students with this score, toal deviation for these 6 scores is -8*6 = -48

Similarly, the deviation of 92 is 92-78 = 14

Total deviation of 3 scores = 14*3=42

Let the deviation of X is y. Total deviation of 5 such scores is 5y.

The deviation of 64 is 64-78=-14.

The sum of all deviations is 15-48+42+5y-14.

This SUM should be Zero.

ie 15-48+42+5y-14=0.

ie 5y=5
is y=1.

Note that y is positive and y is only a deviation not the actual score (the answer).

In fact, the deviation y can be written as X-78.

therefore, X-78=1 or X = 79.

The earlier post did give you a nice table. I have the explanation for it.
Now, you have both. Hopefully it makes sense now.

-Srinivas.

july05 wrote:
Srinivas,

could you please show how to work this formula using this problem?

thanks a lot.
Manager
Joined: 05 May 2005
Posts: 92
Location: Kyiv, Ukraine

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13 May 2005, 14:52
well what do you do with the official answer then? it's 77 and not 79...

thanks.
CIO
Joined: 09 Mar 2003
Posts: 463

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13 May 2005, 15:00
july05 wrote:
do you think balance method will work here?

83-78=5 5*5=25

70-78=-8 -8*6=-48

92-78=14 14*3=42

78-64=14 -14*1=-14

Sum=25-48+42-14=5 5x=5 x=1 and then the average is 79

is there a mistake somewhere?

thanks

There are some pretty complicated approaches to this problem being displayed here. But July05, you have the least complicated. You just made one mistake.

You can fix it by keeping track of the negatives, and remember - everything has to balance to zero.

So you got +25, -48, +42, -14. Combine the positives and the negatives:
+67; -62

So what do you need to balance? An additional MINUS 5. That would make the grand total of the minuses be -67, which makes balance.

There are five students who scored X, and together they make -5 from the average, which means each one must be -1, or 78-1=77
Manager
Joined: 05 May 2005
Posts: 92
Location: Kyiv, Ukraine

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13 May 2005, 15:28
you rock, Ian! thanks a lot. now i can sleep at night.
Intern
Joined: 06 May 2005
Posts: 17

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13 May 2005, 19:01
Sorry, I have a mistake in the earlier post.

It shoudl read 5y = -5 and y = -1. Hence the X-78=-1 and X = 77.

-Srinivas
13 May 2005, 19:01
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