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Re: The scores on a certain history test are shown above. How [#permalink]

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10 Apr 2013, 02:40

Thanks, yes this is the classic way. Is there a shortcut especially for calculating the mean without having to really do it?[/quote]

In this problem , yes you can calculate it in 10 seconds , if you can observe the total number of terms is 10. so since every number is a multiple of 10 you can do it easily .

Re: The scores on a certain history test are shown above. How [#permalink]

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10 Apr 2013, 02:46

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venkat18290 wrote:

Thanks, yes this is the classic way. Is there a shortcut especially for calculating the mean without having to really do it?

In this problem , yes you can calculate it in 10 seconds , if you can observe the total number of terms is 10. so since every number is a multiple of 10 you can do it easily .

If there is a general method do let me know.[/quote]

Thanks, yes this is the classic way. Is there a shortcut especially for calculating the mean without having to really do it?

40, 45, 45, 50, 50, 60, 70, 75, 95, 100

You can use deviations to find the mean here.

Assume that the mean is 60 since it's kind of in the middle.

Now notice that 50 is 10 less than 60 and 70 is 10 more so ignore both. Now 40, 45, 45, 50 together are 60 away from 60 while 75, 95, 100 are 90 away from 60.

So we have a total of 90 - 60 = 30 deviation from 60. Since there are 10 numbers, the average must be 30/10 = 3 more than 60 i.e it must be 63.

Thanks, yes this is the classic way. Is there a shortcut especially for calculating the mean without having to really do it?

40, 45, 45, 50, 50, 60, 70, 75, 95, 100

You can use deviations to find the mean here.

Assume that the mean is 60 since it's kind of in the middle.

Now notice that 50 is 10 less than 60 and 70 is 10 more so ignore both. Now 40, 45, 45, 50 together are 60 away from 60 while 75, 95, 100 are 90 away from 60.

So we have a total of 90 - 60 = 30 deviation from 60. Since there are 10 numbers, the average must be 30/10 = 3 more than 60 i.e it must be 63.

why did you not consider the second 50 in your calculation? Because then 40,45,45,50,50 would be 70 away from 60.

There is a second 50 which is 10 away from 60 and then there is a 70 which is also 10 away from 60. So together, they cancel off each other and give an average of 60 only. Or consider that the numbers on the left are 70 away but then the numbers on the right are 100 away since there is a 70 also which we haven't considered. Hence, overall, we still have 30 extra which will be distributed evenly among the 10 numbers. So the average will be 63.
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I think iam going crazy......I am able to solve 700 level mistake.but fail to do 600-700.Always doing some HOMER(Simpson) mistake.........

;(

Slow down a bit then. When you are done with the question, take another look at the problem and ensure you have answered what was asked. You will catch up on time again after some practice.
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Re: The scores on a certain history test are shown above. How [#permalink]

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12 Sep 2016, 12:29

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score780 wrote:

Bunuel wrote:

score780 wrote:

40, 45, 45, 50, 50, 60, 70, 75, 95, 100

The scores on a certain history test are shown above. How many scores were greater than the median score but less than the mean score?

A. None B. One C. Two D. Three E. Four

The median = (the average of two middle terms) = (50 + 60)/2 = 55. The mean = (the sum)/(# of terms) = 630/10 = 63.

Only 60 is between 55 and 63.

Answer: B.

Thanks, yes this is the classic way. Is there a shortcut especially for calculating the mean without having to really do it?

I think you can use a baseline method. Choose 40 as baseline. So differences between every other number and 40 are: 5, 5, 10, 10, 20, 30, 35, 55, 60. Sum all of them and divide by 10 = 23. Add to 40 = 63, which is the average of all 10 numbers.

Re: The scores on a certain history test are shown above. How [#permalink]

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28 Oct 2017, 09:07

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