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Re: In a class, there are 40 students. In a test, the average (arithmetic [#permalink]
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Bunuel wrote:
In a class, there are 40 students. In a test, the average (arithmetic mean) score of the girls is 30, that of boys is 40, and that of the class is 32. If the score of a boy was incorrectly computed as 30 for 40, what is the correct average (arithmetic mean) score of the class?

(A) 31
(B) 31.50
(C) 32.25
(D) 41
(E) 42


Before we even make any actual computation, we note that since one of the grades was actually higher than the grade used in calculation, the correct average has to be greater than the computed average, so we can eliminate A and B. Furthermore, in a class of 40 people, a difference of 10 points in one of the grades cannot change the average grade by 9 or 10 points, so we can eliminate D and E as well. The only answer choice that makes sense here is C. Let’s verify that.

First, let’s let g = the number of girls and b = the number of boys in the class. We have:

g + b = 40

We create the weighted average equation:

(30g + 40b)/40 = 32

Multiplying both equations by 40, we have:

40g + 40b = 1600

and

30g + 40b = 1280

Subtracting these two equations, we have:

10g = 320

g = 32

So b = 8.

Since one of the boys’ scores should be 40 instead of 30, then the sum of the boys’ scores should be 8 x 40 + 10 = 330. Therefore, the correct average of the entire class should be:

(32 x 30 + 330)/40 = 1290/40 = 32.25

Answer: C
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In a class, there are 40 students. In a test, the average (arithmetic [#permalink]
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Bunuel wrote:
In a class, there are 40 students. In a test, the average (arithmetic mean) score of the girls is 30, that of boys is 40, and that of the class is 32. If the score of a boy was incorrectly computed as 30 for 40, what is the correct average (arithmetic mean) score of the class?

(A) 31
(B) 31.50
(C) 32.25
(D) 41
(E) 42

Since the number of students has not been changed, if the error is corrected to calculate the new average score, the errored score will be evenly distributed among all the student.
The error was =40-30 = 10
the Average was underrepresented by = 10 /40 (Per students)= .25
So, the new Average will be 32+.25 =32.25
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In a class, there are 40 students. In a test, the average (arithmetic [#permalink]
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Kudos
Bunuel wrote:
In a class, there are 40 students. In a test, the average (arithmetic mean) score of the girls is 30, that of boys is 40, and that of the class is 32. If the score of a boy was incorrectly computed as 30 for 40, what is the correct average (arithmetic mean) score of the class?

(A) 31
(B) 31.50
(C) 32.25
(D) 41
(E) 42



Through Allegation method we can easily find out the ratio of girls and boys.

32 - 40 = 8

32 - 30 = 2

Ratio : girls : boys = 8 : 2

we can find out the exact number of girls and boys by ratio.

8x + 2x = 40

x = 4.

Number of girls = 8x = 8*4 = 32

Number of boys = 2x = 2*4 = 8

Average of the class: \(\frac{32 * 30 + 8*40 + 10}{40}\) = 32.5.

Only change we must consider is 10. while everything is fine but in boy's total we need to add 10 more as we made a mistake by including 30 instead of 40.

The best answer is C.

Originally posted by KSBGC on 03 Oct 2018, 05:26.
Last edited by KSBGC on 03 Oct 2018, 06:29, edited 1 time in total.
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Re: In a class, there are 40 students. In a test, the average (arithmetic [#permalink]
Bunuel wrote:
In a class, there are 40 students. In a test, the average (arithmetic mean) score of the girls is 30, that of boys is 40, and that of the class is 32. If the score of a boy was incorrectly computed as 30 for 40, what is the correct average (arithmetic mean) score of the class?

(A) 31
(B) 31.50
(C) 32.25
(D) 41
(E) 42





Don't solve just read the whole question at once.. You will save 2 critical minutes for your test.. Is the bottomline..

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Re: In a class, there are 40 students. In a test, the average (arithmetic [#permalink]
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