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In a class, there are 40 students. In a test, the average (arithmetic
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26 Sep 2018, 04:56
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Re: In a class, there are 40 students. In a test, the average (arithmetic
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26 Sep 2018, 06:21
Given that Total no of students = 40 Avg score of girls = 30 Avg score of boys = 40 Let’s assume total no of girls = Tg Total no of boys = Tb Tg + Tb = 40
As per question , Sum of girls student /Tg = 30 Sum of boys student /Tb = 40
Total avg score = (Sum of girls student + sum of boys student ) / (Tg + Tb) = 32
Now , from above equation We get 30Tg + 40 Tb / ( Tg + Tb) = 32
Now as per question , 10 marks computed more in case of a boy so now the new avg score = 30 Tg + 40 Tb 10 / ( Tg + Tb)
Now while solving we get Avg score = 31.5
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Re: In a class, there are 40 students. In a test, the average (arithmetic
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29 Sep 2018, 18:01
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
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03 Oct 2018, 00:16
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 =4030 = 10 the Average was underrepresented by = 10 /40 (Per students)= .25So, 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
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Updated on: 03 Oct 2018, 06:29
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 selim on 03 Oct 2018, 05:26.
Last edited by selim 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
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03 Oct 2018, 05:55
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 Note that it doesn't matter how many girls and boys are there. That is given just to confuse you. Looking at the options, I would mark (C) in a moment and move on. The previous average was 32. The total increases by 10 points so the average will increase. A small increase of 10 points (from 30 to 40) is no way going to increase the average by 9 points (to 41) over 40 children. Answer must be 32.25. Even if I want to calculate, the total increases by 10 points over 40 students. So this increase the average by 10/40 = .25 points to get 32.25. If you are not sure, look at it this way. This is what we know: There are 40 students. Their average score is 32. Score of a boy was incorrectly computed as 30 instead of the actual 40 (so the total needs to be 10 points extra) Correct average = 10 distributed over 40 students (= .25) added to the average to give 32.25 Check out this post for more details of this method https://www.veritasprep.com/blog/2012/0 ... eticmean/
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Re: In a class, there are 40 students. In a test, the average (arithmetic &nbs
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