My Math Forum Averages

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 October 2nd, 2011, 02:11 PM #1 Newbie   Joined: Oct 2011 Posts: 1 Thanks: 0 Averages I have a problem that I've been working on FOREVER! I CANNOT use Algebra or anything like Trig or Calculus to solve it. I have to describe it in it's basic form. By trial and error, I think the answer is 20, but I'm also not allowed to use trial and error (guess and check). How in the world can it be solved without Algebra? So far noone has been able to help me on other math forums without using Algebra. I already understand the Algebra way to solve it. Any ideas? 3. Two algebra classes took the same exam. The mean grade for the first class was 88. The second class had a mean grade of 79. The mean grade of the two classes combined was 84. If there were 25 students in the first class, how many were in the second class?
 October 2nd, 2011, 03:26 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Averages We know the combined score for the first class is: $25\cdot88=2200$ Letting n be the number of students in the second class, their combined score is: $n\cdot79=79n$ The combined score of the two classes together is: $(25+n)84=2200+79n$ $2100+84n=2200+79n$ $5n=100$ $n=20$ We could also use a weighted average to state: $\frac{25\cdot88+n\cdot79}{25+n}=84$ $25\cdot88+n\cdot79=84(25+n)$ $2200+79n=2100+84n$ $100=5n$ $n=20$ The best way to solve this problem is to use algebra, representing the unknown quantity with a variable, then using given information to solve known relationships for that variable.
October 3rd, 2011, 06:31 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

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Re: Averages

Hello, mswanson502!

Quote:
 3. Two algebra classes took the same exam. The first class had 25 students; the mean grade for the first class was 88. The second class had a mean grade of 79. The mean grade of the two classes combined was 84. How many were in the second class?

I have many questions.

Is this an assignment for a math class?
[color=beige]. . [/color]If not, where did you find this problem?

If so, at what level is the class?
[color=beige]. . [/color]It appears to be at a pre-Algebra level.
[color=beige]. . [/color]Yet playing-with-means seems quite advanced.

Were you explicitly forbidden to use Algebra or guessing?
[color=beige]. . [/color]Those are sadistic restrictions!

I see no way so solve this without Algebra.
[color=beige]. . [/color]MarkFL offered two excellent algebraic solutions.

Without resorting to trial-and-error, I have only one option:
[color=beige]. . [/color]Pray ... and await Divine Inspiration.

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