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The mean of a set of 10 numbers is 40, the mean of a set of 30 numbers is 60, and the mean of a set of 60 numbers is x. If the mean of these 100 numbers is 34, what is the value of x?

What do I do?

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06-05-2014, 07:49 AM
(This post was last modified: 06-05-2014, 07:58 AM by Jonathan Whatley.)
Let's answer an easier question that's really close. That'll give us the skeleton of an equation we can use to solve the question asked.

The mean of a set of 10 numbers is 40, the mean of a set of 30 numbers is 60, and the mean of a set of 60 numbers is 1. What is the mean of these 100 numbers?

( (10 numbers * 40) + (30 numbers * 60) + (60 numbers * 1) ) / 100 numbers total = x = 22.6

Now, instead of plugging in that 1 and solving for the mean, let's plug in the mean and solve for the value we just pretended was 1, the mean of that set of 60.

(The equation above was algebra, we just probably don't think of it as algebra because the unknown x is on its own on the right after the equal sign. It was really easy algebra. The next part is slightly less easy algebra, but still pretty easy!)

( (10 numbers * 40) + (30 numbers * 60) + (60 numbers * x) ) / 100 = 34

( 400 + 1800 + 60x ) / 100 = 34

2200 + 60x = 3400

60x = 1200

x = 20

To check our work, let's plug in 20 in place of x and, yep!, the mean of that whole set of numbers is 34.

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06-05-2014, 08:20 AM
(This post was last modified: 06-05-2014, 08:28 AM by soliloquy.)
Thank you for your help! Were just sticking that number 1 in there to try and show me how to solve an algebra problem? I was like where did the 1 come from? LOL But, the part where you said "Now, instead" makes sense.

I can solve the equation without any problems after it is set up. What I do not understand is how you set up this equation.

Why do you multiply the number of numbers by the mean? I knew before posting that the denominator would be 100 and that the equation would be set equal to 34 but I don't understand how you came up with the numerator of the equation. Can you explain?

My confusion may be coming into play because in algebra "is" usually means equals. But, that's not what you did with "is" in the numerator when you multiplied the number of numbers by the mean.

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How do we solve for a mean for a set if we know the value for each member in the set? With two steps:

1. value for member 1 + value for member 2 + â¦ = sum of values

2. sum of values / number of members in the set = mean

In this question, we aren't given the values for individual members of any of the sets. But because we have the means of each set, we can do step 2 backwards to get the sum of values for each set.

Because

sum of values / number of members = mean

therefore

mean * number of members = sum of values

And that "sum of values" for the first set described is

(10 numbers * 40)

for the second set described is (30 numbers * 60), and for the third set described is (60 numbers * unknown).

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Thank you! I'll probably have to read that again when my brain unfries but that does make sense. I appreciate you taking the time to answer my question.

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This is the first time I've ever read this thread. Jonathan, I suggest you head for the hills. I'm starting a correspondence course for College Alg. soon, and I will hunt you down. You do an excellent job of explaining things.

TESU BSBA - GM, September 2015

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LaterBloomer Wrote:This is the first time I've ever read this thread. Jonathan, I suggest you head for the hills. I'm starting a correspondence course for College Alg. soon, and I will hunt you down. You do an excellent job of explaining things.

You're too kind! :o

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