Algebra problem- Can you solve it.

cj65

cj65

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I cannot. My 7th grade is redoing his test he flunked for homework. He has a problem that goes like this:

The Ramirez family has 3 kids. The sum of the 3 kids age is 10. The product of the 3 kids ages is 20. What are their ages? I for the life of me cannot figure it out. Is there a solution? How did you get there? I tried to solve for A B and C, but seemed to get more complex than his math level, so I stopped. I then tried to do actual sums to 10 with 3 addends. There are 7 combinations. 1,3,6 was as close as I could get. I still could not solve. Is there a solution? If so, what is it?
 
Couldn't it be:

5+4+1 = 10

5*4*1 = 20

I didn't actually solve that as an equation, just ran through some possible combinations in my head. Probably helps that I spend all day working with numbers anyway.
 
cj65 said:
i cannot not. My 7th grade is redoing his test. He has a problem that goes like this:

The ramirez family has 3 kids. The sum of the 3 kids age is 10. The product of the 3 kids ages is 20. What are their ages? I for the life of me cannot figure it out. Is there a solution? How did you get there? I tried to solve for a b and c, but seemed to get more complex than his math, so i stopped. I then tried to do actual sums to 10 with 3 addends. I still could not solve. Is there a solution? If so, what is it?
Click to expand...

1, 4, 5

oops: too slow
 
I'm in my lunch break, so I don't have long, but I'll make a stab. Let x, y and z be the ages of the kids. Then x+y+z=10 and x.y.z=20. Now this is only two equations, but there are three unknowns, so you need some more information or try numerical answers. I could do this in seconds on my own computer, but that will have to wait. Check back later.
 
The problem is pretty simple if you just reduce it to whole numbers that divide into 20.

So 1,2,4, and 5 are the only possibilities (10 and 20 are eliminated since the sum would be too high). 2 is eliminated pretty quickly since there's no combo with it summing to 10. Voila.
 
Alright,

I am an idiot. Thanks, I tried many other combinations, but overlooked the obvious.
 
That's more of a trick question or math puzzle than a legitimate algebra problem in my opinion. Algebra isn't about answers; it's about methods. The answers are usually listed in the back of the book so the students can check their homework, but just writing the answers isn't going to fool the teacher. He needs to see that you understand the method.

If you encounter a problem like that in real life as soon as you notice there are three unknowns you naturally conclude you either need more information or you'll have to resort to trial and error, which computers are good at. (See numbersman's post.)

If it's a puzzle, though, you can take a different approach. You know one of the unknowns is either 0 or 1 because that's the way that kind of trick always works. It can't be 0 because that would make the product 0 and you've been told the product is 20. Therefore one of the unknowns is 1. Now you can write some equations:

A + B + 1 = 10
A * B * 1 = 20

You can solve that pair with legitimate algebra methods.
 
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Thanks very much to mongol8 who deprived me of my chance to redeem myself!!! His approach is correct, look for numbes that multiply together to give 20, and he has summed it up nicely. But I still reckon a MATLAB problem would solved it in a blink.
 
Thanks Cougar! I feel better now. I discounted the 1 right off the bat. I tried to do A plus B plus C equals 10 and ABC equals 20, and then solve for A, but I hit a brick wall there too, and realized, it was not the way to do it. The book (gee, read the chapter I tell the kids) gave an example of trial and error, test and revise. This was the path. I am embarassed that I could not get it however. I thought about all the intelligent blokes on BF, which is only a key stroke away, and had my answer in minutes. Bladeforums.......capture the magic!
 
cj65 said:
Alright,

I am an idiot. Thanks, I tried many other combinations, but overlooked the obvious.
Click to expand...

Just wait for high school. I have one in Algebra II and one in Calculus. Helping with homework gets tricky.:)
 
I will have to brush up on quadratic equations for sure.
 
With two variables, the solution to each equation would be a line segment or a ray. I think that if you compared the two, you could arrive at a solution. A proof, if you will, Anything squared would result in a parabola. That said, I solved it in my head in about 60 seconds or so. Anything more diffficult?
 
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