Anyone Good at Math? Help, please!

[amberfocus]

I have a guess and check story problem I am trying to figure out so I can help my son figure it out and all I am managing to do at the moment is bang my head against the wall.  The problem is this:

Russell had just served up 55 cones, some single scoops, some double scoops, and some triple scoops.  If he served up a total of 103 scoops, how many of the cones were single scoops, how many were double scoops, and how many were triple scoops?

Usually I figure out guess and checks algebraically, but I can't do that here because there is not enough information and I've gone through two pieces of scrap paper trying to figure it out otherwise.  We didn't have bloody problems like this when I was in the fifth grade!

Comments

[janna-hawkins.livejournal.com]

That is a tough one! I tried to work through it but no luck. Wish I could help :/

[amyo67.livejournal.com]

Trevor trying to work it through for you as I type this.

[amberfocus.livejournal.com]

Thank you. I don't know if he'll be able to figure it out or not. It's super hard.

[amyo67.livejournal.com]

Oh, Trev got it too, but he was trying to come up with an equation and spent some time overthinking it and it happy to know there are more than one solution. He got: ( 16x1=16)+(30x2=60)+(9x3=27). 30+9+16=55 16+60+27=103

[frakup.livejournal.com]

I follow and rarely comment, but you piqued my interest here. I'll try to explain.

55x2=110, which is close enough to 103 to use it as a starting point. Then I just worked backwards. I tried 45x2, leaving 13 scoops over 10 cones. That math didn't work, but is very close, so I started working backwards from there. Turns out 44x2, 9x1, 2x3 = 103.

It's not an equation, but it's how I solved it. Does that help?

[amberfocus.livejournal.com]

Yes, it does. Thank you.

[honorh.livejournal.com]

The problem is, without more information, that problem can be solved in many different ways. Put a single scoop on all of them and that leaves 48 scoops to be distributed. So you know you've got at least seven single cones. After that, though, the number of doubles and triples can be worked any which way. If you go for the maximum number of triples, you'll end up with 24 triples and no doubles. So, you'll have anywhere from one to 23 triples, if you have to get all of them in, and anywhere between 46 and two doubles. As for singles, you'll then have anywhere between seven and thirty. So, quite frankly, I think the math problem is deeply flawed.

[ramblinsuze.livejournal.com]

I'm no math whiz, but unless there is some other information, there are a TON of answers to this one. I mean, both of these would be correct:

(2x2) + (2x3) + (1x93) = 103
4 + 6 + 93

(2x5) + (3x5) + (1x78) = 103
10 + 15 + 78

And so on and so forth.

It seems to me like the problem is flawed. 1x + 2x + 3x = 103 doesn't work because you don't come up with one number that works for x. I suppose they could be looking for 1x + 2y + 3z = 103, but you still run into the same problem of multiple answers (they don't have a qualifier like find the maximum number of single, double, and triple scoops that equals 103).

[annissamazing]

(2x5) + (3x5) + (1x78) = 103
10 + 15 + 78

I think you should go with this answer, but only because 10/15/78 is my birthday.

[bratflorida.livejournal.com]

I am disgustingly happy to see there's different ways to answer this, or that info might be missing. I was a book keeper for ten years and I was thinking WTF??? *starts planning for future nervous breakdown with daughter's homework*

*edited because I just saw your son is in the 5th grade.

Kill me now....

[annissamazing]

Wow. You weren't kidding when you said this is a "guess and check" problem. This is a "systems of equations" problem without enough info to actually make the system. Plugging it into a matrix on my graphing calculator doesn't work and neither does the substitution method. You literally have to guess and check the answer. I'm not sure how that's math, but maybe it's meant to get the fifth graders to practice their addition and multiplication.

x + y + z = 55
x + 2y + 3z = 103
There should be a third line here to make this a true system.

You'd be fine if they were only two variables. It's that third one that throws off the whole system.

That's my answer and I'm sticking to it. Otherwise I'll be entirely too nervous about my Modeling Algebra final on Tuesday.

[draconin.livejournal.com]

Strange problem for such a young student. I suspect that there is no intention that the student should find ALL possible solutions, merely that they find one possible one.

The earlier comments are correct in that this is two equations in a system that, in order to have a unique solution, requires three.
x + y + z = 55
x + 2y + 3z = 103

With a bit of algebra you can rewrite this as

x = z +7 and
y = 48 - 2z

In other words, you can at least get x and y (one and two scoops) in terms of z (the number of three scoops). This may not seem to help but it does. Remember that x, y and z are all required to have non-negative and integer values (no fractions or decimals). Therefore z can't be smaller than zero, nor can it be larger than 24 (without forcing y to be negative).

A bit of work with a spreadsheet lists all possible solutions deriving from these (with z running from 0 to 24 and x and y derived from it). There may well be others - I'm two glasses of red wine down at the moment and not working to full capacity! *g*

[honorh.livejournal.com]

Heh. Amber, you ought to send your boy to school with this as his result. Blow his teacher's mind!

[arnica.livejournal.com]

Wow, that is very open ended for 5th grade math, isn't it? I shudder to think about the homework looming on the horizon when my kid gets into school

[develish1]

for once I think I've rather happy I got here late. Just a glance at the answer others have given was enough to give me a headache, and I though I was good at math

[yavie-namarie.livejournal.com]

Just looking at some of the comments to this entry is making my brain short circuit, so I'm sure I won't be of any help. But I can head up the Empathy Department, if you'd like... ;)