Another Algebra Question -- Help!

[DemonLlama]

What is the EASIEST way to solve the following problem:

26 fifth graders collected 500 canned goods. If some students collected one more than the rest, how many collected 19 and how many collected 20?

I know we're just missing some simple algebraic formaula that would make it easier than the way we've come up with, but I can't figure it out.

Thanks!

 

[KingsFanInRI]

as a master of improvising the answer to algebra questions (much to the chagrin of my math teacher, Mrs. Dunagan), what I quickly saw was that you have a round number to deal with (500). So there are only so many groups of 19 you could have (like 10 students would have 190 cans). Double that, you have 20 students with 19 cans, or 380 cans. 120 cans and 6 students left. Simple division shows that you have found your answer.

In short, with a round total of 500 cans, you have to have a multiple of ten AND 19 as part of the equation. The 19 has to be multiplied until the total ends in a zero.

This is much harder to describe than it is to do in my head, I hope it was clear enough.

 

[damo]

First of all, I would assume that each student either collected 19 or 20 items (but that is not stated in the problem).

Think of it this way, some number times 19 plus some number times 20 must equal 500. And those two numbers must add up to 26. Using guess and check, that is trial and error, you would get 20 students collecting 19 items and 6 students collecting 20 items.

Using an algebraic equation:

Let x be the number collecting 20 items
Let 26-x be the number collecting 19 items

20x + 19(26-x) = 500
20x + 494 - 19x = 500
x = 6

So 6 students collect 20 items and 20 students collect 19

 

[KarenC]

What grade is your child in who has this for homework? This problem type sounds VERY familiar to me...is it from Everyday Mathematics?

If your child is not yet taking algebra I suspect they want him or her to get the answer via the trial and error method. Not easy or quick, but the point is to build the blocks toward understanding algebra. So when he or she gets to 7th or 8th grade and gets to algebra they understand the point.

If this is algebra, never mind.

 

[tkd lisa]

To do this in algebra, you have to say that the number of students who collected 19 is x, the number of students who collected 20 is y. You know that x+y=26 (the total number of students). This is then a classic 2 equations, 2 unknown problem. To solve it you put one variable in terms of the other.

The formula would be:
500=19x+20y
x+y=26

Solving formula 2 for x:
y=26-x

Plugging in to the first formula:
500=19x+20(26-x)
500=19x+520-20x
500-520=-x
x=20

Plugging x back into the original equation to find y:
y=26-20 (the value for x)
y=6

So this means that 20 people sold 19, and 6 people sold 20

Double checking that this is right:
500 total were sold
20x19 cans + 6x20 cans =500

Sorry, it's the engineer in me coming out!

 

[browneyes]

I bet KingsFanInRI ruined all the curves when it came test time.