What school?

At the 3rd grade, students are not yet exposed to algebra. However, many answers provided here are algebra based. What is going on?

(Not the PP.)

Draw a bar. Give it a value of 96. Divide it somewhere. Label one part M and one part D. Stick another bar onto the end of it. Give it a value of 93. Divide it somewhere. Label one part M and one part K. Now you have a long bar, divided in four parts, labeled M, D, M, K, with a value of 96+93=189.

Now you know what D+K is, namely 81. So shuffle the pieces around in the bar so that you have four parts, labeled M, M, D, K. The two parts labeled D and K, combined, have a value of 81. So the two remaining parts, labeled M and M, have a value of 189-81=108.

If M+M=108, then M=54.

And then you can figure out the other ones.

(I'm the PP who's a fan of bar models, not the PP who came up with the bar-model solution.)

Anonymous wrote:
Easy to solve if you learn Singapore bar modeling! I was typing on an iPhone and it didn't come out that well. Let me try on the computer.

.........96..................93
--------------l---------------- 189
......D +M...............M + K



..............?..............81
---------------l----------------189
M + M D+K


---------------108
......M+M


M= 54

This is a typical singapore math type word problem. No need for algebra.

Ugh it still didn't format correctly.

Easy to solve if you learn Singapore bar modeling! I was typing on an iPhone and it didn't come out that well. Let me try on the computer.

96 93
Step 1. --------------l---------------- 189
D +M M + K



Step 2 ? 81
---------------l----------------189
M + M D+K


Step 3 ---------------108
M+M


Step 4 M= 54

This is a typical singapore math type word problem. No need for algebra.

OP, does the problem sheet say "Continental Math League" on it?

This is the engineer who solved the problem using algebra. My husband solved it differently; he feels the problem is more "logic" than math.

Add the totals up:
D + K = 81
D + M = 96
K + M = 93
----------------
= 270

Since each kid is double counted above, divide by 2 to find the total number of cards. Once you have the total, subtract out the known amounts above to find the missing amount:
D + K + M = 135

We already know that D + K = 81
So subtract 135 - 81 = 54
Mike = 54

You can do the rest....

We have wicked smart kids

Anonymous wrote:Pp here- I forgot to add you can solve this problem using a singapore math bar modeling approach without using algebra based on the solution I just wrote.
--------- ------------189
D+M. M+k

-------------------189
81 M+M.

---------108
M +M


M=54

I love love love the Singapore Math bar modeling approach.

Anonymous wrote:You don't need to use any negative numbers.
D+M =96
M+k=93
So D+M+M+k= 96+93
And you know D+k=81

So substitute
M+M +81 =189
M+M=108
M=54

I understand this better than multiplying by the -1!

Pp here- I forgot to add you can solve this problem using a singapore math bar modeling approach without using algebra based on the solution I just wrote.
--------- ------------189
D+M. M+k

-------------------189
81 M+M.

---------108
M +M


M=54

Please do not teach algebra to a 3rd grader. You will cause more harm than good.

Let them guess and check, then refine their answer. Understanding what they need to find is the biggest achievement.

Anonymous wrote:I don't think they want trial and error There is an easier way to solve this problem -

Solve two equations simultaneously:
M+K = 93
D+K = 81

You want to get rid of the Ks so you multiply the last equation by - 1
then have:

M+K = 93
-D-K = -81
then add together the two equations (the Ks are eliminated)

M-D = 93-81
M-D = 12
D = M-12

Plug this in to the first equation (D+M = 96):

M-12 + M = 96
2M = 108
M = 54

Then plug this figure in to get the other numbers. I'm an engineer and this is how it should be explained.

Thanks.

9:50's version easier to follow for a kid I think.

You don't need to use any negative numbers.
D+M =96
M+k=93
So D+M+M+k= 96+93
And you know D+k=81

So substitute
M+M +81 =189
M+M=108
M=54

Anonymous wrote:They should start by teaching kids to set up equations and then teaching kids to solve equations with two variables only.

And they do. But not in third grade math. Which is why one appropriate way for third graders to solve this problem is through guess-and-check.

Anonymous wrote:I'm the math teacher PP (actually, ex-math teacher). I haven't taught in this country, though (but in other English-speaking countries). I want to add that I would never have given that problem to 3rd graders - apart from the level of it, it has absolutely no context and is a million miles from a real life problem that someone might come across - and not even interesting to boot.

It's such a shame that US math has lost its way in recent years. When I was training as a math educator, the US had some of the best resources. Now I'd look elsewhere for inspiration.

I think that US Math Textbooks are the best. I routinely send it back to my home country for my nieces and nephews.

Unfortunately - the way Math is taught -without textbooks - and in a random selection of units - it makes it harder for anyone to understand the concepts.

Pick a textbook and stick with the way the chapters have been organized. Share the textbooks with the parents as well. Make kids either have 1) a copy of the textbook at home or 2) Ask them to keep it in the backpack.