Anonymous wrote:(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:Anonymous wrote:Anonymous wrote:Anonymous wrote:At the 3rd grade, students are not yet exposed to algebra. However, many answers provided here are algebra based. What is going on?
The problems can be solved using formal algebra -- and if you know formal algebra, that's probably how you'd do it. But they don't have to be solved using formal algebra, as other PPs have demonstrated.
That's true, but it seems to be the case that they're not teaching any method to solve it. I think 3rd graders that could solve this without having some method explained would be few and far between.
I agree. It would have been a good Problem of the Week for my kid in fourth grade at the HGC last year, though.
Anonymous wrote:Anonymous wrote:Anonymous wrote:At the 3rd grade, students are not yet exposed to algebra. However, many answers provided here are algebra based. What is going on?
The problems can be solved using formal algebra -- and if you know formal algebra, that's probably how you'd do it. But they don't have to be solved using formal algebra, as other PPs have demonstrated.
That's true, but it seems to be the case that they're not teaching any method to solve it. I think 3rd graders that could solve this without having some method explained would be few and far between.
Anonymous wrote:Anonymous wrote:At the 3rd grade, students are not yet exposed to algebra. However, many answers provided here are algebra based. What is going on?
The problems can be solved using formal algebra -- and if you know formal algebra, that's probably how you'd do it. But they don't have to be solved using formal algebra, as other PPs have demonstrated.
Anonymous wrote:At the 3rd grade, students are not yet exposed to algebra. However, many answers provided here are algebra based. What is going on?
Anonymous wrote:What school?
Anonymous wrote: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:OP, does the problem sheet say "Continental Math League" on it?