Anonymous wrote:OP,
This is a big issue many parents have with MCPS right now. They pretend to give "enrichment" by giving 3rd graders these challenging problems. I cut and pasted one of DS's math problems in Google and came up with a complete PDF of the question, possible answers and methodology commentary, where the author explained that this problem was given to 5th and 6th graders.
My concern is that the rest of the math curriculum is very simple (ie, not as advanced as I would want). The math problems are too complex compared to the rest. There is nothing in the middle to "train" the kids to stretch to these complex multistep problems. Everyone wants their children to learn how to formulate and strategize in math. But here the burden is placed on the parent to support their child, not the school, which is ALL WRONG.
So why does MCPS give out these problems?
Either MCPS genuinely believes the effort to think things through is worth it even if students don't fully grasp the strategy involved. This I strongly disagree with: my kid could learn this so much better with a different approach, ie step-by-step instruction in class to solve simple to complex word problems;
Or MCPS is just fobbing off parents with talk of "enrichment" (one problem a week!) so that no one can accuse them of lack of differentiation/acceleration. I believe the latter.
Either way, it's quite disheartening. Parents have had their children cry because they cannot understand these problems, which undermines the whole premise of encouraging the students to stretch their minds. We have alerted the principal and 3rd and 4th grade teachers, but she probably cannot do anything about it, since it is mandated by the curriculum 2.0.
Anonymous wrote:
Either way, it's quite disheartening. Parents have had their children cry because they cannot understand these problems, which undermines the whole premise of encouraging the students to stretch their minds. We have alerted the principal and 3rd and 4th grade teachers, but she probably cannot do anything about it, since it is mandated by the curriculum 2.0.
Anonymous wrote:At that age, I expect they want the kids to do something by trial and error.
Anonymous wrote:
Then plug this figure in to get the other numbers. I'm an engineer and this is how it should be explained.
Anonymous wrote:I don't think they want trial and errorThere 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.
Anonymous wrote:Anonymous wrote:DD brought home this problem. They have not been taught how to solve it in class, and I can't remember how.
Dustin and Michael together have 96 baseball cards.
Michael and Kevin to together have 93 cards.
Dustin and Kevin together have 81.
How many cards does each boy have?
There is a bonus question too, but I can't recall what it is.
Any ideas how to figure this out?
D + M = 96
M + K = 93
D + K = 81
You can multiply the last line by -1 on both sides, so:
D + M = 96
M + K = 93
-D - K = -81
Add all of them together and you get:
-D + D = 0
M + M = 2M
K + -K = 0
96 + 93 - 81 = 108
So, 2M = 108, M = 54.
Substitute back and you get D = 42 and K = 39.
This is how I would do it. I have NO idea of the process that they would teach to a 3rd grader. These are problems similar to the ones my daughter got in an HGC last year in 4th.
Anonymous wrote:Anonymous wrote:DD brought home this problem. They have not been taught how to solve it in class, and I can't remember how.
Dustin and Michael together have 96 baseball cards.
Michael and Kevin to together have 93 cards.
Dustin and Kevin together have 81.
How many cards does each boy have?
There is a bonus question too, but I can't recall what it is.
Any ideas how to figure this out?
D + M = 96
M + K = 93
D + K = 81
You can multiply the last line by -1 on both sides, so:
D + M = 96
M + K = 93
-D - K = -81
Add all of them together and you get:
-D + D = 0
M + M = 2M
K + -K = 0
96 + 93 - 81 = 108
So, 2M = 108, M = 54.
Substitute back and you get D = 42 and K = 39.
This is how I would do it. I have NO idea of the process that they would teach to a 3rd grader. These are problems similar to the ones my daughter got in an HGC last year in 4th.
Anonymous wrote:DD brought home this problem. They have not been taught how to solve it in class, and I can't remember how.
Dustin and Michael together have 96 baseball cards.
Michael and Kevin to together have 93 cards.
Dustin and Kevin together have 81.
How many cards does each boy have?
There is a bonus question too, but I can't recall what it is.
Any ideas how to figure this out?