Another Solve Singapore Word Problem

[danybug]

Here is another word problem I would be thrilled to see how you would show your student how to solve this:

 

There are three pieces of rock, X, Y and Z. Y is 2463g heavier than X. Z is 1914g heavier than Y. The total weight of Y and Z is 8628g. How much does X weigh?

 

Thanks again for your help!

[Dana]

With the bar models, I'd use three bars. One for X, one for y, one for z.

 

I'd have the y bar shown at same length as x; then 2463 past it.

The z bar shown as y-length, then 1914 past it.

 

Then a brace linking y and z with 8628 listed.

 

[don't know that models here will work...]

 

X [............]

 

Y [............][ 2463 ]

 

Z [.............][ 2463 ][ 1914 ]

 

(can't draw the 8628 but it would be over y & z)

 

This gives the eqns:

Y + Z = 8628

(x + 2463) + (x + 2463 + 1914) = 8628

 

Hope this helps!

[KAR120C]

There are three pieces of rock, X, Y and Z. Y is 2463g heavier than X. Z is 1914g heavier than Y. The total weight of Y and Z is 8628g. How much does X weigh?

Y + Z = 8628 and Z is 1914 more than Y... so

<-----------Z----------->

<-----Y-------> <-1914->

We could add 1914 to the total (8628) to get twice Z... I think that that would be 10542 (check the math - I'm doing it in my head!) so Z would be half that, or 5271. That means Y has to be 5271-1914, or 3357. Then we know that Y is 2463 more than X, so X has to be 3357-2463, or 894.

[Truscifi]

Here is another word problem I would be thrilled to see how you would show your student how to solve this:

 

There are three pieces of rock, X, Y and Z. Y is 2463g heavier than X. Z is 1914g heavier than Y. The total weight of Y and Z is 8628g. How much does X weigh?

 

Thanks again for your help!

 

Ok, I'll take a stab at this one too. I don't know if you are supposed to be using algebra though.

 

Y + Z = 8628, & Z = Y + 1914, so

Y + Y + 1914 = 8628

solve for Y, get 3357

X = Y - 2463

X = 894

 

Did I do that right?

[mommy&herboys]

Weight of X is Ag, of Y is Bg and Z is Cg.

So B = A + 2463 g and C = B + 1944 g and B + C = 8628

B + C = 8628

B + (B+1944) = 8628

2B = 8628-1944

2B = 6684

B = 3342

Now B = A + 2468

So, A = B – 2468 = 3342 – 2468 = 874

You can get value is C by subtracting B from 8628

[NicksMama-Zack's Mama Too]

Here is another word problem I would be thrilled to see how you would show your student how to solve this:

 

There are three pieces of rock, X, Y and Z. Y is 2463g heavier than X. Z is 1914g heavier than Y. The total weight of Y and Z is 8628g. How much does X weigh?

 

Thanks again for your help!

 

Again, I go back to "what do you know?"

 

Y is heavier than X

 

so...

X|--------|

 

Y weighs the same as X, plus 2463g

Y|--------||--2463g--|

} Y & Z = 8628g

Z is 1914g heavier than Y

Z|--------||--2463g--||--1914--|

 

 

Since we are looking for the weight of X, I would point out to the student that we can use what we know of Y & Z to find the weight of X since X shares that same weight up to a point...

 

Add 2463+2463+1914 and subtract from 8628, then you will be left with two units, so you will need to divide the result by 2 to find X.

[Mama2Three]

Just curious, from which SM book do these problems come?

[forty-two]

There are three pieces of rock, X, Y and Z. Y is 2463g heavier than X. Z is 1914g heavier than Y. The total weight of Y and Z is 8628g. How much does X weigh?

|----------------Z--------------|---------Y---------|

[1814g-][--------------------][--------------------]

[----------------------8628g-----------------------]

 

Y is one unit; Z is one unit plus 1814g; Y and Z together are 8628g, which equals two units plus 1814g. Two units equal 6814g, so one unit equals 3407g. X is one unit minus 2463g, so X equals 944g.

Edited November 30, 2009 by forty-two
clarity

[danybug]

Again, thanks for taking the time to answer! So far we have not hit any true algebraic teaching of solving word problems. And so I sometimes struggle on how to help my ds solve them otherwise. But I think using the bars will help.

 

I did not order the HIG for Singapore up until this point in 3B and maybe I missed teaching the bar part as a solving method. We also use the Standards Edition, which I am not sure if that makes a difference.

 

Anyway, thanks again!

[danybug]

Just curious, from which SM book do these problems come?

 

These came out of the Challenging Word Problems Primary 3 from Singapore. We are using them for more work along with our 3B Standards Edition.

[Capt_Uhura]

Ok, I got 894. For the first time, I felt comfortable using the bar diagrams and didn't revert to algebra!!!!! You can teach an old dog new tricks! We're in CWP 3 right now but haven't come across this problem yet.

 

I did it similar to someone above. I made X my unit and then drew the bars for Y, Z, and Y+Z. using Y+Z and the total, I got the unit Y (after dividing by 2), then based on the unit Y, I subtracted 2463 to get X=894.

 

Simply lovely...... ok gotta go back and do this with algebra just for fun. :lol:

 

Capt_Uhura

[NicksMama-Zack's Mama Too]

These came out of the Challenging Word Problems Primary 3 from Singapore. We are using them for more work along with our 3B Standards Edition.

 

Many SM users prefer to use a CWP from a year lower than they are currently teaching. Often the problems in CWP need to be worked out through with much discussion between the teacher and the student.

 

hth

K

[Dana]

I've found using the iExcel books (which I've only found at the Singapore site) to be helpful in setting up and seeing the bar models. The first half of each book is just general approaches to solving math problems/brain teasers. But the second half shows bar models for the arithmetic itself, then for word problems. They've been very helpful when we do the CWP books.

[danybug]

I never noticed the i-Excel books before. Maybe we will get one and try them out. Thanks for the suggestion!