I don't understand this word problem (Singapore)

[sbgrace]

We've seen these supposition type problems over and over. I can follow the procedure, which always works, but I don't understand how/why it works. Can anyone help me get the why of this type of problem?

 

There are 23 chickens and cows at a farm. They have 60 legs altogether. How many of each type of animal are there at the farm? (or cars/motorcycles, 20 and 50 cent coins, whatever...I don't really understand any of them).

 

I understand that we suppose all are chickens (so 46 feet, or 14 less than we need).

Then we take 14/2 (the 2 extra legs the cow has) and get 7

There are 7 cows and (23-7) 16 chickens.

 

 

Ok, yes, it works. It works every time.

 

Just memorizing the steps isn't going to stick long term, so why does this work? And how important is this type of problem solving?

 

 

[wapiti]

Does it help if you think of it from an algebra perspective (set up systems of equations)?

[Snickerdoodle]

I like these kind of problems bc it forces the kid to think about whether it makes sense or not and to avoid the plug and chug.    A similar problem is the "how many blocks can you fit into this box" type problems where you can't just calculate the volumes;  you have to stop and ask whether you could actually fit the blocks into that space.  

[MASHomeschooler]

Here's a crazy thought:

 

Imagine they are magic chickens - when you add two legs, they become cows (if you add 1 leg or more than 2 legs, they will die, so don't do that). So, you know how many animals you have (23), and they all start as chickens, so they have 46 legs. But you also know that you have 60 legs total. So you have a pile of 14 extra legs that can't just be left there. So you add them, 2 at a time, to chickens, which then turns those chickens into cows. 14 extra legs, divided by 2 legs per animal, gives you 7 animals that you will turn into cows. So, of your original 23, 7 have become cows, and 16 remain chickens. Does that help?  :lol:

 

BTW, I tend to think of and solve these algebraically as well.

[Btervet]

It's much easier to see with algebra, but I'll try to explain without since it's meant to be solved without algebra. Sorry this came out sounding like a messed up version of taxidermy. 

 

1. I understand that we suppose all are chickens (so 46 feet, or 14 less than we need).

Here you have all the animals you can, but left over legs. You need to add some legs to the animals without changing the number of animals. 

2. Then we take 14/2 (the 2 extra legs the cow has) and get 7

For each animal you change (from chicken to cow) you are adding two legs to the total (you had assumed two legs, and are changing the assumption to four legs). So you are trying to figure out how many chicken -> cow conversions you can make you take the extra leg number divided by the amount of legs you are adding per conversion. Which is 14/2 in this instance. This is how many conversions you can make.

3. There are 7 cows and (23-7) 16 chickens.

You made 7 conversions of chickens to cows, so you have 7 cows and 23-7 chickens.

[Tsuga]

I'll try to explain the relationship between the way you are doing it--plug-and-chug--and algebra.

 

Here's the "easy" way, algebra. Your son is in 5th grade, a bit young:

 

c=cows, h=chickens

 

4c + 2h = 60 (four times the number of cows = # cows legs, two times the number of chickens = # chicken legs, and we know total legs = 60)

 

c + h = 23 (cows + chickens = 23)

 

c = 23 - h

 

Substitute that into the first equation so that you can solve for h in the first equation. Then you have the number of chickens and can solve for cows.

 

What you are doing is basically saying, we don't want to do a whole system of equations, so let's pretend # cows = zero. So 23 is chickens.

 

h=23.

 

Plug it in to the other equation and you get 4c (four times the number of cows = 0 because we have zero cows) + 23*2 legs each = 60.

 

0 + 46 = 60

 

46 = 60

 

Well, that's not right.

 

46 is fourteen than 60. So we need at least fourteen more legs. Now, we can't have any more chickens because the maximum number of chickens is 23. You can't have more chickens than animals.

 

We get two extra legs for each cow we substitute, as you wrote. So to get 14 more legs, we need 14/2 legs per cow = 7.

 

I am guessing that this is all to make algebra a relief since it's much easier to solve algebraically than to do all that thinking.

 

Systems of equations are really important.

 

But I would not say that the guessing game is critical except to prove the need for algebra.

[maize]

I think of the legs in sets if two. If there are sixty legs, that is thirty sets of two. If there are only 23 animals, 7 of the sets of two legs must be attached to other sets of two, making up a four-legged animal. Since we now know there are 7 4-legged animals, the remaining 16 must be two-legged animals.

[sbgrace]

Thank you all. I think I can explain it now (I'm going to use the magic chickens with one child, and I think I'll use algebra with the other). I should have asked this last year, we've been working without really understanding the whole time.

[Farrar]

Honestly, I could not explain this without algebra and it was one of the reasons that I became disillusioned with the Singapore problems. Like, I get a lot of them - most of them really - and have learned to use the bar diagrams and one of my boys is pretty decent at them and I see their usefulness and I like the whole multi-step, really think about it aspect of having complex word problems in general... BUT there were a few like this where understanding it with algebra was SO much simpler. Like so much simpler. It's the reason we have algebra in the first place is to solve this stuff. So why in the world was I trying to make a poor 7 yo (I'm just guessing, but it sounds like one from the 2nd grade book from when we used it way back when) understand this? Why oh why? Why as a society do we think this is a good idea?

 

Sorry... just... yeah, I couldn't see the point of this one without algebra.

[nansk]

Honestly, I could not explain this without algebra and it was one of the reasons that I became disillusioned with the Singapore problems. Like, I get a lot of them - most of them really - and have learned to use the bar diagrams and one of my boys is pretty decent at them and I see their usefulness and I like the whole multi-step, really think about it aspect of having complex word problems in general... BUT there were a few like this where understanding it with algebra was SO much simpler. ...

Sorry... just... yeah, I couldn't see the point of this one without algebra.

But this one is not solved using the bar method. Apart from the bar method, Singapore Maths teaches a number of heuristics (problem solving methods) including listing, guess-and-check, and assumption method.

 

In Primary 1 and 2, students are taught to solve this type of problem, first by drawing each object (chicken/car/whatever) and later by the guess-and-check method. And from Primary 3 onwards, the guess-and-check method segues into the assumption method (which is what the OP described as 'supposition'.)

 

And while using Algebra may be intuitive to you, it will be too abstract for most primary school kids.

[kiwik]

If I were trying to explain it to a 7 year old I would draw a picture or use toothpicks or rods.

[Ausmumof3]

Here's a crazy thought:

 

Imagine they are magic chickens - when you add two legs, they become cows (if you add 1 leg or more than 2 legs, they will die, so don't do that). So, you know how many animals you have (23), and they all start as chickens, so they have 46 legs. But you also know that you have 60 legs total. So you have a pile of 14 extra legs that can't just be left there. So you add them, 2 at a time, to chickens, which then turns those chickens into cows. 14 extra legs, divided by 2 legs per animal, gives you 7 animals that you will turn into cows. So, of your original 23, 7 have become cows, and 16 remain chickens. Does that help? :lol:

 

BTW, I tend to think of and solve these algebraically as well.

This is the best explanation I have heard... Normally I have to use algebra to figure these out.

[Farrar]

But this one is not solved using the bar method. Apart from the bar method, Singapore Maths teaches a number of heuristics (problem solving methods) including listing, guess-and-check, and assumption method.

 

In Primary 1 and 2, students are taught to solve this type of problem, first by drawing each object (chicken/car/whatever) and later by the guess-and-check method. And from Primary 3 onwards, the guess-and-check method segues into the assumption method (which is what the OP described as 'supposition'.)

 

And while using Algebra may be intuitive to you, it will be too abstract for most primary school kids.

The book made it clear you were not supposed to guess and check - that all of us could could have gotten - but rather to do the algorithm the OP showed. But none of us could understand the why behind it. And finally I just decided it was too complex for most 7 yos. Why have kids this age memorize an algorithm for this? It's time much better spent on other things. The point isn't that 7 yos should do algebra, but rather why have 7 yos learn this particular algorithm for this particular problem. My boys are in 5th grade now. One of them would be able to understand this with algebra now relatively easily. I'm convinced some of the time (not all) we spent trying to do those Singapore problems at such a young age was really a waste.

[fdrinca]

Charts are a strategy, too. Making a chart exposed a simple algorithm that my DS could then use to get his answer.

[Momling]

We called those "leg problems" and struggle yearly with them. We ended up making a whole series of them (legs at a Beatles and spider farm, fingers at a polydactyly conference, heads of mythological dogs, cherries that come in bunches of twos or threes). I recall at first giving my daughter marshmallows as bodies and toothpicks for legs and having her solve and write easy versions of the same problem. She really got the idea of distributing the legs for the fewer legged animal first then when everyone had them, distributing the remainder of the legs. I think she'd solve it now with algebra, though the marshmallows were yummy.

[Heidi]

Here's a crazy thought:

 

Imagine they are magic chickens - when you add two legs, they become cows (if you add 1 leg or more than 2 legs, they will die, so don't do that). So, you know how many animals you have (23), and they all start as chickens, so they have 46 legs. But you also know that you have 60 legs total. So you have a pile of 14 extra legs that can't just be left there. So you add them, 2 at a time, to chickens, which then turns those chickens into cows. 14 extra legs, divided by 2 legs per animal, gives you 7 animals that you will turn into cows. So, of your original 23, 7 have become cows, and 16 remain chickens. Does that help? :lol:

 

BTW, I tend to think of and solve these algebraically as well.

I totally get it bc of this explanation. I'm not joking!

[nansk]

..at first giving my daughter marshmallows as bodies and toothpicks for legs and having her solve and write easy versions of the same problem....

This is a fab idea. :)

[Lucy the Valiant]

Here's a crazy thought:

 

Imagine they are magic chickens - when you add two legs, they become cows (if you add 1 leg or more than 2 legs, they will die, so don't do that). So, you know how many animals you have (23), and they all start as chickens, so they have 46 legs. But you also know that you have 60 legs total. So you have a pile of 14 extra legs that can't just be left there. So you add them, 2 at a time, to chickens, which then turns those chickens into cows. 14 extra legs, divided by 2 legs per animal, gives you 7 animals that you will turn into cows. So, of your original 23, 7 have become cows, and 16 remain chickens. Does that help?  :lol:

 

BTW, I tend to think of and solve these algebraically as well.

 

But can you MILK a magic chicken?!?

[hellen]

Here's a crazy thought:

 

Imagine they are magic chickens - when you add two legs, they become cows (if you add 1 leg or more than 2 legs, they will die, so don't do that). So, you know how many animals you have (23), and they all start as chickens, so they have 46 legs. But you also know that you have 60 legs total. So you have a pile of 14 extra legs that can't just be left there. So you add them, 2 at a time, to chickens, which then turns those chickens into cows. 14 extra legs, divided by 2 legs per animal, gives you 7 animals that you will turn into cows. So, of your original 23, 7 have become cows, and 16 remain chickens. Does that help? :lol:

 

BTW, I tend to think of and solve these algebraically as well.

That's not crazy. It is absolutely brilliant!

[Tsuga]

I know nothing about Singapore as a country. Are children in Singapore more familiar with farm animals than our children, so this is a more tangible story to imagine for them? Like do they regularly visit grandparents in the Singaporean countryside, if there is such a thing (I thought it was a city state)?

 

 

[nansk]

I know nothing about Singapore as a country. Are children in Singapore more familiar with farm animals than our children, so this is a more tangible story to imagine for them? Like do they regularly visit grandparents in the Singaporean countryside, if there is such a thing (I thought it was a city state)?

There is no countryside in Singapore.  :lol:  There are no cows that I know of and only a few tiny farms where chickens are bred.

 

Word problems like these start in Primary 1. Sometimes it is cows and chicken, sometimes motorcyles and cars, and (Primary 4 onwards) 10-cent, 20-cent, and 50-cent coins.

 

In Primary 1, the number of heads (or number of vehicles) are under 20. So the children can draw 20 circles and give them 2 legs each first. It starts exactly like Momling's idea of using marshmallows and toothpicks, but on paper.

 

Then, in the latter part of Primary 1 and in Primary 2, the students draw a chart and use the guess-and-check method to solve the problems with bigger numbers (up to 50 or 100 heads, for example). So the columns look like:

 

No. of Chickens | No. of Cows | Total Animals | Chicken legs | Cow legs | Total legs  | Difference

 

With practice (and a bit of guidance) the students learn to guess the right answer in 2 steps.

For example:

 

There is a total of 12 cows and ducks on a farm. They have 38 legs altogether. How many cows are there on the farm?

 

Assume 6 cows and 6 ducks first.

Guess #| Ducks |  Cows | Animals | Duck legs       | Cow legs          | Total legs  | Difference

 #1         |      6   |        6  |    12       | 6 x 2 =  12      | 6 X 4 = 24        |        36       |  38-36 =   2

 

2 more legs needed => 1 more cow needed

 

Guess #| Ducks |  Cows | Animals | Duck legs       | Cow legs          | Total legs  | Difference

 #2         |      5   |        7  |    12       | 5 x 2 =  10      | 7 X 4 = 28        |        38       |  38-38 =   0 (correct)

 

Ans: There were 7 cows.

 

This method thus naturally leads to the assumption/supposition method in Primary 3 and 4.

[Farrar]

The CWP book never presented it as a chart though. A chart is easy. Guess and test is actually how I ended up teaching it because the method in the OP didn't make sense to us. I was thrilled when we started using the Process Skills in Problem Solving books much more recently and found that they do teach guess and test and make a chart strategies.

[fdrinca]

The CWP book never presented it as a chart though. A chart is easy. Guess and test is actually how I ended up teaching it because the method in the OP didn't make sense to us. I was thrilled when we started using the Process Skills in Problem Solving books much more recently and found that they do teach guess and test and make a chart strategies.

 

My bad - you're right, they didn't present a chart  for this specific problem, but in CWP 3 it uses a chart as a way to solve the "Joan collected three buttons for every two that John found" sort. For a while, my son was making a chart for everything, including leg problems. 

[chelsmm]

Here's a crazy thought:

 

Imagine they are magic chickens - when you add two legs, they become cows (if you add 1 leg or more than 2 legs, they will die, so don't do that). So, you know how many animals you have (23), and they all start as chickens, so they have 46 legs. But you also know that you have 60 legs total. So you have a pile of 14 extra legs that can't just be left there. So you add them, 2 at a time, to chickens, which then turns those chickens into cows. 14 extra legs, divided by 2 legs per animal, gives you 7 animals that you will turn into cows. So, of your original 23, 7 have become cows, and 16 remain chickens. Does that help?  :lol:

 

BTW, I tend to think of and solve these algebraically as well.

 

 

This is awesome.  I am SO using this with my kids the next time this kind of question comes up!  I remember this question when I did this level with dd, and I think it's coming up soon with ds.  LOVE this!