Solutions to SingaporeWordProblemSampler3

Primary 3: This one is easy to solve with algebra, but I think Singapore math intends it to be done with bar diagrams at this age. If you have a 3rd or 4th grader working on this, it might be good to try doing this with them using a bar diagram. This paper by Sybilla Beckman (Catherine just posted about it) will show you the basics of how to do bar diagrams.

Here's the algebra solution.

If Margo has p pears and a apples, then

p = 3 a

says that there are 3 times as many pears and apples, and

a + p = 84

says that there are 84 pieces of fruit altogether. Substituting 3_a_ for p in the second equation gi ves

a + 3 a = 84, or 4 a = 84.

Therefore there are 21 apples, and so 63 pears.

Primary 4: This is the kind of problem that Christopher and Ben would refuse to do because it's too easy to do in your head.

Jim and Tom together ate 6 pieces out of 12, therefore leaving 6 pieces out of 12, or 1/2 of the cake.

Primary 5: The bag of potatoes weighs 7/8 kg., and the bag of yams weighs 4/5 as much.

Therefore the bag of yams weighs (4/5)(7/8) kg., or 7/10 kg..

(Note: a good thing to teach is that the word 'of' in word problems, as in '4/5 of 7/8 kg.', often means 'multiply the numbers.' This is actually a tip from Wickelgren).

To get the combined weight, add 7/8 and 7/10 kg., giving 63/40 kg..

Primary 6: This is a problem that I would tackle with algebra, though the Primary 6 text does set it up with bar diagrams. My reasoning is that if a kid is ready to tackle a problem that's this sophisticated, then they're ready to do it with algebra.

Let e be the money that Eric has, j be the money that Joshua has, and c be the money that Carl has.

The first statement, Eric has 75% as much money as Joshua has , translates to: e = .75 j.

The second statement, Carl has 60% as much money as Eric and Joshua have together, translates to: c = .6( e + j ).

The third statement, Eric has 36 dollars less than Carl, becomes: e = c -36.

It's three equations in three unknowns. There are many ways to solve such a problem: here is one.

I would substitute expressions in j for the variables e and c in the second equation. This is nice because, once you've solved that second equation for j, you're done: you don't have to substitute into the other equations to get the specific answer to the problem.

I have an equation that expresses e in terms of j: equation 1. How do I turn the third equation, that doesn't involve j, into an expression in j ? Fortunately, I have the first equation. Plugging e = .75 j into equation 3 gives

.75 j = c -36.

Solving for c gives c = .75 j + 36. I already have e = .75 j, so substituting for e and c in equation 2 gives:

(.75 j + 36) = .6(.75 j + j) = .6(1.75 j).

Now you take out your handy calculator and figure out what .6 times 1.75 is. (JOKE!) It's 1.05, so

.75 j + 36 = 1.05 j, and subtracting .75 j from both sides gives

36 = .3 j.

So 120 = j: Joshua has 120 dollars.

-- CarolynJohnston - 05 Jul 2005

Comments

Back to: Main Page.