### An algebra problem

This question was set in one of the renowned primary school from Singapore. Given to me by "anonymous" to solve.

Solution:

Let A be the initial amount Andy has, and P the initial amount Peter has.

Then we know that A = P + 200. We're going to use that later, but for now I'm going to write it all in terms of A and P.

Andy gives 60% of his money or 0.6A to Peter.

Peter has now P + 0.6A.

Andy has now 0.4A.

Peter then gives 25% of his money to Andy. But this isn't 0.25P because Peter doesn't have P dollars anymore because Andy already gave him some. It's 0.25 (P + 0.6A) that he gives back to Andy.

Peter has now 0.75(P + 0.6A)

Andy has 0.4A + 0.25 (P + 0.6A)

In the end, Peter has $200 more than Andy. That gives us a way to write an equation:

0.75(P + 0.6A) = 200 + 0.4A + 0.25 (P + 0.6A)

It's now simple to solve this equation of two variables by first substituting A = P + 200. The rest of it is just basic manipulations.

0.75(P + 0.6(P + 200)) = 200 + 0.4(P + 200) + 0.25(P + 0.6(P + 200))

0.75(P + 0.6P + 120) = 200 + 0.4P + 80 + 0.25(P + 0.6P + 120)

0.75(1.6P + 120) = 280 + 0.4P + 0.25(1.6P + 120)

0.75(1.6P + 120) = 280 + 0.4P + 0.25(1.6P + 120)

1.2P + 90 = 280 + 0.4P + 0.4P + 30

0.4P = 220

P = 550

and so A = 750

Check:

First Peter had $550, Andy $750.

Then Andy gives $450 to Peter.

Peter has now $1000, Andy $300.

Then Peter gives $250 to Andy.

Peter has now $750, Andy $550.

Andy has $200 more than Peter. Andy gives 60% of his money to Peter. Peter then gives 25% of his money to Andy. In the end, Peter has $200 more than Andy. How much did Andy have at first?This is a great problem to solve with algebra. Why don't you try it first, before reading further? It sounds kind of interesting... first one guy has $200 more than the other, and in the end it's reversed.

Solution:

Let A be the initial amount Andy has, and P the initial amount Peter has.

Then we know that A = P + 200. We're going to use that later, but for now I'm going to write it all in terms of A and P.

Andy gives 60% of his money or 0.6A to Peter.

Peter has now P + 0.6A.

Andy has now 0.4A.

Peter then gives 25% of his money to Andy. But this isn't 0.25P because Peter doesn't have P dollars anymore because Andy already gave him some. It's 0.25 (P + 0.6A) that he gives back to Andy.

Peter has now 0.75(P + 0.6A)

Andy has 0.4A + 0.25 (P + 0.6A)

In the end, Peter has $200 more than Andy. That gives us a way to write an equation:

0.75(P + 0.6A) = 200 + 0.4A + 0.25 (P + 0.6A)

It's now simple to solve this equation of two variables by first substituting A = P + 200. The rest of it is just basic manipulations.

0.75(P + 0.6(P + 200)) = 200 + 0.4(P + 200) + 0.25(P + 0.6(P + 200))

0.75(P + 0.6P + 120) = 200 + 0.4P + 80 + 0.25(P + 0.6P + 120)

0.75(1.6P + 120) = 280 + 0.4P + 0.25(1.6P + 120)

0.75(1.6P + 120) = 280 + 0.4P + 0.25(1.6P + 120)

1.2P + 90 = 280 + 0.4P + 0.4P + 30

0.4P = 220

P = 550

and so A = 750

Check:

First Peter had $550, Andy $750.

Then Andy gives $450 to Peter.

Peter has now $1000, Andy $300.

Then Peter gives $250 to Andy.

Peter has now $750, Andy $550.

## Comments

All primary schools students are only taught simple algebra. My teacher say that this can be done without using algebra and she strongly discourage her students to use algebra to solve the sums.

She has hinted that this problem can be solved using Singapore model drawing.

Unfortunately, i am unable to solve this problem as i do not know how to draw a bar model.

Kindly help me by Singapore model drawing method. Thanks

Reflected below depicts my working:

let the the amount of money Peter has at first be $x.

We ignore the dollar sign for easy computations.

Solution:At first: Peter: x

Andy: x + 200

Firstly, Andy gives 60% to Peter:

Andy's left: 0.4(x + 200)

Peter left after receiving from Andy : x + 0.6(x + 200)

Thereafter, Peter gives 25% of what he has to Andy.

Peter's left after giving to Andy :0.75(x + 0.6(x + 200))

= 0.75x + 0.45x + 90

= 1.2x + 90 (simplify it for easy computation)

Andy's receive : 0.4(x + 200) + 0.25(x + 0.6(x + 200))

= 0.4x + 80 + 0.25x + 0.15x + 30

= 0.8x + 110 (simplify it for easy computation)

Therefore:

P = A + 200 or P - A = 200

1.2x + 90 = 0.8x + 110 + 200

0.4x = 220

x = 550

Peter has $550 at first.

Andy : x + 200

= 550 + 200

= 750

Andy has $750 at first.

PS:

However, my level does not permit the students to use algebra as mentioned eariler from my previous thread that they have not been taught how to use algebra to solve tedious and difficult problems like the above.

Other than algebra, i sincerely hope that you can use model drawing which was taught and used in all Singapore primary school to help me better understand and solve the problems.

Thanks

Thanks for your compliment. Care to share with us your method of doing this questions. For your information, majority of Singapore students who attempted this questions during their Prelimary exam score ZERO for this question. Most students were in tears as they spent most of the time trying to solve this cranky but killer question. Not to mention that i am one of the cohort as tension overwhelmed me.

Students were stumped by the way the question was being phrase and the most disturbing part is they are not allowed to use

algebra.Students are advised to use model drawing or any of the heuristic to facilitate them in solving such tough question above.

Thanks

anyalgebra... Sorry.Maybe I could if I devoted much more time to it, but I can't.

Surely if this was in an exam, there should be solutions available from the exam makers.

Also, maybe these kind of problems are present in the curriculum. Seeing one example might help me to figure it out without using algebra, but I haven't seen such.

Personally I feel it's too tough for elementary students... but the exam makers in Singapore must feel differently.

Maybe someone else will find this post and present a solution without algebra.

And, it can be solved using one variable only; essentially with that solution you substitute x + 200 (or P + 200) for Andy's amount in the beginning, and not in the end.

But using two variables while developing the equation can be clearer for others to follow.

We can solve this without using algebra too. We can solve this base on the model approach which is commonly widely used in Singapore Primary School.

The question is:

Andrea has $200 more than Bala. Andrea gives 60% of his money to Bala. Bala then gives 25% of his money to Andrea. In the end, Bala has $200 more than Andrea. How much did Andrea have at first?

Solution:

At first ----- ?

Andrea ----- 1 unit + 200

Bala ----- 1 unit

Andrea gives 60% ----- ?

Andrea ----- 1 unit + 200 – 0.6 unit – 120

Bala ----- 1 unit + 0.6 unit + 120

Andrea ----- 0.4 unit + 80

Bala ----- 1.6 units + 120

Bala gives 25% ----- ?

Bala ----- 1.6 units + 120 – 0.4 unit – 30

Andrea ----- 0.4 unit + 80 + 0.4 unit + 30

Bala ----- 1.2 units + 90

Andrea ----- 0.8 unit + 110

Bala now has $200 more ----- ?

Bala – Andrea ----- 200

1.2 units + 90 – 0.8 unit –110 ----- 200

0.4 unit – 20 ---- 200

0.4 unit ----- 200 + 20 = 220

1 unit ---- 220 divided by 0.4 = 550

Andrea at first -----

1 unit + 200 ----- 550 + 200 = 750

Your approach to solving the problem with Andrea and Bala is nice, but it's just algebra in disguise.

Instead of x, you used "unit", and instead of = signs, you put ----

But it's identical to the algebraic approach, where you use an unknown and an equation.

I'm not saying it's bad, by the way. I do like using algebra to solve these kind of problems.

I solved this problem by writing out the equations in one variable also. That said, I agree with you that the model method demonstrated by anonymous is algebra! I am interested in these problems as a homeschooling mom because I am currently using Singapore Math with my 5 year old who is just finishing the primary 1A book. It gives me much to think about as we move forward.

See http://model4maths.blogspot.com/, hope it can add values as what you have been doing.

Great blog!

Model approach