Math question for kids

[Weemum883]

Please help, how do you solve this?

12

Probably by trial and error, and realising the number has to be divisible by 2 (to get half) and 3 (to get third). Which even numbers satisfy this and only have 2 between half and a third?

Melissa has X coins.

x/2 - 2 = x/3

Multiply both sides by 6

3x - 12 = 2x

Add 12 to both sides

3x = 2x + 12

Takes 2x off each side

x = 12

2 coins is 1/6, so 12.

Yup, and it has to be a number smaller than 20. So ...

The algebraic way would be more GCSE level. What age?

You know the total number of coins is an even (because 'half the coins' is an integer) number less than twenty that's also divisible by three (because 'one third of the coins' is an integer). So to begin with that narrows it down to 6, 12, or 18.

You also know that half the coins, minus two, is one third of the coins.

Half of 6 = 3. One third of 6 = 2.
Half of 12 = 6. One third of 12 = 4.
Half of 18 = 9. One third of 18 = 6.

This only works if there are 12 coins.

If slightly older you can do it with algebra and equations and substituting in terms.

(First of all I would cross out fewer and write less……..) sorry.

think about what you know about the coins as she lays them out.
you know that there are less than 20 coins.
you know that they can be divided into half.
therefore the answer is an even number below 20.
you also know that if you change two of the coins you can divide them into a third and two thirds, so the number needs to be divisible by three as well as by two.

so you need an even number below 20 that you can also divide by three .

There are three possibilities but only one works by changing two coins.

I think if I were teaching this and they were completely stuck I’d first get them to work out the total number of coins it could - 6, 12 or 18 (as the only numbers divisible by 2 and 3). I’d them get them to try these out maybe using real coins (depending on age) or drawing the the different combinations.

For primary they will be expecting a blend of logic and trial and error.

I’d get my class to use trial and error, knowing there must be an even number of coins that’s also divisible by 3.
So the possibilities are 18, 12, or 6.
1/3 or 18 is 6, 1/2 is 9 so it can’t be 18.
1/3 of 12 is 4, 1/2 is 6, 6 - 2 = 4 so it must be the right answer.

I think ‘fewer’ is correct @viques because coins are countable. We use ‘less’ for uncountable nouns like butter or water for example.

It has to be an even number which is also in the 3 times table, and is under twenty.

So it's either 18, 12, or 6.

Then I'd try each of them out with some actual coins.

Just text Melissa and ask.
And then suggest she gets a hobby. All this fiddling around with coins is dull.

It should be fewer. Not less.

Yes, sorry, my bad, brain freeze!

I did wonder because normally when people note the difference between fewer and less it’s because they know the difference. Don’t worry about it, I thought the answer the the problem was 6.

2 coins is the difference between 1/2 and 1/3 so a year 6 could find 1/2-1/3 =1/6.

So now you know 1/6 = 2, 6/6 = 12.

Or trial and error.

Thanks.