Division Topic Challenge

Ok so the way I was taught at school to do division was using BODMAS and splitting each equation into blocks of problems so with this challenge my answer would be something along the lines of (Just as a side note I have no higher level math’s training, Only what I Was taught at school 20+ years ago):

9 - 3 / 0.33 + 1 = x
9 - 3 = 6
6 / 0.33 = 2
2 + 1 = 3
x = 3

Please feel free to explain if I’m wrong or you wish to make addendums to my calculation as I’ve seen some wildly different answers here.

Hi @UltraFRS, here’s how I would approach this calculation.

Starting with 9 - 3 ÷ 1/3 + 1 and following BODMAS…

Deal with the division and multiplication first.
We’re dividing by a fraction here, so I find it easier to treat it as a multiplication by the reciprocal of the fraction to give us:

9 - (3 ÷ 1/3) + 1
9 - (3 * 3) + 1
9 - 9 + 1

Now we can finish up with the addition and subtraction:
9 - 9 + 1 = 1

I hope that helps.

1 Like

Hello Gary,

Thanks for your calculation and showing how you wanted the answer to be worked out. Maybe my understanding of BODMAS is completely wrong (Like I said I haven’t done anything involving numbers on a regular basis for some 20 years and my brain is slightly more addled than my teenage brain from back then haha), But I’m confused where the multiplication comes from specifically the 9 - (3 * 3) + 1, shouldn’t you multiply by 0.33 seem’s the equation asked for a fraction of a whole number assuming that whole number is 1, 1/3 should be in theory 0.33 although is just an assumption.

Many Thanks

P.S Maybe I should go over that video again

This is a little math trick that we can use when dividing by fractions.
What we’re doing is converted the division by a fraction to a multiplication by the reciprocal of the fraction.

That’s a lot of math words but you can really just think of the reciprocal as a fraction being flipped on its head - So the reciprocal of 1/3 = 3/1
For reference, the defining trait of the reciprocal is that any number multiplied by its reciprocal always 1.

If we look at it in the context of how I’ve used it here, we’re saying that 3 ÷ (1/3) = 3 * (3/1).
I’d encourage you to run both of these statements through a calculator to confirm that they both equal 9 (and maybe try some other examples as well, like 4 ÷ 3/5 = 4 * 5/3).

As for BODMAS, I prefer to think of it as BO(DM)(AS), since the order of multiplication/division and addition/subtraction are always solved left to right. You also have many tricks that allow you to treat division like multiplication and subtraction like addition, so they’re not really separate things.

To explain how your calculation differed from mine, I solved it as 9 - (3 ÷ (1/3)) + 1 whereas you solved it as ((9 - 3) ÷ (1/3)) + 1.
So, you’ve done a subtraction first, then a division, then an addition.

My best advice for tackling calculations like this is to try and add in some extra parentheses to help make things as clear as possible.
Thankfully, most mathematicians will do this for you when initially writing the equation, since they want you to actually be able to solve it without adding unnecessary ambiguity to the solution (math is hard enough as it is without making it worse!)

Privacy & Terms