Maths question.

xfabgalx

anyone know how to solve 2004 question 3 c) ii) ?
In the marking schemes, they only give the answer.
http://examinations.ie/archive/exampapers/2004/JC003ALP1EV.pdf

xfabgalx

That is big
Thank you so much !

xfabgalx

I have another probably easy question::
3000-2568 = 432
for some reason I get 532.

3000
-2568

what I do:
^23^10^90^10
2568
0532

Indiego

xfabgalx wrote: »

I have another probably easy question::
3000-2568 = 432
for some reason I get 532.

3000
-2568

what I do:
^23^10^90^10
2568
0532

That should be 9 because you 'borrowed' a one to make the next digit 10, but a one was borrowed from that afterwards so its a 9 too, and the last digit is 10 because no numbers 'borrowed' from it

so it goes:
3000
32^10^0^0
32^109^10^0
32^109^109^10
= 299(10)

spurious

xfabgalx wrote: »

I have another probably easy question::
3000-2568 = 432
for some reason I get 532.

3000
-2568

what I do:
^23^10^90^10
2568
0532

Without knowing exactly what method you are using, when you are borrowing from the 3, it becomes 2 (great), but it looks like you have left that 1 on the second column, without moving it over to the next one.
So, you end up taking the 5 from 10, instead of from 9 which it should be.
Does that make sense to you?

**Edit** what Indiego said.

xfabgalx

Thanks so much guys!
My dad showed me the old way of doing it and then I remembered I used to do the old way when I was in primary school! That's why I didn't cross off the 0 and make it a 9 because I didn't know that method haha.
I guess that's what 3 years of calculators do to your brain :L