My answer is incorrect as it’s based on 15% of the reduced price of £391. I know the answer but can’t figure out how to work to it, can anyone lead me there?
I would divide 391 by 0.85 to get the original amount. If there was a 15% discount on the telly, 391 is 85% of what amount?
4 Likes
Yes 391/0.85 (85%) is 460 Which is the correct answer.
It’s very obvious now you’ve pointed that out. I’ve been doing these all morning really well so far.
I think it’s either the way it’s worded, more likely I need a break from this maths course, I’ve been doing loads over the last few days.
Thanks Dave
391 / 85 * 100 would be my formula
1 Like
Take a break from it Ian. Maybe try diagramming sentences, instead. 
2 Likes
Too much coffee?
I’m doing an apprenticeship at work. The course requires Maths and English to proceed through the main parts. And guess what, I don’t have my exam results anymore (which I passed) so I’m doing them again now. My English is pretty strong, but some of the maths stuff is really catching me out.
I am enjoying getting through it though, it’s a good refresher.
2 Likes
2 Likes
Since math is an universal language Discourse should understand your sentences?
1 Like
1 Like
391+15=406
15=9+6
9x6=54
406+54=460
1 Like
But how do you choose those two numbers? 9+6
Mike, Abbott and Costello would be SOOOOO proud.
1 Like
Where’s @TheOnlyAaron on this one? Just kidding Aaron.
Look at crossed out price on tag. Done.
Actually, the correct answer is £391. The store just told you they’d reduced the price by 15% to make you buy it…
2 Likes
To find the original price, let the original price be X. The reduced price is X - 15%. Set X - 15%X = £391 and solve for X to find the original price, usually people who study maths have high IQ.
Did it take you four years to work that out?
2 Likes
Ive been out of math a long time but as I recall:
If the price has been reduced to 391 and it was reduced by 15% then to get the normal original value you simply need to multiply
3.91x1.15 (or the sale number plus 15%) = 449.65 as it states.