How do you round decimal figures?

[cakedup]

So how would you round, to one one significant figure, 0.46 for example? Because we've tried 0.5, 0, and 1 and none of them seem to be right!

0.5

it depends on the format of the question. How many decimal places is it asking for?

Rounding to a whole number would be 1. To 1 decimal place would be 0.5 etc

Rounding to a whole number would be 0 surely?

I think that 0.5 would be one significant figure though. Either way I’m stumped if it’s definitely not 0 or 0.5

0.5

www.bbc.co.uk/schools/gcsebitesize/maths/number/roundestimaterev3.shtml

To one significant figure would be 0.5

With significant figures I think it needs to be the same number of digits in total as the original so 0.50 not just 0.5

It should be 0.5 as 0.50 suggests the accuracy is to 2 decimal places.

The question is:
By rounding each number to 1 significant figure, estimate the answer to:
8.2 x 6.7 divided by 0.46

So we've done it like this:

8 x 7 = 56
60 ÷ 0.5 = 120

But it's incorrect! (online instant marking)

Perhaps don't round up the in between step, from 56 to 60, I'd say that was more open to interpretation than 0.46 becoming 0.5.

surely the answer is 100?

Yes don't round the 56, it should be 56/0.5 which is 112.

Both 112 and 120 to 1 sf are 100.

Yeah I dont think 56 needs to be rounded.

6.7 might round to 6.5

I'd round each of the numbers, then calculate without further rounding.
So 8*7÷0.5=112

I would calculate it as 112 as PP's have done, does that work!?

yes it doesn't say to round the answer, only the numbers in the question, so I'd try 112 as well.

Yes, it's 112! Weird really, I would have thought you kept on going with the rounding off, it does say 'estimate' the answer and 112 doesn't sound like an estimate

Erm....so how would you round 0.18 to a significant figure? Would that be 0 then??

One Sig fig? 0.2

0.18 to 1 sig fig is 0.2.

With estimation questions in maths, the official GCSE ‘procedure’ is to round each number in the calculation to 1 sig fig and then do it in your head. You wouldn’t round interim figures, so you’d do 56/0.5 which is just doubling.

And 0.018 would go to 0.02.
0.0018 would go to 0.002

Ok, last one i promise, to get my head around this.

So 23.4 x 13.9 divided by 0.18

So wouldn't that be:
23 x 14 = 322
322 ÷ 0.2 = 1610

Apparently not! What am I getting wrong??

Erm....so how would you round 0.18 to a significant figure? Would that be 0 then??

It depends on how many significant figures.