If x is the decimal 8.1d5...

[SuzanR808]

If x is the decimal 8.1d5, with d as an unknown digit, and x rounded to the nearest tenth is equal to 8.1, which digits could not be the value of d?

[note: no multiple choice options; it's a general question for practicing rounding to the nearest place value]

The book states that d cannot be 5, 6, 7, 8, or 9 -- but that d can be all lesser numbers including 4.

I would think that d cannot be 4 either, and am struggling with this answer because...

My thinking is that rounding begins from the far most right digit - in this case, the "5" in the thousandths place.

Given that x = 8.1d5 —> if d=4 and the resulting number were 8.145, then wouldn’t that number rounded become 8.2?

Since the 5 in the thousandths place would round the hundredths digit from 4 to 5, and then that 5 in the hundredths place would round up the tenths place digit to “2”, resulting in a final number of x=8.2

[RonPurewal]

8.145 rounds down to 8.1... for the same reason 645 rounds down to 600, and not up to 700.

for some reason, you're doing that rounding in two different steps: first, you're rounding to the hundredths place, and THEN you're rounding to the tenths place.
that's not how it works. it's one step -- you round directly to the tenths place.

what you're doing is like first rounding 645 up to 650, and then rounding 650 up to 700. you understand why that's not "rounding 645 to the nearest hundred", right?

[RonPurewal]

by the way -- this is the wrong folder for this question. please be sure to post questions in the correct folder from now on (including any further questions about this one, if you have any). thanks.