HI. My name is cj

how do you divide 0.21 by 0.33

it is very challenging. thank you

HI. My name is cj

how do you divide 0.21 by 0.33

it is very challenging. thank you

Tutors, please sign in to answer this question.

I don't see any reason to do this in decimal form. Like many fraction problems, it's done most easily as a common fraction.

Begin by putting the .21 over the .33. Now you have .21/.33. Get rid of the decimals by multiplying by 100/100. that gives you 21/33. 21 and 33 are both multiples of 3, so reduce by dividing top and bottom by 3. This gives you 7/11, which is the answer. No need for decimals repeating to infinity.

Let's move divisor's decimal point to the right until we will get the whole number and we
* must* move the dividend's decimal point same places to the right.

0. 33

First, in the quotient we will write "0" with the decimal point (33 goes into 21 zero times)

Second, let's write 2-3 zeros (it does not matter) .

Now, we are ready to do division:

0. 33

–

198

120

–

99

210

–

198

12 ......

0.21 ÷ 0.33 =

CJ,

First move your decimal to places to the right. Then, set up the problem just like any other long division problem. Then you simply keep adding zeros to the end of the number your are diving into until you have solved the problem to the required decimal place. So, for your problem, you are asking the question, "How many times does 33 go into 21."

33 goes into 21.0 0.6 times with a remainer of 0.12

33 goes into 0.120 0.03 times with a remainder of 0.021

From here you should realize that the answer is 0.6363 repeating.

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## Comments

Well, Mr. David U, I don't see any reason to do all those procedures (above and under of this comment) either, but if you are 6th or 7th grade student, all those steps are a requirement.

Is cj a 6th or 7th grader? I see nothing in the original question to indicate this. And while the material may be considered 6th or 7th grade level (or even younger), I've encountered more than a few college students who find such problems challenging. So I wouldn't want to make any assumptions of that kind.

Furthermore, in my experience students often misunderstand exactly what is required and what isn't. In this case, for example, even if the answer is required to be in decimal format, I think it's far easier to do as I've suggested, and then convert the 7/11 into decimal format by dividing 11 into 7.0000...

Of course there are many different ways to do almost any problem, and individual preferences as to which way is best can vary.