# What is the value of .03 *.50?

What is the value of 0.03 *0.50 The answer is “0.015” Once again I’ve no idea how that comes about??? =S

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If you multiply .03 and .5 you get .015 Come on these are easy!

Easy on a calculator, sure. What’s the procedure?

0.5 = a half, so a half of 0.03 is 0.015. Say if you multiplied 0.03 by 1000, you’d have 30 and a half of 30 is 15 yeah? Divide 15 by a thousand and you get 0.015.

By the way, I’m happy to help and welcome to Fluther but it is not a website where people do your homework for you. Oh, I’m glad you’re asking for the procedure though :)

0.03×0.5 can be changed to 0.03 / 2

Cheers for that, explained it perfectly.

By the way I didn’t mean to be rude or anything, I’m just warning you. You can certainly ask for help and a nudge but people here get testy if you ask them to do your homework for you. I can see you are not but are asking to be shown how to do it yourself so you can continue, so I apologise :)

The procedure is to start by multiplying as if the decimal point were not there. 0.03 * 0.50 => 3 * 50 = 150.

Then you add the number of decimal places in both terms and put that number of places in the result. In this case, they both have 2 decimal places, so you need 4 decimal places in the result. 150 => 0.0150.

Didn’t you take sixth grade math?

It’s absolutely worthless.

LoLz no actually I didn’t. =S

0.03 * 0.50

=(3/100)*(5/10) ‘Get tens (or powers of tens) in denominators

=(3*5)/(100*10) ‘Separate numerator and denominator

=15/1000 ‘Proceed

=0.015 ‘Answer

There is a general method for determining where the decimal point goes. This problem can be used as an example.

..03 has the decimal point moved two places to the left from 3, which means that .03 = 3/(10*10). Similarly .50 has the decimal point moved two places to the left from 50 so .50 = 50/(10*10).

Then .03 * .50 = 3/(10*10) * 50/(10*10) = 3*50/(10*10*10*10) = 150/(10*10*10*10), which is 150 with the decimal point moved 4 times to the left, which is .0150, which is the same as .015.

In general, to multiply two numbers with decimal points, first multiply the numbers as if they did not have decimal points. Then add the number of times the decimal point had to be moved for each of the numbers and that tells how many places to move the decimal point in the answer.

Yeah, as you said.

If you have difficulty for counting decimals, then first remove right-most zeros; e.g. 0.50 would then be 0.5.

Then, count the numbers to the right of decimal point. And add those many zeros in the tens in the denominator (0.5—>one count->10 in denominator, 0.03—>two count->100 in denominator, and so on).

Multiply all the terms in the numerator and denominator. Keep in mind that for multiplication of tens, just add number (count) of zeros and write 1 before (10*100—>one and two zeros, so total three zeros->1000).

Now, as you arrive with 15 divided by 1000, again count number of zeros in the denominator (three zeros here). The decimal point in the division would come those many digits left (start from right-most digit) from the number in the numerator. Add zero if the count is more than the number of digits in the numerator (as it is in this case, 15—>count is two and there are three zeros in denominator, start from right of 15, take first 5 (of 15), then 1(of 15), add one zero, and place the decimal point. So, the result is 0.015).

This technique works for as many numbers you want to multiply, two, three, or more.

Keep practising. All the best!

I guess that since the equation states 0.03*0.50, the decimal accuracy is 2 digits.

Otherwise it could have stated 0.5 (neglecting accuracy) or 0.500 (more accuracy).

Accuracy of the result cannot exceed the accuracy of any of the factors so although the answers above are correct, it should actually be taken into account that there is a difference in multiplying by 0.50 versus dividing by two or multiplying by 0.5 .

Taking accuracy into account the answer to this calculation should be 0.02

@whitenoise: That’s true in science and engineering, but not in mathematics; and in science and engineering, significant digits count for more than decimal places.

@whitenoise “Accuracy of the result cannot exceed the accuracy of any of the factors…”

This may not always be correct, e.g. 0.3 * 0.4 = 0.12.

However, your answer is right with rounding it to two decimal places.

Well, in our math classes, we would loose points if we would not answer

0.30*0.4 = ? with: 0.30*0.4=0.12 -> 0.30*0.4=01.

We were instructed to keep accuracy in mind if we were confronted with formats with extra trailing zeros that indicate accuracy, or when the instruction explicitly asked for keeping accuracy in mind.

Maybe we had an overzealous teacher. It however stuck with me that 0.5, 0.50 and ½ are all of different accuracy.

@whitenoise: That’s an engineering approach to arithmetic, which isn’t **wrong**, but it’s not **universal**.

@cwilbur May also be that your math teachers teach math a little different from my European.

In our part of Europe, our engineers’ math is governed by the same rules of mathematics as that of the rest of us.

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