How do you express #0.0001/0.04020# as a decimal?

4 Answers
Feb 11, 2016

Answer:

#1/402#

Explanation:

Take #0.0001/0.04020# and multiply top and bottom by 10000.
#{0.0001 xx 10000}/{0.04020 xx 10000}.#
Use the "move the decimal" rule. ie. #3.345 xx 100=334.5# to get:
#1/402.# This is the answer in fraction form.

If the goal was to covert the decimal directly to fractions and then solve, in #0.0001#, the #1# is in the ten thousandth column, making it the fraction #1/10000# and the 2 in 0.0402 is also in the ten thousandth column so #0.0402=402/10000#.

#0.0001/0.04020= {1/10000}/{402/10000} =1/10000-:402/10000 #
#=1/10000 xx 10000/402 =1/402#.

Feb 11, 2016

Answer:

Multiply numerator and denominator by #10^4# to get #1/402#, then long divide to get:

#1/402 = 0.0bar(0)2487562189054726368159203980099bar(5)#

Explanation:

To calculate #0.0001 / 0.04020# first multiply both numerator and denominator by #10^4# to get #1/402#

Assuming we want a decimal expansion of the quotient, let's use long division.

First write out the multiples of #402# we will use:

#0:color(white)(XX000)0#
#1:color(white)(XX0)402#
#2:color(white)(XX0)804#
#3:color(white)(XX)1206#
#4:color(white)(XX)1608#
#5:color(white)(XX)2010#
#6:color(white)(XX)2412#
#7:color(white)(XX)2814#
#8:color(white)(XX)3216#
#9:color(white)(XX)3618#

Then our long division starts:
enter image source here

Write the dividend #1.000# under the bar and the divisor #402# to the left. Since #402# is somewhat less than #1#, there are several zeros for the quotient before it 'gets going'. Once we have brought down 3 #0#'s from the dividend our initial running remainder is #1000# and the first non-zero digit of the quotient is #color(blue)(2)# resulting in #2 xx 402 = 804# to be subtracted from the remainder to yield the next remainder.

Bring down another #0# from the dividend alongside the remainder #196# to give #1960# and choose the next digit #color(blue)(4)# for the quotient, etc.

Notice that with the running remainder having arrived at #10# we are essentially back to dividing #1/402# again - that is we have found the recurring decimal expansion:

#1/402 = 0.0bar(0)2487562189054726368159203980099bar(5)#

Feb 11, 2016

Answer:

I want to capitalize on George C. answer and give my version of #1/402#!!!

Explanation:

have a look:
enter image source here

Feb 11, 2016

Answer:

Just for fun I thought I would add a solution as well. I am going to considerably limit the number of decimal places!!

#color(blue)( 0.0001/(0.04020)" "~=" "0.00024)#

Explanation:

Given:#" " 0.0001/(0.04020)#

#color(purple)("Making them into more mentally manageable numbers")##color(purple)("and apply a correction at the end!")#

Multiply the numerator by #10^7# giving: 1000 so the correction is#xx10^(-7)#

so # 0.0001/(0.04020)" "=" "1000/0.0402xx10^(-7)#

Multiply the denominator by #10^4# in the form of

#1/0.0402xx1/10^4 ->1/402# so the correction for this bit is #xx10^4#

Putting this all together gives:

#1000/402 xx (10^(4-7))" "=" "1000/402color(green)(xx10^(-3))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Step 1")#
#" "underline(" ")#
Write as:#" " 402|1000#

Consider just the hundreds: #10-:4=2+"Remainder"#
Do not worry about the remainder!

#" "underline(" 2 ")#
Now write:#" " 402|1000#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#

#2xx402=color(brown)(804)#

#" "underline(" 2 ")#
Now write:#" " 402|1000#
#" "color(brown)(underline(804 -))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

subtract the 804 from the 1000
#" "underline(" 2 ")#
#" " 402|1000#
#" "color(brown)(underline(804 -))#
#" "196#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 5")#

402 > 196 so put a decimal place to the right of the 2 and put a
#color(red)(0)# to the right of 196

#" "underline(" 2"color(red)(.)" " )#
#" " 402|1000#
#" "underline(804 -)#
#" "196color(red)(0)#

#402xx5=2010 >1960# so too big
#402xx4=color(magenta)(1608)<1960# so we pick this one

so #1960-:402=color(green)(4) +"Remainder"#

So now we write:

#" "underline(" "2"."color(green)(4)" " )#
#" " 402|1000#
#" "underline(804 -)#
#" "1960#
#" "underline(color(magenta)(1608-))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 6")#

#" "underline(" "2"."color(green)(4)" " )#
#" " 402|1000#
#" "underline(804 -)#
#" "1960#
#" "underline(1608-)#
#" "352#

352 < 402 so put #color(red)(0)# to the right of 352 and we repeat step 5. This cycle go on for ever if the number is irrational!
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So far we have 2.4. Applying the correction this becomes:

#2.4 color(green)(xx10^(-3))" "->" "2.4/1000" "=" "0.00024#

# 0.0001/(0.04020)" "~=" "0.00024#

Look at the beginning to see where #color(green)(xx10^(-3))# comes from.