How do you simplify #9.594 div 0.06#?

2 Answers
Apr 9, 2017

Answer:

A way to get round the decimal place in the initial division expression.

#159.9#

Explanation:

Note that #9.594# is the same as #9594xx1/1000#

Note that #0.06# is the same as #6xx1/100#

So #9.594-:0.006# is the same as #[9594-:6]xx[1/1000-:1/100]#

#(9594-:6)xx1/10#

I will do the division first then multiply by the #1/10# at the very end

#" "9594#
#color(magenta)(1000)xx6->ul(6000) larr" subtract"#
#" "3594#
#color(magenta)(color(white)(1)500)xx6->ul(3000) larr" subtract"#
#" "594#
#color(magenta)(color(white)(10)90)xx6->ul(color(white)(0)540)larr" subtract"#
#" "color(white)(0)54#
#color(magenta)(color(white)(100)9)xx6->" "ul(54) larr" subtract"#
#" "0#

As we have 0 we may now stop the division

#9594-:6 = color(magenta)(1599)#

Now we multiply by the #1/10# giving:

#159.9#
~~~~~~~~~~~~~~~~~~~~~~~~~~~

If the division had not finished with a 0 at that point then we would have gone into decimal values.

Aug 11, 2017

Answer:

#= 159.6#

Explanation:

The rule with dividing BY a decimal is simple ... DON'T!!

You can change any decimal into a whole number by multiplying by a power of #10#

Write the division as a fraction:

#9.594/0.06#

Multiply the numerator and the denominator by the same number

#9.594/color(blue)(0.06) color(red)(xx 100/100)" "larr [color(red)(100/100 =1)]#

This changes the denominator into a whole number:

#=959.4/color(blue)(6)" "larr# divide as usual

#6|ul(9" "^3 5" "^5 9." "^5 4)#
#color(white)(xx)1color(white)(x.x)5color(white)(xxx)9. color(white)(x.)6" "larr# details below

#= 159.6#

#9 div 6 = 1 " carry " 3#
#35 div 6 = 5 " carry " 5#
#59 div 6 = 9 " carry " 5#
#54div 6 = 9# , no remainder