How do you evaluate #1.2g# if g=77?

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Answer 1 · 2016-12-26T12:34:06

A slightly different approach and not using a calculator
92.4

If it is the decimal that is giving you trouble try this approach.

Let get rid of the decimal for the first part of the solution then put it back afterwards.

1.2 is the same as #12xx1/10#

So instead of #1.2xxg# we have #12xxgxx1/10#

We are told that #g# is worth 77 so this becomes

#color(green)(12xxcolor(red)(g)xx1/10" "->" "12xxcolor(red)(77)xx1/10#

12 can be split into 10+2

#" "color(green)(10xx77=770)#
#" "color(green)(color(white)(1)2xx77=ul(154) larr" add")#
#" "color(green)(924)#

So now we have:

#color(green)([12xxcolor(red)(77)]xx1/10" "->" "[924]xx1/10#

#" "color(green)(92.4)#