Question #97f62

Question

1323 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-04T15:56:27

For 1 gallon you have #1/2# of the something else

Assumption: the given ratio is of 2 gallons : 1 of something else
and that you require to determine how much there is 'something else for 1 gallon.

Let 'something else be represent by #s#

Express initial condition ratio in fractional form #(2g)/(1s)#

Let the unknown count or volume of #s# be #x#

#(2g)/(1s)-=(1g)/(xs)#

Turn the whole thing upside down

#(1s)/(2g)-=(xs)/(1g)#

Multiply both sides by #1g larr # gets the #xs# on its own

#color(brown)((1color(green)(s))/(2color(green)(g))xx1color(green)(g)-=xcolor(green)(s))#

Separating numbers from units of count/measurement:

#color(brown)(1/2 xx 1" ")color(green)(s/(cancel(g)) xx cancel(g))#

Numbers left #1/2" units left"-> s#

So for 1 gallon you have #1/2# of the something else