scoobydooby wrote:the mixture now has 1g of ethanol and 19g gasoline. let x g of ethanol be added to the mixture to so that the ethanol % is 10% (optimum mixture)

=>(1+x)/(20+x)=10/100
=>9x=10
=>x=10/9

hence, C

mehravikas wrote:Hey @doclkk,

I hope this clears your doubt -

Total quantity = 20
5% ethanol = 1
95% mix = 19

you have to add ethanol to make it 10%. Let's say you add x gallons of ethanol. therefore

10 % of (20 + x) should be equal to = x + 1

(20 + x ) * 10 / 100 = x + 1

therefore, x = 10/9

doclkk wrote:Bob just filled his car's gas tank with 20 gallons of gasohol, a mixture consisting of 5% ethanol and 95% gasoline. If his car runs best on a mixture consisting of 10% ethanol and 90% gasoline, how many gallons of ethanol must he add into the gas tank for his car to achieve optimum performance?

9/10

1

10/9

20/19

2

[spoiler](C) [/spoiler]

Can someone explain the process of this particular mixture problem?

I wanted to try .05X + .y(1-x) = .1x

MGMAT is really big on making the charts for these but I learned on this forum to run .X + Y(1-X) = .Zx for mixture / weighted average problems but this one doesn't seem to really apply?

sreak1089 wrote:I would do it this way, ethanol and gasolene are in ratio of 1:19
New ratio, you would want is 1:9, you can form the equation as below:

e/g = 1/19
(e+x)/g = 1/9 can be expressed as
e/g + x/g = 1/9
x/19 = 1/9 - 1/19
x = 19 (10)/(19*9) = 10/9
Hence, C

ssuarezo wrote:

sreak1089 wrote:I would do it this way, ethanol and gasolene are in ratio of 1:19
New ratio, you would want is 1:9, you can form the equation as below:

e/g = 1/19
(e+x)/g = 1/9 can be expressed as
e/g + x/g = 1/9
x/19 = 1/9 - 1/19
x = 19 (10)/(19*9) = 10/9
Hence, C

mmm .. I would take as a rule of three:

if 19 should be 90%, how much is 10%

19 --- 90
x ---- 10
x= 2.1, but, one gallon is already there so, I need 1.1 gallon, that's 10/9

too easy to function in all mix problems I think ...