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Mixture problem -

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by doclkk » Tue Jul 21, 2009 6:42 pm
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?
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by scoobydooby » Tue Jul 21, 2009 7:18 pm
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
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by doclkk » Tue Jul 21, 2009 8:43 pm
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
Can you explain the logic behind the equation.

Thanks
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by mehravikas » Tue Jul 21, 2009 8:55 pm
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
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by doclkk » Wed Jul 22, 2009 7:21 am
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
Thanks for this - very clear
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by sreak1089 » Fri Jul 24, 2009 2:43 am
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
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Re: Mixture problem -

by shahdevine » Tue Aug 18, 2009 12:02 pm
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?

you've got to understand the concept behind what you're doing...forget about formulas they will trip you up.

basic concept is at start rob has 20 gallons which is .05 ethanol or 1 gallon ethanol. Translate further and you have 1/20 of the gas is ethanol. The question is how much ethanol would you need to add to make the fraction 1/10 of the gas ethanol so set up:

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

you got this man!
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by ssuarezo » Wed Aug 19, 2009 8:34 pm
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 ...
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What is the flaw in my logic?

by gkumar » Fri Oct 02, 2009 12:00 am
I had the equation:
Gasahol = E + G
20 = 5/100 * 20 + 95/100 * 20
20 gal of gasahol = 1 gal of Ethanol + 19 gal of ethanol

To increase the percentage from 5% to 10% for ethanol, x gallons of ethanol will need to be added
Gasahol = E + G
20 + x = 10/100 * (E+x) + 90/100 * (G)
20 + x = 1/10 * (1+x) + 9/10 * 19
200 + 10x = (1+x) + 9*19
200 + 10x = 172 + x
9x = -28
x = -28/9

x should be POSITIVE and not negative
What is the flaw in my logic?
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by bondguy » Fri Oct 09, 2009 6:26 pm
Great thinking ...
liked the concept..
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 ...
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by everything's eventual » Wed Aug 15, 2012 6:58 pm
From the original mixture you get that Gasoline is 19 ltr and Ethanol is 1 ltr.

Since the quantity of gasoline doesn't change in the new mixture you write the following :

19 = 0.9 * X ( Where X is the new quantity of mixture)

X = 190/9

Total Ethanol in new mixture = 190/9 - 9 = 19/9

Out of this 1 ltr is already present.

So Ethanol to be added = 19/9 - 1 = 10/9.
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