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# Bob just filled his car's gas tank with 20 gallons of gasohol, a mixtu

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Re: Bob just filled his car's gas tank with 20 gallons of gasohol, a mixtu  [#permalink]

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14 Jan 2019, 10:20
above720 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?

A. 9/10
B. 1
C. 10/9
D. 20/19
E. 2

$$\frac{Ethnol}{Gasoline} = \frac{1}{19}$$

Or,$$\frac{Ethnol}{Gashol}= \frac{1}{20}$$

Or,$$\frac{1 + x}{20 + x} = \frac{10}{100}$$

Or, $$10 + 10x = 20 + x$$

Or, $$9x = 10$$

So, $$x = \frac{10}{9}$$ , thus Answer must be (C) $$\frac{10}{9}$$
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Re: Bob just filled his car's gas tank with 20 gallons of gasohol, a mixtu  [#permalink]

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22 Jan 2019, 15:50
above720 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?

A. 9/10
B. 1
C. 10/9
D. 20/19
E. 2

My reasoning if it helps anyone:

We have one mixture that is 5% ethanol and another that is 100% ethanol. We want to mix those together to get a mixture that is 10% ethanol.

As such we need to mix them in a ratio of 18 parts gasohol to 1 part pure ethanol.

From here we can solve for the amount of ethanol that must be added:

$$\frac{18}{1} = \frac{20}{x}$$

$$x= \frac{10}{9}$$
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Re: Bob just filled his car's gas tank with 20 gallons of gasohol, a mixtu  [#permalink]

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29 May 2019, 04:20
Easy way to solve this (for me) was:

Given Mixture

Ethanol Gasoline Total quantity (Gallons)
1 19 20 (5% of 20 =1 and 95% of 20 = 19)

Needed some quantity to make Optimum Gasoline (10% of E and 90% Gasoline) = X

Hence,

Ethanol Gasoline Total quantity (Gallons)
1+X 19 20+X

= 10% of (20+X) = (1+X)
= 10/100*(20+X) = (1+X)
= (20+x) = 10*(1+X)
= 20+X = 10+10X
=20-10=10X-X
=10 =9X
X =10/9

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Re: Bob just filled his car's gas tank with 20 gallons of gasohol, a mixtu  [#permalink]

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28 Jul 2019, 20:55
Bunuel wrote:
gmatpapa 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?
a) 9/10
b) 1
c) 10/9
d) 20/19
e) 2

Is it that we dont't need to have 20 litres of gasohol? Because if we do, I think the answer will be B.

Currently there is 5% of ethanol out of 20 litres of fuel that means the mixture contains 1 litre of ethanol. To have 10% ethanol, Bob needs to have 2 litres of ethanol, therefore he needs to add 1 more litre of ethanol.

19 gallons of gasoline which is now in the tank must comprise 90% of the whole fuel after adding some amount of ethanol --> thus whole amount of fuel after adding must be 19/(9/10)=190/9 gallons --> amount of the ethanol which Bob must add is 190/9-20=10/9 gallons.

We already have 19L of gasoline and the car runs best at 10% ethanol. So we need 10% of 19 ie 1.9L of ethanol. And since we already have 1L of ethanol, we need an additional of 1.9 - 1 = 0.9L of ethanol. Hence A. Can you please tell me where am I going wrong?
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Re: Bob just filled his car's gas tank with 20 gallons of gasohol, a mixtu  [#permalink]

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28 Jul 2019, 21:09
1
1
Rajeet123 wrote:
Bunuel wrote:
gmatpapa 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?
a) 9/10
b) 1
c) 10/9
d) 20/19
e) 2

Is it that we dont't need to have 20 litres of gasohol? Because if we do, I think the answer will be B.

Currently there is 5% of ethanol out of 20 litres of fuel that means the mixture contains 1 litre of ethanol. To have 10% ethanol, Bob needs to have 2 litres of ethanol, therefore he needs to add 1 more litre of ethanol.

19 gallons of gasoline which is now in the tank must comprise 90% of the whole fuel after adding some amount of ethanol --> thus whole amount of fuel after adding must be 19/(9/10)=190/9 gallons --> amount of the ethanol which Bob must add is 190/9-20=10/9 gallons.

We already have 19L of gasoline and the car runs best at 10% ethanol. So we need 10% of 19 ie 1.9L of ethanol. And since we already have 1L of ethanol, we need an additional of 1.9 - 1 = 0.9L of ethanol. Hence A. Can you please tell me where am I going wrong?

The mixture should consist of 10% ethanol and 90% gasoline. So, ethanol should be 10% of the whole mixture NOT 10% of the amount of gasoline as you did.
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Re: Bob just filled his car's gas tank with 20 gallons of gasohol, a mixtu  [#permalink]

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29 Jan 2020, 01:43
V1 = 20 gallons; E = 5% = 1 gallon; G = 95% => E/G = 5/95= 1/19;
V2 = ? gallons; E = 10%; G = 90% => E/G = 1/9;

1st approach:
Ethanol right now is 1/20 of the whole mix and we want to become 1/10 so:
(1+X)/(20+X) = 1/10 => X = 10/9

2nd approach:
20/V2 = (100 - 10)/ (10-5) = 90/5 => V2 = 20*5/90 = 10/9
Re: Bob just filled his car's gas tank with 20 gallons of gasohol, a mixtu   [#permalink] 29 Jan 2020, 01:43

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