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

In this post above: https://gmatclub.com/forum/bob-just-fill ... l#p1253824
I have already shown how to use scale method here.

We can let x = the number of gallons of ethanol to be added to the gas tank.

Note that since the 20-gallon mixture of gasohol is 5% ethanol, there is 20 x 0.05 = 1 gallon of ethanol already in the tank. Since we want the gasohol to be 10% ethanol, we have the following equation:

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

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

10(1 + x) = 20 + x

10 + 10x = 20 + x

9x = 10

x = 10/9

Answer: C
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Besides some easy methods we can also do this question with the help of allegation approach:-
Initially, there was 5% ethanol and we have to add pure ethanol i.e 100% ethanol and the final concentration we need to obtain is 10%.
5%-------------------------100%
------------10%----------
90%-------------------------5% (Use allegation method)

so we have final ratio 90:5 or 18:1
We know we have 5% ethanol 20ml so the amount of 100% ethanol required assume y is:- [18/20 = 1/y] => y = 10/9 (Simple ratios)

Bob just filled his car's tank with 20 gallons of gasohol, a mixture containing of 5% ethanol and 95% gasoline.
5% of 20 gallons is 1 gallon.
So, PRESENTLY, there is 1 gallon of ethanol in the car's tank.


We want to add some PURE ethanol to the tank in order to get a 10% mixture of ethanol.
Let x = number of gallons of pure ethanol we ADD to the tank.

FACT #1: Once we add x more gallons of pure ethanol, the car's tank contains 1+x gallons of ethanol.

FACT #2: Once we add x gallons of ethanol, the car's tank contains a TOTAL of 20+x gallons of mixture.

We want the tank to have a 10% mixture of ethanol. In other words, we want the mixture in the tank to contain 1/10 ethanol.
So, we can write the equation: (1+x)/(20+x) = 1/10
Cross multiply to get: 10(1 + x) = 1(20 + x)
Expand: 10 + 10x = 20 + x
Rearrange: 9x = 10
Divide both sides by 9 to get: x = 10/9

Answer: C

Cheers,
Brent

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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Amount of gasoline stays the same. So using gasoline, Ci*Vi = Cf * Vf

95% of 20 gallons = 90% of X gallons

\(\frac{19}{18} * 20 = X\)

X = 190/9

The volume changed from 20 gallons to 190/9 gallons i.e. an addition of 190/9 - 20 = 10/9 gallons was done. So 10/9 gallons of ethanol was added.

Answer (C)

RajatGMAT777

19/(9/10)=190/9 gallons

can you explain the calculation?

Sure:

\(\frac{19}{(\frac{9}{10})}=19*\frac{10}{9}=\frac{190}{9}\).
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by punching options
20=1+19
check the options a and c are inverted so likely answer either of the two
lets check c
1+10/9 +19
19/9+19
2.1+19
21.1 so 10% of this is 2.1 so c is right