Yes, alligation uses a diagram (with a cross with average in the middle) which leads to this formula. This formula is very effective in solving most weighted average/mixture problems.
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Let the total gallons = 100.
therefore, it has 5 gallon ethanol and 95 gallon gasoline and we need to make the mixture of 10:90 instead.
Let the volume of ethanol added be 'x' in order to make the ratio 10:90.
therefore, (5+x)/95 = 10/90.
x = 50/9.

Now, if in 100 gallons of fuel we need to add, 50/9 gallons of ethanol to make the ratio 10:90
In 20 gallons we would add = (50/9)/100 X 20 (simple unitary method)
= 10/9 (Ans)

Let x is the gallons of ethanol to be added in the mixture => x has 100% ethanol.
Totally ignore the information about gasoline because we're dealing with ethanol now.

The new mixture should be 20 + x (gallons) and have 10% ethanol => (20+x)*10%
Add x gallons of 100% ethanol to 20 gallons of 5% ethanol => x*100% + 20*5% = x+1

We have the equation: (20 + x)*0,1 = x + 1 => x = 10/9

initial 20 = 1E + 19G

new 20+E = (20+E)/10 +19 (as amount of G same only E added)

=> 18 +9E/10 = 19

HENCE E = 10/9

Ethanol ............... Gasoline .............. Total

1 ............................. 19 ..................... 20 (Current fill)

1+x ........................... 19 ................... 20+x (Say "x" litres of ethanol is added to make it 10%)

Equation can be setup in two ways

\(\frac{10}{100} (20+x) = 1+x\)

OR

\(\frac{90}{100} (20+x) = 19\)

For both, computation of x would be the same

\(x = \frac{10}{9}\)

Answer = C

I solved it using algebra, for concepts sake, is it possible to use alligation or the scale method here?

In this post above: bob-just-filled-his-car-s-gas-tank-with-20-gallons-of-45790-20.html#p1253824
I have already shown how to use scale method here.
For more on the formula, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
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Using the cross method works out the best way to solve mixtures problems. But in this problem the final conversion from 18 parts to the asked 20 parts was something that tricked me. The silliness struck me after i solved it algebraically.

\(Soln:\)
Total amount: 20
Ethanol = \(\frac{5}{100}*20=1gl\)
Gasoline = \(\frac{95}{100}*20=19gl\)

Let \(x\) be the amount of ethanol that will be added to get 10% of ethanol in the final solution.

=> (1 + \(x\)) = \(\frac{10}{100}\)\(*(20+x)\)
=> 100 + 100\(x\) = 200 + 10\(x\)
=> 90\(x\) = 100
=> \(x\) =\(\frac{10}{9}\)

Therefore, \(\frac{10}{9}\) would be the amount of ethanol that needs to be added to \(20gl\) of fuel.

Regards,
Pratik

I solved it a different way. It took me about 3 mins so I think I might be adding an extra step but this is what worked for me.

I started by solving how many gallons of ethanol and gas are in the tank already. .05 * 20=1 (gal of e); .95 * 20=19 (gal of g)

The current ratio of e:g is 1/19. The ideal ratio is 10/90 or 1/9. This question asks how much more ethanol we need to get to the 1/9 ratio.

Thus, e/19 = 1/9. --> 9e=19. e=19/9 total. As there is already 1 gal of e in there, 19/9 - (9/9)= 10/9 more gal of e.

how about this:

E:G
5%:95% >> 1:19

19 is 90% of 190/9. 10% of 190/9 is 19/9.
so we need : 19/9-9/9 = 10/9 gallons more.

Let x gallon ethanol is added, making the total solution= 20+x

5% of 20 + x = 10%(20+x)
1+x= 2 + x/10
x-x/10= 1
x= 10/9

C is the answer
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This problem involves weighted averages, but there isn't a shortcut per-se. Instead, you should take the the current amount (1 gallon) plus x, the incremental amount of gas, and set this quantity equal to (x+20)/10, a tenth of the new total once the incremental gas has been added. This righthand quantity is the ideal amount.

After this point, it's easy algebra to determine that x = 10/9.

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)

an easy way to do this question
remember question says that one has to add ethanol to make it 10% and original quantity of ethanol is 1 liter out of 20 (5% of 20). so it mean 19 liters of gasoline is fixed and it will be 90% of the new quantity.
Simply make an equation saying: 19 = 0.9 of x (this x is the total new quantity)
19/0.9 = x and that is equal to 21.1
New quantity - Old quantity = 21.1 - 20 = 1.1 and that is equal to 10/9
C it is...one has to add 1.1 (10/9) of ethanol to the old mixture
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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
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5% of 20= 1g of ethanol


1+x = 10%(20+x), simplifying we get 10+10x=20+x => x = 10/9.

Hence answer C.

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+1 for kudos please

Current,
Ethanol= 20*.05=1
Gasoline= 20*.95=19
Let, x gallons of Ethanol to be mixed.
Therefore,
(1+x)/19=10/90
(1+x/19=1/9
x=10/9

Let gasohol = G & Ethanol = E

Using Alligation method

G...........M......................E
5...5 .....10.......90.........100

G/E = 90/5= 18/1 ....then


20/E = 18/1......E = 20/8 =10/9

Answer: C