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Re: Bob just filled his car's gas tank with 20 gallons of
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Updated on: 02 Jan 2013, 11:18
In the original mixture we have 95% G= 19 gallons and 5% E= 1 gallon We have to add X amount of Ethanol to make it 90:10 mixture, hence: (1+x/20+x)*100=10 (1+x)*10=20+x 9x=10; x=10/9 thus the answer is C
Originally posted by nk9285 on 02 Jan 2013, 11:01.
Last edited by nk9285 on 02 Jan 2013, 11:18, edited 1 time in total.



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Re: Bob just filled his car's gas tank with 20 gallons of
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20 May 2013, 14:17
I got 1 cause i got 5% to be 1 and the reason i forgot that is because i forgot to 95% gasoline



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Re: Bob just filled his car's gas tank with 20 gallons of
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05 Aug 2013, 21:54
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 Responding to a pm: Using Weighted average formula: w1/w2 = (A2  Avg)/(Avg  A1) You need to mix 5% ethanol mixture with 100% ethanol to give a 10% ethanol mixture. w1/w2 = (100  10)/(10  5) = 18/1 For every 18 parts of gasohol, you need to put 1 part of pure ethanol. So if you have 20 gallons of gasohol, you need to put 20/18 (= 10/9) gallons of pure ethanol. Answer (C)
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Re: Bob just filled his car's gas tank with 20 gallons of
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14 Aug 2013, 04:49
VeritasPrepKarishma wrote: 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 Responding to a pm: Using Weighted average formula: w1/w2 = (A2  Avg)/(Avg  A1) You need to mix 5% ethanol mixture with 100% ethanol to give a 10% ethanol mixture. w1/w2 = (100  10)/(10  5) = 18/1 For every 18 parts of gasohol, you need to put 1 part of pure ethanol. So if you have 20 gallons of gasohol, you need to put 20/18 (= 10/9) gallons of pure ethanol. Answer (C) Hi, Is this alligation rule method? Regrds, Rrsnathan.



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Re: Bob just filled his car's gas tank with 20 gallons of
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15 Aug 2013, 03:38
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 Ethanol now = 5% of 20 = 1 gallon Let, we need x gallons of ethanol . so, (20+x)/(1+x) = 100/10 or, x = 10/9 (Answer)
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Re: Bob just filled his car's gas tank with 20 gallons of
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15 Aug 2013, 21:42
rrsnathan wrote: VeritasPrepKarishma wrote: 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 Responding to a pm: Using Weighted average formula: w1/w2 = (A2  Avg)/(Avg  A1) You need to mix 5% ethanol mixture with 100% ethanol to give a 10% ethanol mixture. w1/w2 = (100  10)/(10  5) = 18/1 For every 18 parts of gasohol, you need to put 1 part of pure ethanol. So if you have 20 gallons of gasohol, you need to put 20/18 (= 10/9) gallons of pure ethanol. Answer (C) Hi, Is this alligation rule method? Regrds, Rrsnathan. 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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Re: Bob just filled his car's gas tank with 20 gallons of
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19 Sep 2013, 03:50
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 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)



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Re: Bob just filled his car's gas tank with 20 gallons of
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19 Nov 2014, 04:17
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



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Bob just filled his car's gas tank with 20 gallons of
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01 Dec 2014, 17:08
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



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Re: Bob just filled his car's gas tank with 20 gallons of
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02 Dec 2014, 01:46
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
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Re: Bob just filled his car's gas tank with 20 gallons of
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11 Dec 2014, 06:40
VeritasPrepKarishma wrote: Emaco wrote: Guys, I made the following formula, could anyone of you please tell me what's wrong with it ? 0.95 (20) + X = 90/100 (20+x) 1+X = 18+0.9x 0.1x = 17 ??? .95 of 20 is the amount of gasoline already in his car. He adds more ethanol and not gasoline so you cannot add x (amount of ethanol added) to .95 of 20. Instead, you need to find the current amount of ethanol and add x to it. .05(20) + x = .10 (20 + x) .9x = 1 x = 10/9 I solved it using algebra, for concepts sake, is it possible to use alligation or the scale method here?



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Re: Bob just filled his car's gas tank with 20 gallons of
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14 Dec 2014, 03:15
saadis87 wrote: I solved it using algebra, for concepts sake, is it possible to use alligation or the scale method here? In this post above: bobjustfilledhiscarsgastankwith20gallonsof4579020.html#p1253824I 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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Re: Bob just filled his car's gas tank with 20 gallons of
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21 Aug 2015, 12:32
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 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
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Re: Bob just filled his car's gas tank with 20 gallons of
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21 Aug 2015, 18:49
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.



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Bob just filled his car's gas tank with 20 gallons of
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23 May 2016, 20:22
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 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 : 19/99/9 = 10/9 gallons more.



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Re: Bob just filled his car's gas tank with 20 gallons of
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24 May 2016, 05:34
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 Let x gallon ethanol is added, making the total solution= 20+x 5% of 20 + x = 10%(20+x) 1+x= 2 + x/10 xx/10= 1 x= 10/9 C is the answer
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Re: Bob just filled his car's gas tank with 20 gallons of
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23 May 2017, 20:01
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 This problem involves weighted averages, but there isn't a shortcut perse. 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.



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Re: Bob just filled his car's gas tank with 20 gallons of
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26 May 2017, 10:47
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 We can let x = the number of gallons of ethanol to be added to the gas tank. Note that since the 20gallon 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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Re: Bob just filled his car's gas tank with 20 gallons of
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21 Jul 2017, 20:05
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)



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Re: Bob just filled his car's gas tank with 20 gallons of
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21 Jul 2017, 21:07
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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