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2 Subtle mixture problems

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2 Subtle mixture problems

by bblast » Sat Jan 08, 2011 1:38 am
My concern is why is my logic flawed while solving the second problem below.

1>Source MGMAT cat :
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?

initial ratio of ethanol to gasoline = 1/19

required ratio = 10/90

therefore

(1+x)/19 = 10/90

x = 10/9 answer is correct.

2>Source -https://www.algebra.com/algebra/homework ... 89319.html

A spice mixture is 25% thyme. How many grams of thyme must be added to 12g of the mixture to increase the thyme content to 40%?

initial ratio of thyme to X = 1/3

required ratio = 2/3

therefore

(1+x)/3 = 2/3

x = 1.

But answer is X= 3 and logically 3 should be correct.

can anyone highlight my mistake in the second problem?

I usually take this approach to solve mixture problems.
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by Anurag@Gurome » Sat Jan 08, 2011 2:04 am
bblast wrote:A spice mixture is 25% thyme. How many grams of thyme must be added to 12g of the mixture to increase the thyme content to 40%?
Amount of thyme on 12g of the mixture = 25% of 12 g = 3 g

Say, x g of thyme must be added.
Then, amount of thyme in new mixture = (3 + x) g
and, total amount of new mixture = (12 + x) g

According to the question,
  • .... (3 + x)/(12 + x) = 40/100 = 0.4
    => (3 + x) = (0.4)*(12 + x)
    => (3 + x) = (4.8 + 0.4x)
    => 0.6x = 1.8
    => x = 3
Hence 3 g of thyme must be added to increase the thyme content to 40%.
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by Anurag@Gurome » Sat Jan 08, 2011 2:40 am
bblast wrote:A spice mixture is 25% thyme. How many grams of thyme must be added to 12g of the mixture to increase the thyme content to 40%?

initial ratio of thyme to X = 1/3
required ratio = 2/3
therefore
(1+x)/3 = 2/3
x = 1.
But answer is X= 3 and logically 3 should be correct
You cannot write that.
In the first problem you got lucky as the numerical values of the amount of ethanol (1 L) and gasoline (19 L) in the given amount of mixture (20 L) was same as the numerical values of the ratio (1:19). But in this case it is not so. Here the numerical values of the amounts (3 and 9) is different from that of the ratio (1 and 3).

Thus yous should write, (3 + x)/(9) = 2/3
=> (3 + x) = 6
=> x = 3
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by bblast » Sat Jan 08, 2011 2:49 am
Exactly, thanks a ton Anurag.

I suspected that I had got lucky in the first problem reading your last post. But this post confirms my misconception.

Thanks a lot. Wont err anymore in mixture problems now :)


In fact there was another error in my concept which I uncovered through this problem :


If something is added to a ratio between 2 things the new ratio cannot be determined until the exact amount of the total mixture is known.

There is an official PS(179) question based on this concept from OG 12 I remember.
Last edited by bblast on Sun Jan 09, 2011 1:46 am, edited 1 time in total.
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by gmat7202011 » Sat Jan 08, 2011 1:15 pm
Thanks Anurag, Nice problems Pushkhar,

However, i doubt the first solution too.

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?


The initial value of ethanol is 1 in 20 gallons of gashol i.e 1/20

now we need to add x gallons of ethanol to the mixture to get the optimum ratio of 1/9

so, (1+x)/(20+x) = 1/9

solving gives 9+9x = 20+x

x = 11/8

Isnt this the correct answer ?

I doubt if 10/9 is correct.
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by GMATGuruNY » Sat Jan 08, 2011 3:15 pm
bblast wrote: A spice mixture is 25% thyme. How many grams of thyme must be added to 12g of the mixture to increase the thyme content to 40%?
This sort of problem is easily solved by plugging in the answer choices, one of which would say that 3 grams of thyme must be added.

Answer choice: 3 grams of thyme must be added
Total thyme = original amount + 3 = 3+3 = 6.
Total mixture = original amount + 3 = 12+3 = 15.
Thyme/Mixture = 6/15 = 40%.
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by gmat7202011 » Sat Jan 08, 2011 6:23 pm
Any thoughts on the problem 1 ?

Thank You
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by GMATGuruNY » Sat Jan 08, 2011 7:07 pm
gmat7202011 wrote:Any thoughts on the problem 1 ?

Thank You
We can plug in the answer choices for problem 1 as well. Here is the problem along with answer choices:
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
Answer choice C: 10/9 ethanol must be added
New amount ethanol = old amount + 10/9 = 1 + 10/9 = 19/9
New amount of gasohol = old amount + 10/9 = 20 + 10/9 = 190/9
Percentage ethanol = (19/9)/(190/9) * 100 = (19/9)*(9/190)*100 = 10.

The correct answer is C.
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by gmat7202011 » Sat Jan 08, 2011 7:42 pm
Oh, realised my mistake... I was looking at the optimum ratio as 1/9 where it actually is 1/10

10/9 is correct.

Thank You so much !!!
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by Target2009 » Sat Jan 08, 2011 7:49 pm
My Approach :

1. Find Tyme in 12g mix = 3g (as 25% of mix is tyme) & other substance = 9g
2. lets say we added x gram tyme to 12g mix . so equation will be

New tyme = 40% of Mix
(3+x) = .40 of 12X - Solve it we will get X = 3
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by nadib002 » Wed Apr 27, 2011 9:39 pm
@bblast

How did you get the ratio as 2/3 for the thyme problem.

Could you please explain

Thanks
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by Ryandmitri » Wed Apr 27, 2011 11:51 pm
For Problem 1 I had the following approach :

20 gallons has 5% ethanol and 95% gasoline i.e. 1 gallon ethanol and 19 gallons gasoline.
Now since only ethanol is added the amount of gasiline remains same and this amount has to be 90% of the new mixture.

Therefore, 90 - 100
19 - x

Therefore x = 190/9 = new mixture

Ethanol is 10% of this mixture = 19/9.

1 gallon is already present. There additional to be added = 1-(19/9) = 10/9



For problem 2,

12g mixture has 25% thyme i.e. 3 g thyme. We need to thyme so that it becomes 40% of the new mixture.Difference between 25% and 40% isnt much so I just kept on adding one till i got 40%

3 - 12 ( 25%)
4 - 13 ( not 40%)
5 - 14 ( not 40%)
6 - 15 ( 40%)

Bingo !!! :p

Thyme to be added = 6-3 = 3g
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