Problem Solving

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.

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.

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.

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%.

Any thoughts on the problem 1 ?

Thank You

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.

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 !!!

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

@bblast

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

Could you please explain

Thanks

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