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I am trying to pick good questions from the GMAT forum on a [#permalink] New post 09 Jun 2006, 05:32
I am trying to pick good questions from the GMAT forum on a specific topic for easy reference. I am having difficulty with mixture problemes so my first set is about mixture problems. ( I will paste the problems directly if there explanation is easy to follow or else I will rework the problem for better understanding)


Reference to Mixture Principlal
http://www.gmatclub.com/phpbb/viewtopic ... e&start=40





Problem 1: http://www.gmatclub.com/phpbb/viewtopic.php?t=29156


Prolem 2: http://www.gmatclub.com/phpbb/viewtopic.php?t=29116
Mixture A is 15% alcohol, and mixture B is 50% alcohol. If the two are poured together to create a 4 gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

A. 1.5
B. 1.7
C. 2.3
D. 2.5
E. 3


Soln 1 : Traditional way
let A be x gallons, B is (4-x) gallons

0.15x + 0.5(4-x) = 0.3 * 4
0.15x + 2 - 0.50x = 1.2

Solving for x,
x = 2.3

Answer is C



Soln 2 : Exellent trick by Yurik79
Find parts
Number of parts of A in final solution = | 30 - 50 | = 20
Number of parts of B in final solution = | 30 - 15 | = 15
Total Parts = 20 + 15
Size of each part = 4/35
So size of A in final solution = 4/35 * 20 = 2.3

[/url]


Problem 3 : A can contains a mixture of two liquids A & B in proportion 7:5. When 9 litres of mixture is drawn off and the can is filled with liquid B, the proportion of A & B becomes 7:9. How many litres of liquid A was contained by the can initially?

A. 25
B. 10
C. 20
D. 21
E. 22

Answer :
A:B = 7:5

Total = 12 units

When 9 litres of mixture is removed, ((7/12) * 9) litres of A is removed, and ((5/12) * 9) litres of B is removed

Then 9 litres of B is added so that new ratio is 7:9

In new mixture, Total volume of A is 7x - ((7/12) * 9) = 7x - 21/4
And total volume of B is 5x - ((5/12) * 9) + 9 = 5x - 3 3/4 + 9 = 5x + 21/4

So, (7x - 21/4)/(5x + 21/4) = 7/9

Solving for x:
x=3
Original Volume of A = 7x = 21

Answer is D
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 [#permalink] New post 09 Jun 2006, 08:21
Problem 3 : A can contains a mixture of two liquids A & B in proportion 7:5. When 9 litres of mixture is drawn off and the can is filled with liquid B, the proportion of A & B becomes 7:9. How many litres of liquid A was contained by the can initially?

A. 25
B. 10
C. 20
D. 21
E. 22
on real exam i would go with D not really solving))
where were 7 parts of liquid A in first mixture
looking on the answer choices 7 is multiple only off 21
Not sure this is a good idea to make a blind shot but D it is))
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 [#permalink] New post 09 Jun 2006, 14:30
Exellent input Yurik79. I am going to continue collecting mixture problems in this Set.

Problem 4 : Two vessels have petrol, diesel and kerosene mixed in the ratio 1:2:4 and 3:5:6 if the quantities in the two vessels are mixed in the ratio 1:1,
what is the ratio of petrol, diesel and kerosene in the resultant mixture?

a. 5:9:14 b.3:6:8 c. 4:8:11 d. 4:7:10
A and heres why


They are mixed in 1:1 ratio

so M1 = 1:2:4
and M2 = 3:5:6
to make sure they are in 1:1

mix M1 = 2:4:8 with M2 3:5:6 , now both have volume of 14= 2+4+8 = 3+5+6

so the new mixture is 2+3 : 4+5 :6 +8 = 5:9:14



Problem 5: A 5-liter jug contains 4 liters of a saltwater solution that is 15 percent salt. If 1.5 liters of the solution spills out of the jug, and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt?

My choice is 7,5%....

Simply did the following...

2.5 liters of which 15% is salt = 0.375 l.

The jug refilled to 5 liters.... so salt remains at 0.375 and total mixture is 5 liters...

so 0.375/5=0.075.... 7.5%

Though believe it is not that simple Smile



Problem 6: A mixture contain 17% spirit and rest water. 10 liters are drawn and the vessel is filled up with water. The proportion of spirit now is 15 1/9%. Whats the vessel capacity?

Original volume of spirit = 17V/100
After removing 10 L, it becomes 17(V-10)/100
so 17(V-10)/100=136V/(9*100)
Solve for V=90L

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 [#permalink] New post 09 Jun 2006, 15:04
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 [#permalink] New post 10 Jun 2006, 19:54
briozeal,
I have question, I am weak at these kind of questions as well and I am glad that you are talking the time to do this... :lol:

My question is...

In problem 5:
I got a different answer and I was wondering if you could help me understand were I missed my logic.
5 liters of total solution = 1 (h2o) and 4(15% salt h2o)

I first tried to find the salt percentage for one liter.
(4*0.15)/5 = 0.12 of salt per liter.
Now we have lost 1.5 liters of solution which means that we have lost 0.18 of salt from the solution.
Which leads to (0.60-0.18)=0.42 of salt remaining before we add more.
Finally (0.42/5) = 0.0804 of salt in total solution...8.4%
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 [#permalink] New post 11 Jun 2006, 00:01
acfuture wrote:
briozeal,
I have question, I am weak at these kind of questions as well and I am glad that you are talking the time to do this... :lol:

My question is...

In problem 5:
I got a different answer and I was wondering if you could help me understand were I missed my logic.
5 liters of total solution = 1 (h2o) and 4(15% salt h2o)

I first tried to find the salt percentage for one liter.
(4*0.15)/5 = 0.12 of salt per liter.
Now we have lost 1.5 liters of solution which means that we have lost 0.18 of salt from the solution.
Which leads to (0.60-0.18)=0.42 of salt remaining before we add more.
Finally (0.42/5) = 0.0804 of salt in total solution...8.4%

Hi I think the problem is that where was 4 liters of saltwater in 5 liter jag
and not 5 litters
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 [#permalink] New post 11 Jun 2006, 09:56
Yurik79 wrote:
acfuture wrote:
briozeal,
I have question, I am weak at these kind of questions as well and I am glad that you are talking the time to do this... :lol:

My question is...

In problem 5:
I got a different answer and I was wondering if you could help me understand were I missed my logic.
5 liters of total solution = 1 (h2o) and 4(15% salt h2o)

I first tried to find the salt percentage for one liter.
(4*0.15)/5 = 0.12 of salt per liter.
Now we have lost 1.5 liters of solution which means that we have lost 0.18 of salt from the solution.
Which leads to (0.60-0.18)=0.42 of salt remaining before we add more.
Finally (0.42/5) = 0.0804 of salt in total solution...8.4%

Hi I think the problem is that where was 4 liters of saltwater in 5 liter jag
and not 5 litters


Thank you for clarifying that ... :oops:
  [#permalink] 11 Jun 2006, 09:56
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