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A 30 litre mixture of Bacardi and Coke contains 20% Coke in

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A 30 litre mixture of Bacardi and Coke contains 20% Coke in [#permalink]

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11 Jan 2004, 11:25
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A 30 litre mixture of Bacardi and Coke contains 20% Coke in it. How many litres of the mixture should be replaced with pure Bacardi to bring down the percentage of Coke in the mixture to 5%?

a.90 litres
b.15
c.22.5
d.18
e.11
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11 Jan 2004, 13:09
A) 90 liters of Bacardi

Original mixture has 24L Bacardi/ 6L coke
Let x = amount of pure Bacardi to be added
New mixture ratio : 6 / [ 30 + x ] = 5 / 100
600 = 150 + 5x
5x = 450
x = 90
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11 Jan 2004, 13:21

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11 Jan 2004, 13:33
90 lts.

20% of 30 = 6

we need to decrease coke's % to 5%
6/(30+x) = 5/100
600 = 150 + 5x
5x = 450
x=90

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11 Jan 2004, 13:38
Oops, didnt even read the question asking for how many liters of original mixture should be "replaced". I just calculated the # of new bacardi to be added to original mixture
Answer should be C
take out 22.5L
You're left with 7.5 liters of orginal
it means 1.5L Coke and 6L Bacardi
New mixture: 1.5 Coke / 6+22.5 Bacardi + 1.5 Coke
= 1.5 / 30 = 5%
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11 Jan 2004, 13:39
DJ, we did same mistake!
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11 Jan 2004, 13:40
We only have 30 litres to work with. How can we replace 90 litres if we only have 30 litres to begin with. I am having difficulties solving this problem, but the above fact would eliminate answer choice (a) right off the bat...wont it
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11 Jan 2004, 13:42
Paul wrote:
DJ, we did same mistake!

shoo..

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11 Jan 2004, 13:47
Paul your right. It is (C)
But, can I ask, How did you get 22.5?
Can you work the problem from start to finish please, so I can underatdn your logic.
Cheers
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11 Jan 2004, 13:47
to make 5% ok coke's mixture, amount of coke should be 1.5 lt.

original amount was 6. we decreased this to 1.5. means 1/4 th.

1/4 of 24 (bacardi) = 6

hence, we have 6 (bacardi) + 1.5 (coke)....original amount.
added to it = 30-7.5 = 22.5 bacardi.

damn ...!

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11 Jan 2004, 13:53
Sunniboy, I think the easiest way around this problem is just to backsolve or do it methodically as dj did. Honestly, figuring out the algebraic problem will just drain all your time. dj explained it well enough I guess
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11 Jan 2004, 14:03
To get it fastest, I just backsolved. I was anxious to be first one to post to correct my mistake!
For instance, if you pick D) 18L of original to be replaced. What you're left with is 12L of original mixture. You know that orginal mixture is 20% Coke. Therefore 20%*12L = 2.4L coke and the rest, 9.6L is Baccardi. Now, what you removed (18L), you will have to replace it back with pure Baccardi and the total mixture should be 5% Coke. Calculating the %Coke for the new mixture, you get 2.4 / 9.6 + 18 + 2.4 ( You have to add the 2.4L Coke to know the TOTAL mixture volume ). This gives 2.4 / 30. This is not right because if we want something 5% Coke, then final Coke volume should have been 5%*30 = 1.5L Coke in then end. Therefore, you have to take out more of the original mixture. 18L of original mixture removed was not sufficient. Only left is C with 22.5. This means that you have 7.5 of orginial left after removing 22.5. 20% of 7.5 gives 1.5L coke which is right what we're looking for... This is a lot of verbiage for something you can do much more quickly mentally though. Hope you did not get lost in my wording
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11 Jan 2004, 14:13

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11 Jan 2004, 18:03
Here is what I did:

C=Coke
B=Bacardi

Info given: C/B=2/8 and C+B=30 then C/B=6/24

Suppose x is the amount to be replaced:

( 6-x*0.2 ) / ( 24 -x*0.8 + x*1 ) = 5/100

x=160/7=20+20/7

Answer is C.

Hope it helps.

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11 Jan 2004, 18:06
Direct caculation:
A 5% mixture contains 1,5 l Coke. If we add just pure Bacardi, we will have 1,5 l Coke, which is from the old mixture. So that means that 1,5 l is 20% of previous mixture which remains in the "bottle" .
1,5l = 20%
100 %= ? => 1, 5 x 5 = 7,5 l

Conclusion : 30 l of a new mixture ( 5% Coke) cointain 7,5 l of an old mixture (20%) and 22,5 l of pure Bacardi.

Answer is 22,5 l.

It took me about 30 sec...I think it is faster than backsolving...

Cheers!!!

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22 Feb 2005, 12:15
Zeka,
Excellent reasoning.

Good discussion here.One lesson learned here is dont drink too much of baccardi

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22 Feb 2005, 13:36
Just think it this way. You have 20% and you only want it 5%. So you divide them into 4, each have 5%. You replace three of the four with pure whatever, you now are only left with 5%. So 30 divided by two is 15, 15 divided by two is 7.5. The three share is 15 plus 7.5 = 22.5.

This may seem long when I write it out, but is very straight forward in thinking. Don't even need to write a number in your scratch paper.

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22 Feb 2005, 16:29
any practise problems links for solutions in the net..

Sunniboy....that was a cameo

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23 Feb 2005, 08:46
"C"

If X is the mixture to be replaced, then

(30-X)80/100 + X = 30*95/100

X = 22.5

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23 Feb 2005, 12:33
20% * 30 (initial volume of coke) - 20% * x (volume that must be drained off) + 0 * x (bacardi that must be replaced; 0 because it has no concentration of coke) = 5% * 30 (resulting volume of coke)

22,5 !

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23 Feb 2005, 12:33

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