An interesting mixture problem from the archives. This will

Question

An interesting mixture problem from the archives. This will be good practice. Please show your timing.

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

jeremy02 · Answer

did this in 30 seconds. hope i'm right.

I've taken some time off after my gmat A#@$ beating, and then the presidential A#$% beating.

So we have 30 liters of Rum & Coke, no problems so far. We want the mixture to go from 20% coke to 5% coke, makes perfect sense to me. 

We need to first figure out of much of the mixture is Coke, 20% of 30 liters equals 6 liters. 
We want the final mixture to be 5% Coke, this means that we need to find the total mixture where 6 liters = 5%.

5% = 1/20 6 liters * 1/20 = 120 liters, we currently have 30 liters, we need an additional 90 liters of RUMto reach a solution in which 6 liters of Coke = 5%.

Now how many swizzle sticks do we need?

smcgrath12 · Answer

Took me around 30 seconds. I assumed x litres Bircardi added. So now 6 litres of (30+x) equals 5%. Find x = 90.

Paul · Answer

Guys, re-read the question. You are missing something :wink:

smcgrath12 · Answer

Oops, 22.5 is the answer.

Say, x litres taken out. So x/5 litres of Coke and 4x/5 litres of Bicardi taken out. Then x litres of bicardi added back. So total litres remains at 30.

So, equation becomes (6-x/5)/30 * 100 = 5.

Solve for x.

venksune · Answer

I have two approaches - one is intuitive and the other fomula based.

1.Intuitive method:
Given that the mixture has 6 litres of coke and 24 litres of Rum/some alchohol. If we take x litres away it will still have the same percents of coke and rum in the x litres. So we have (6-x/5). We add x litres back with alcohol making the mixture now have only 1.5litres of coke (5% of 30).

So, we get 6-x/5 = 1.5
=> x=22.5

2. Formula method: (good to remember if a slightly difficult situation. For ex. x litres taken away, mixed with xlitres of pure alchohol and say if we repeat this proceedure more than once - say thrice. Yes, the above method can become painful). So, here's the formula

%of coke left/%of coke originally present = [1-r/m]^y.
note r= here x for us. m=30 litres. y=number of times we do this taking away and putting business. here y=1.

So, we have 5/20 = [1-x/30]^1
=>1/6 = 1-x/30
=>1/6 = (30-x)/30
=>7.5=30-x
=> x= 22.5. 8-)

NEWKID · Answer

You are awesome Venks. :wave

jeremy02 · Answer

Damn you Paul with those tricky questions, thanks for keeping me on my toes.

I think the quickest way is to determine that we want 5% of 30 liters to be coke, this is 1.5 liters, how much of the original solution do we need to have 1.5 liters of coke, we know that for every liter of mixture we have 1/5 liters of coke, so we need 5 * 1/5 liters or 7.5 liters of the original solution, 30 - 7.5 = 22.5 

Thanks to all for showing your work, helps me realize why i know the answer.

Paul · Answer

Good job all. OA is 22.5 :-D

An interesting mixture problem from the archives. This will

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