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5 liters are taken off from a vessel full of water and

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5 liters are taken off from a vessel full of water and [#permalink] New post 20 Oct 2005, 20:25
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5 liters are taken off from a vessel full of water and replaced with pure milk, again, five more liters of the mixture is taken and replaced with pure milk. After this process, if the vessel contains water and milk in the ratio 9:16 what is the capacity of vessel?
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 [#permalink] New post 20 Oct 2005, 21:11
I think the answer is 125/4 (31.25 litres)

I get 25 litre when water : milk is 16: 9, but the question says other way around.

If my answer is correct, then I will explain my method.
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 [#permalink] New post 20 Oct 2005, 21:12
please guys explain, I hate those problems, I would like to understand
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 [#permalink] New post 20 Oct 2005, 21:29
I did by number picking:

Assume that x be the capacity of the vessel. I picked a number to be 25.

Initial we have 25:0 for water:milk

Removed 5 lites from it and the new ratio is: 20:5 (in terms of x it is x - 5:5)

Removed 5 litres for second time. Newest ratio is: 16:9.
The above in terms of x is ( (x-5) - (x-5)/5: 5 - 1 + 5)

(5x - 25 -x + 5)/5 -> (4x - 20)/5 : 9 = 9 : 16

(4x - 20)/9*5 = 9/16

[Oops I made a mistake in my calculation in the prev post]

Solving for X, I get 11.33 (approx)

Note: I am not sure whether this approach is correct though
Note2: I edited the post for proper calculation.
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 [#permalink] New post 20 Oct 2005, 21:45
Ok this is a little more calculation than I would like.

(5-5/x*5+5)/x=16/25 (equation is established for milk)
10-25/x=16/25x
16x^2-250x+25^2=0
(4x-125/4)^2-(125/4)^2+25^2=0
(4x-125/4)^2=25^2/4^2(5^2-4^2)
(4x-125/4)^2=(25*3)^2/4^2
4x-125/4=75/4
4x=200/4=50
x=12.5

Verification:
If x=12.5, after the first milk added the ratio of milk is 5/12.5=2/5. So when you get 5 of mixture out you get 2 milk out. Then you add another 5 milk. So you've got 5-2+5=8 milk left. 8/12.5=16/25. The answer seems to be correct.

Only thing is there should be an easier way to go about this.
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Last edited by HongHu on 25 Oct 2005, 08:22, edited 1 time in total.
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 [#permalink] New post 21 Oct 2005, 13:54
I don't have OE for this but 12.5 is OA, Honghu thanks for explaining this
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 [#permalink] New post 21 Oct 2005, 17:12
HongHu wrote:
Verification:
If x=12.5, after the first milk added the ratio of milk is 5/12.5=2/5. So when you get 5 of mixture out you get 2 milk out. Then you add another 5 milk. So you've got 5-2+5=8 milk left. 8/12.5=16/25. The answer seems to be correct.

Only thing is there should be an easier way to go about this.

in your example, the water remained constant but when you take out 5 lt and another 5 lt the the water also take out not only milk, right? i am little confused here............

x = 12.5
1st take out = 5 lt water.
now the vessel has = 12.5-5 = 7.5 water
add 5 lt milk, then vessel contains = 7.5+5 = 12.5 water and milk.
again take out 5 lt, now vessel has 7.5 water and milk

now, milk = 5 - 2 = 3 lt.
water = 7.5 -3 = 4.5

now add 5 lt milk, then total = 12.5 water and milk (water = 4.5 and milk 8)

ratio of water/milk = 4.5/8 = 9/16...hmmmmmmmm ok....... thats correct. thankxxxxxxxxxxxxxxxxxxxxxxxxxxx

i did otherway round: milk:water.

so "read and understand the question correctly. doing so is 1/2 problem solved".
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 [#permalink] New post 21 Oct 2005, 17:21
Yes I didn't use water, I used the entire volume because this might be easier. However you'll have to remember to use 16/25 instead of 16/9. :)
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 [#permalink] New post 21 Oct 2005, 17:26
Let's see this may be a little easier to do then the earlier approach, if you are good at manipulating ratios...

(5-5/x*5+5)/x=16/25 (same as before, equation is established for milk over total volume)
(10x-25)/x^2=16/25
(x^2-10x+25)/x^2=(25-16)/25
(x-5)^2/x^2=9/25
(x-5)/x=3/5
5/x=2/5
x=25/2

At least this gets rid of square of big numbers ...
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Last edited by HongHu on 25 Oct 2005, 08:23, edited 2 times in total.
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 [#permalink] New post 21 Oct 2005, 17:27
Excellent explanation, HongHu!! Thanks!
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 [#permalink] New post 25 Oct 2005, 00:08
HongHu wrote:
Let's see this may be a little easier to do then the earlier approach, if you are good at manipulating ratios...

(5+5/x*5+5)/x=16/25 (same as before, equation is established for milk over total volume)
(10x-25)/x^2=16/25
(x^2-10x+25)/x^2=(25-16)/25
(x-5)^2/x^2=9/25
(x-5)/x=3/5
5/x=2/5
x=25/2

At least this gets rid of square of big numbers ...


Very clever trick to use only milk's percentage here.
Is this typo (5+5/x*5+5)/x=16/25 should it be a minus sign like
(5-5/x*5+5)/x=16/25.
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 [#permalink] New post 25 Oct 2005, 00:53
(9x)/x=(1-5/x)^2 => x=1,25 => solution must be a multiple of 1,25. hence i would choose it among the answers...
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 [#permalink] New post 25 Oct 2005, 01:46
Quote:
(5+5/x*5+5)/x=16/25 (same as before, equation is established for milk over total volume)


Hong, can you explain how you arrived at this first equation?
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 [#permalink] New post 25 Oct 2005, 06:23
There is a very handy formula to remember for these kind of questions..

If initially 'a' liters of liquid A is present in a vessel; if 'b' liters is taken from this vessel and substituted with liquid B, then 'b' liters is again taken and substituted with liquid B and this process is repeated 'n' times then after n times:

(Liquid A left in vessel / Liquid B left in vessel) =
[(a-b)/a]^n/{1-[(a-b)/a]^n}

Here, Liquid A left in vessel / Liquid B left in vessel = 9/16
b = 5, n = 2 Solve to get a = 12.5
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 [#permalink] New post 25 Oct 2005, 08:30
jainvineet wrote:
Is this typo (5+5/x*5+5)/x=16/25 should it be a minus sign like
(5-5/x*5+5)/x=16/25.


Yes, thanks! I copied the first typoed equation for my second post, which explains why I had the typo for both posts. :oops: I've corrected them both. :)

BumblebeeMan wrote:
Hong, can you explain how you arrived at this first equation?


Considering milk. The first time you added 5 liter milk to the vessel. So now the ratio of milk to entire volume is 5/x. Then you take 5 liter of mixtures out, there's 5/x liter milk in each liter of mixture, so 5/x*5 milk in the 5 liter mixture you take out. Then you added another 5 liter in. Now you have 5-5/x*5+5 liters of milk in the x liter mixtures. And you know the real ratio is 16/(9+16)=16/25.
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 [#permalink] New post 25 Oct 2005, 08:35
mohish wrote:
(Liquid A left in vessel / Liquid B left in vessel) =
[(a-b)/a]^n/{1-[(a-b)/a]^n}



Yes, good equation to remember. If you understand the reasoning it wouldn't be hard to remember it either.
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 [#permalink] New post 25 Oct 2005, 08:37
HongHu wrote:

Yes, thanks! I copied the first typoed equation for my second post, which explains why I had the typo for both posts. :oops: I've corrected them both. :)

BumblebeeMan wrote:
Hong, can you explain how you arrived at this first equation?


Considering milk. The first time you added 5 liter milk to the vessel. So now the ratio of milk to entire volume is 5/x. Then you take 5 liter of mixtures out, there's 5/x liter milk in each liter of mixture, so 5/x*5 milk in the 5 liter mixture you take out. Then you added another 5 liter in. Now you have 5-5/x*5+5 liters of milk in the x liter mixtures. And you know the real ratio is 16/(9+16)=16/25.


We cannot get better explanation than this. I committed this method for this kind of problems. Thanks a ton.
  [#permalink] 25 Oct 2005, 08:37
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