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A container has 3L of pure wine. 1L from the container is taken out an

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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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New post 07 Feb 2013, 01:57
y7214001 wrote:

Kudos +1 Karishma

Is there an fast way to compute the result of the multiplacation series like we have for \(Cf\)? I actually did the long way . :oops:


It would have taken forever!

\(C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * ....... (\frac{18}{20}) * (\frac{19}{21}) * (\frac{20}{22})\)

You need to observe here that other than first two numerators and last two denominators, all other terms will cancel out.
First term's denominator will cancel out third term's numerator.
Second term's denominator will cancel out fourth term's numerator.
The last two denominators will have no numerators to cancel them out.
The first two numerators have no denominators to cancel them out.
Usually, in such expressions (where terms have a pattern), things simplify easily. You just need to observe the pattern.
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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New post 05 Jul 2013, 05:47
Hi Bunuel and VeritasPrepKarishma,

Would it be possible to use this formula in this case?

New Concentration of wine= Old concentration of wine * (V1/V2)^n

n= number of iterations (it is 19 in this case)
v1 = volume of liquid withdrawn
v1 = initial volume of liquid

I noticed a similar formula being used here:
a-20-litre-mixture-of-milk-and-water-contains-milk-and-water-22212.html

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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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New post 16 Aug 2013, 10:56
I made a guess on this question.

If we are left with 4L of mixture which has 2L of wine and 2L of water after 1st process, the ratio of wine is about 1/2. So after 19 successive processes, ratio must be significantly less than 1/2.

Option B is little less than 1/2 so can't be the answer and we are left with option A and C. At least this helped me narrowed down to two options in 15 sec.
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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New post 16 Aug 2013, 17:20
1
The trick is to identify there should be a pattern as it is not possible to carry out all the calculations.

1. Initially wine was 3L.
2. After first operation wine was 2L
3. After second operation wine was 1.5L
4. After third operation wine was 1.2L

Now we can see the pattern (2) is 2/3 of (1), (3) is 3/4 of (2), (4) is 4/5 of (3) and so on

So in 3 operations wine left is 3 * 2/3 * 3/4 * 4/5 , after cancelling out of numbers we have 3* 2/5 = 1.2 L

So in 19 operations after cancelling out of numbers 3* 2/21 = 2/7 L of wine left
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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New post 19 Aug 2013, 04:33
emailmkarthik wrote:
Hi Bunuel and VeritasPrepKarishma,

Would it be possible to use this formula in this case?

New Concentration of wine= Old concentration of wine * (V1/V2)^n

n= number of iterations (it is 19 in this case)
v1 = volume of liquid withdrawn
v1 = initial volume of liquid

I noticed a similar formula being used here:
a-20-litre-mixture-of-milk-and-water-contains-milk-and-water-22212.html

Regards,


Actually, it is a play on the same formula.

Cf = Ci * (V1/V2)*(V3/V4).....

Usually, in replacement questions, you remove n lts and put back n lts. So initial and final volume in each step is the same. That is why you get (V1/V2)^n.

In case V1 and V2 are different in subsequent steps, you use those volumes V1/V2 * V3/V4 *.....
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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New post 29 May 2019, 08:46
CFfinal= (100%*20!3!)/22!= (100%*2)/(7*22)
Amount= CFfinal * Vfinal
Vfinal= 22
Amount= 2/7
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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New post 08 Jul 2019, 22:24
I still didn't get the logic to solve this. Could anyone please explain logically how to deduce a formula for this.

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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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New post 02 Jan 2020, 02:09
This is a really high-level "sequence" question that's built on some "math steps" that you'll see on Test Day, but not likely in this configuration.

In these types of "repeating steps" sequence question, the key is to figure out the "pattern" behind the math, so that you can avoid most of the calculation.

Here, the repeating step is "remove 1L of mixture and add 2L of water." Most people can't figure out the math pattern off of the top of their heads, so you have to do enough of the math to figure out what the pattern actually is. Here's how I deduced the pattern:

Start: 3L wine
1st: -1L mix + 2L water = 2L wine + 2 L water = 4L total

2nd: -1L mix = 1/4 of total removed = (.5L wine + .5L water removed) + 2L water added = 1.5L wine + 3.5L water = 5L total

3rd: -1L mix = 1/5 of total removed = (.3L wine + .7L water removed) + 2L water added = 1.2L wine + 4.8L water = 6L total

This pattern will continue on, slowly removing wine and quickly adding water to the mixture. Rather than do ALL of that math (for 19 operations!?!), here's the pattern:

Each operation is really about multiplying the remaining amount of wine by (1 - 1/n).

After the first operation, we multiply by 3/4, after the second 4/5 and so on. After 19 operations, we'd end up with…

2(3/4)(4/5)(5/6)…….(20/21) = 6/21 = 2/7

Final Answer: A

Don't worry about this question. It's really not worth your time.

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Re: A container has 3L of pure wine. 1L from the container is taken out an   [#permalink] 02 Jan 2020, 02:09

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