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Is this a GMAT Question ?

by Uva@90 » Sat Nov 30, 2013 12:59 am
A container has 3L of pure wine. 1L from the container is taken out and 2L water is added.The process is repeated several times. After 19 such operations, qty of wine in mixture is
A. 2/7 L
B. 3/7 L
C. 6/19L

OA A

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by [email protected] » Sat Nov 30, 2013 3:32 pm
Hi Uva@90,

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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by Uva@90 » Sat Nov 30, 2013 6:24 pm
[email protected] wrote:Hi Uva@90,

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.

GMAT assassins aren't born, they're made,
Rich
Rich,
Thanks for explaining so clearly.And Mentioning some key things for this kind of question,
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.


However,
I could not get one thing, How you arrived the pattern of (1-1/n)
It should be of (1 - 1/(n+2)) only na ?

For example,
2 is the amount of wine we got from the first process,
From the second process,
we can use the pattern,
2*(1 - 1/(n+2)),
N= 2 => 2*3/4 = 1.5
For the third process,
N=3 => 1.5*(4/5) = 1.2

So pattern should be of (1 - 1/(n+2)).
Am I wrong Rich ?

Regards,
Uva.
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by GMATGuruNY » Sat Nov 30, 2013 6:25 pm
Uva@90 wrote:A container has 3L of pure wine. 1L from the container is taken out and 2L water is added.The process is repeated several times. After 19 such operations, qty of wine in mixture is
A. 2/7 L
B. 3/7 L
C. 6/19L

OA A

Regards,
Uva.
An alternate approach:

1st process:
Here, the total volume is 3 liters.
After 1 liter is removed, 2 liters remain.
Since 2/3 of the total volume remains, 2/3 of the wine also remains.
Thus, the volume of wine is multiplied by 2/3:
3 * (2/3).
After 2 liters of water are added, the new total volume = 4 liters.

2nd process:
Here, the total volume = 4 liters.
After 1 liter is removed, 3 liters remain.
Since 3/4 of the total volume remains, 3/4 of the wine also remains.
Thus, the preceding volume of wine is multiplied by 3/4:
3 * (2/3) * (3/4).

By now we can see the pattern.
In the 1st process, the volume of wine is multiplied by 2/3.
In the 2nd process, the volume of wine is multiplied by 3/4.
Following this pattern:
In the 3rd process, the volume of wine will be multiplied by 4/5.
In the 4th process, the volume of wine will be multiplied by 5/6.

Thus, after the 4th process, the volume of wine = 3 * (2/3) * (3/4) * (4/5) * (5/6).
Notice that the values in red all cancel out.
Implication:
After the nth process, the volume of wine = (3*2)/(denominator yielded by the nth process).

Looking at the first 4 processes, we can see that the nth process yields a denominator of n+2:
1st process --> 2/3 --> denominator of 3.
2nd process --> 3/4 --> denominator of 4.
Thus:
19th process --> 20/21 --> denominator of 21.

Thus, after the 19th process, the volume of wine = (3*2)/21 = 2/7.

The correct answer is A.
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by Uva@90 » Sat Nov 30, 2013 6:38 pm
GMATGuruNY wrote:
Uva@90 wrote:A container has 3L of pure wine. 1L from the container is taken out and 2L water is added.The process is repeated several times. After 19 such operations, qty of wine in mixture is
A. 2/7 L
B. 3/7 L
C. 6/19L

OA A

Regards,
Uva.
An alternate approach:

1st process:
Here, the total volume is 3 liters.
After 1 liter is removed, 2 liters remain.
Since 2/3 of the total volume remains, 2/3 of the wine also remains.
Thus, the volume of wine is multiplied by 2/3:
3 * (2/3).
After 2 liters of water are added, the new total volume = 4 liters.

2nd process:
Here, the total volume = 4 liters.
After 1 liter is removed, 3 liters remain.
Since 3/4 of the total volume remains, 3/4 of the wine also remains.
Thus, the preceding volume of wine is multiplied by 3/4:
3 * (2/3) * (3/4).

By now we can see the pattern.
In the 1st process, the volume of wine is multiplied by 2/3.
In the 2nd process, the volume of wine is multiplied by 3/4.
Following this pattern:
In the 3rd process, the volume of wine will be multiplied by 4/5.
In the 4th process, the volume of wine will be multiplied by 5/6.

Thus, after the 4th process, the volume of wine = 3 * (2/3) * (3/4) * (4/5) * (5/6).
Notice that the values in red all cancel out.
Implication:
After the nth process, the volume of wine = (3*2)/(denominator yielded by the nth process).

Looking at the first 4 processes, we can see that the nth process yields a denominator of n+2:
1st process --> 2/3 --> denominator of 3.
2nd process --> 3/4 --> denominator of 4.
Thus:
19th process --> 20/21 --> denominator of 21.

Thus, after the 19th process, the volume of wine = (3*2)/21 = 2/7.

The correct answer is A.
Thanks Mitch,
I Completely got your method.

Regards,
Uva.
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