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Reserve tank 1 is capable of holding z gallons of water. Water is pump

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Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post Updated on: 05 Apr 2018, 05:42
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Reserve tank 1 is capable of holding z gallons of water. Water is pumped into tank 1, which starts off empty at a rate of x gallons per minute. Tank 1 simultaneously leaks water at a rate of y gallons per minute (x > y).The water that leaks out of tank 1 drips into tank 2,which also starts off empty. If the total capacity of tank 2 is twice the number of gallons that remains in tank 1 after 1 minute, does tank 1 fill up before tank 2?

(1) \(zy < 2x^2-4xy+2y^2\)

(2) Total capacity of tank 2 is less than one half that of tank 1.

Originally posted by JCLEONES on 23 Jan 2008, 12:18.
Last edited by Bunuel on 05 Apr 2018, 05:42, edited 2 times in total.
Edited the question.
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 14 Mar 2011, 01:15
19
9
Time to fill tank 1= z/(x-y)
Tank 1 will be (x-y) full after 1 minute
Thus, capacity of "tank 2"=2(x-y)
Time to fill "tank 2"=2(x-y)/y

Q: is z/(x-y) < 2(x-y)/y
OR
is zy< 2 (x-y)^2

1. zy<2x^2-4xy+2y^2
zy< 2(x^2-2xy+y^2)
zy< 2(x-y)^2
Sufficient.

2. 2(x-y) < (1/2)z
4 < z/(x-y)
z/(x-y) > 4
But is;
z/(x-y) < 2(x-y)/y
Can't simplify any further.
Not sufficient.

Ans: "A"
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 23 Jan 2008, 13:29
2
capacity of tank 2 is u = 2*(x-y)

time it takes for tank 2 to get filled is 2*(x-y)/y
time it takes for tank 1 to get filled is z/(x-y)

z/(x-y) - 2*(x-y)/y < 0? equivalent (since y > 0 , x-y > 0)

zy < 2*(x-y)^2 = 2*x^2-4*x*y + 2*y^2 (1) is sufficient
2 is irrelevant
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 14 Mar 2011, 00:44
I'd say A.

1 tells us that if Z was 100, x and y were 10 and 5 respectively, 1 would not be true. for 1 to be true, x and y will be 10 and 1 or around that area.

B tells us nothing about the rate at which water is flowing into the second tank.

Any other thoughts?
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 14 Mar 2011, 04:07
1
1
Another "lone wolf". Here the complex equation is A

Quote:
# The Lone Wolf

A lone wolf question almost always has a free standing number(or numbers), and a more complex looking equation as the other option. For e.g.

"On a loan, evil necromonger charges X% interest in the first year, and Y% interest in the second. If he loaned Rhyme 20,000$ in 2006, how much Rhyme pay by interest in 2008?"
A) X = 10
B) (X + Y + XY/100) = 100

You can almost be certain, that in such questions, your equations to the stem will reduce to a form that looks like (B), so (A) is actually redundant. Be careful of lone wolves because they will bite you in the posterior if you choose (C).

If you notice a lone wolf question, and you have no clue on how to solve the problem, choose (B) (or whichever is the complex equation).
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 02 Oct 2011, 00:51
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After one minute there is \(y\) gallons in the tank 2, so the capacity of the tank 2 is \(2(x-y)\) gallons.
Obviously, each minute the tank 1 is filled with \(x-y\) gallons of water, and tank 2 is filled with \(y\) gallons.

Let A is the number of minutes after which the 1 tank is full, and B is the number of minutes after which 2 the 2 tank is full.
Then:
\(A(x-y)=z\)
\(By=2(x-y)\)

\(A=\frac{z}{x-y}\)
\(B=\frac{2(x-y)}{y}\)

We need to compare A and B, so we are comparing \(\frac{z}{x-y}\) and \(\frac{2(x-y)}{y}\)

\(\frac{z}{x-y}\) ... \(\frac{2(x-y)}{y}\)
\(yz\) ... \(2(x-y)(x-y)\)
\(yz\) ... \(2x^2-4xy+2y^2\)

If (1) is true, then \(yz< 2x^2-4xy-y^2\)
Since \(2x^2-4xy+2y^2> 2x^2-4xy-y^2\), then \(yz <2x^2-4xy+2y^2\) and we are able to compare two time periods. The statement (1) alone is susfficient.

If (2) is true, then \(2(x-y)<0.5z\)
\(\frac{z}{x-y}>4\)
This means that \(A>4\)
However, \(B=\frac{2(x-y)}{y}\), so there is no z and we only know that \(x>y\), but nothing could be said to compare \(x-y\) and \(y\). For example, if \(x=2y\), then \(B=2\) and \(A>B\). However, if \(x=5y\), then \(B=8\) and we could not compare A and B.

So, the answer is (A)

NOTE: You should post DS problems in the other forum.
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 16 Dec 2011, 07:00
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Nice problem. Got to the answer but took more than 3 mins... :( Problem statement itself took very long to read and understand
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 10 Dec 2016, 08:41
bagrettin wrote:
After one minute there is \(y\) gallons in the tank 2, so the capacity of the tank 2 is \(2(x-y)\) gallons.
Obviously, each minute the tank 1 is filled with \(x-y\) gallons of water, and tank 2 is filled with \(y\) gallons.

Let A is the number of minutes after which the 1 tank is full, and B is the number of minutes after which 2 the 2 tank is full.
Then:
\(A(x-y)=z\)
\(By=2(x-y)\)

\(A=\frac{z}{x-y}\)
\(B=\frac{2(x-y)}{y}\)

We need to compare A and B, so we are comparing \(\frac{z}{x-y}\) and \(\frac{2(x-y)}{y}\)

\(\frac{z}{x-y}\) ... \(\frac{2(x-y)}{y}\)
\(yz\) ... \(2(x-y)(x-y)\)
\(yz\) ... \(2x^2-4xy+2y^2\)

If (1) is true, then \(yz< 2x^2-4xy-y^2\)
Since \(2x^2-4xy+2y^2> 2x^2-4xy-y^2\), then \(yz <2x^2-4xy+2y^2\) and we are able to compare two time periods. The statement (1) alone is susfficient.

If (2) is true, then \(2(x-y)<0.5z\)
\(\frac{z}{x-y}>4\)
This means that \(A>4\)
However, \(B=\frac{2(x-y)}{y}\), so there is no z and we only know that \(x>y\), but nothing could be said to compare \(x-y\) and \(y\). For example, if \(x=2y\), then \(B=2\) and \(A>B\). However, if \(x=5y\), then \(B=8\) and we could not compare A and B.

So, the answer is (A)

NOTE: You should post DS problems in the other forum.



I did not get why did you use yz< 2x^2-4xy-y^2 in the above solution. The solution is sufficient without the usage of this equation.

+1 Kudos if you like the post :)
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 20 Jul 2017, 09:29
T1 - Water is being filled at the rate of X Gallons/Minute and leaking at the rate of Y G/M
In one minute T1 is getting filled at (X-Y) G/M
Z Gallons would get filled in Z/(X-Y)

Capacity of T2 = 2(X-Y) G/M
T2 water is being filled at Y G/M
Therefore, T2 would get filled = 2(X-Y)/Y

A) 2 (X-Y)^2 > ZY
=> 2(X-Y)/Y > Z/(X-Y)
Therefore A is sufficient

B) The total capacity of tank 2 is less than one half that of Tank 1
This doesn't specify the relationship between the rates at which the tanks are being filled. Only the relationship between the capacities.

Therefore, Only A is sufficient.

Hope this helps.
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 05 Apr 2018, 04:34
HOW DO WE SOLVE SUCH QUES IN 2 MIN . IT TOOK ME MORE THAN 2 MIN TO UNDERSTAND AND SOLVE THE SAME.CAN SOME PLS SUGGEST ME WITH HIS OR HER APPROACH
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 01 May 2018, 05:59
Bunuel - Please can you help suggest if is there any method in which we can approach to solve in 2 mins ??

Thanks in advance
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 16 Dec 2018, 09:49
What is the difficulty level of this question.
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 16 Dec 2018, 16:02
JCLEONES wrote:
Reserve tank 1 is capable of holding z gallons of water. Water is pumped into tank 1, which starts off empty at a rate of x gallons per minute. Tank 1 simultaneously leaks water at a rate of y gallons per minute (x > y).The water that leaks out of tank 1 drips into tank 2,which also starts off empty. If the total capacity of tank 2 is twice the number of gallons that remains in tank 1 after 1 minute, does tank 1 fill up before tank 2?

(1) \(zy < 2x^2-4xy+2y^2\)

(2) Total capacity of tank 2 is less than one half that of tank 1.


Dear GMATGuruNY

Can you share your thoughts on this question?
Thanks in advance
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 18 Dec 2018, 11:36
1
Mo2men wrote:
JCLEONES wrote:
Reserve tank 1 is capable of holding z gallons of water. Water is pumped into tank 1, which starts off empty at a rate of x gallons per minute. Tank 1 simultaneously leaks water at a rate of y gallons per minute (x > y).The water that leaks out of tank 1 drips into tank 2,which also starts off empty. If the total capacity of tank 2 is twice the number of gallons that remains in tank 1 after 1 minute, does tank 1 fill up before tank 2?

(1) \(zy < 2x^2-4xy+2y^2\)

(2) Total capacity of tank 2 is less than one half that of tank 1.


Dear GMATGuruNY

Can you share your thoughts on this question?
Thanks in advance


Statement 1: zy < 2x² - 4xy + 2y²
To see the implications of this inequality, plug in values for x and y and solve for z.
Let x=10 and y=2.
Then:
z(2) < 2(10²) - 4(10)(2) + 2(2²)
2z < 128
z < 64.
Here, the capacity of tank 1 is LESS than 64 gallons.

Tank 1:
Since tank 1 receives x=10 gallons per minute and loses y=2 gallons per minute, the net gain for tank 1 = 10-2 = 8 gallons per minute.
Since the capacity of tank 1 is LESS than 64 gallons, the time to fill tank 1 at a rate of 8 gallons per minute must be LESS than 64/8 = 8 minutes.

Tank 2:
After one minute, the volume in tank 1 = 8 gallons.
Since the capacity of tank 2 is twice the volume in tank 1 after one minute, the capacity of tank 2 = 2*8 = 16 gallons.
Time to fill tank 2 at a rate of y=2 gallons per minute = 16/2 = 8 minutes.

While tank 1 requires LESS than 8 minutes, tank 2 requires EXACTLY 8 minutes.
The case above illustrates that tank 1 will fill up before tank 2.
SUFFICIENT.

Statement 2: The total capacity of tank 2 is less than one-half that of tank 1.
In statement 1 above, it is possible that the capacity of tank 2 = 16 gallons, while the capacity of tank 1 = 63 gallons.
These values also satisfy statement 2.
As we saw above, the result will be that tank 1 fills up before tank 2.
But if we increase the capacity of tank 1 to 1000 gallons and leave all of the other values the same, tank 2 will fill up before tank 1.
INSUFFICIENT.


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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 16 Aug 2019, 14:53
JCLEONES wrote:
Reserve tank 1 is capable of holding z gallons of water. Water is pumped into tank 1, which starts off empty at a rate of x gallons per minute. Tank 1 simultaneously leaks water at a rate of y gallons per minute (x > y).The water that leaks out of tank 1 drips into tank 2,which also starts off empty. If the total capacity of tank 2 is twice the number of gallons that remains in tank 1 after 1 minute, does tank 1 fill up before tank 2?

(1) \(zy < 2x^2-4xy+2y^2\)

(2) Total capacity of tank 2 is less than one half that of tank 1.



Do we have any other approach to this question
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Re: Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 29 Aug 2019, 11:12
I manage to get this problem correct by pluging numbers, can someone please try to explain me differently with algebra ? I don't really understand the solutions above
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Reserve tank 1 is capable of holding z gallons of water. Water is pump  [#permalink]

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New post 29 Aug 2019, 12:17
JCLEONES wrote:
Reserve tank 1 is capable of holding z gallons of water. Water is pumped into tank 1, which starts off empty at a rate of x gallons per minute. Tank 1 simultaneously leaks water at a rate of y gallons per minute (x > y).The water that leaks out of tank 1 drips into tank 2, which also starts off empty. If the total capacity of tank 2 is twice the number of gallons that remains in tank 1 after 1 minute, does tank 1 fill up before tank 2?

(1) \(zy < 2x^2-4xy+2y^2\)

(2) Total capacity of tank 2 is less than one half that of tank 1.


Tank 1:
Since water is PUMPED IN at x gallons per minute but LEAKS OUT at y gallons per minute, the net gain per minute = x-y.
Since the z-gallon tank is filled at a net rate of x-y gallons per minute, we get:
Time fill tank 1 \(= \frac{capacity}{rate} = \frac{z}{x-y}\)

Tank 2:
The total capacity of tank 2 is twice the number of gallons that remains in tank 1 after 1 minute.
After 1 minute, the number of gallons in tank 1 = (net gain per minute)(one minute) = (x-y)(1) = x-y.
Since the capacity of tank 2 is twice this number of gallons, we get:
Capacity of tank 2 = 2(x-y) = 2x-2y.

The water that leaks out of tank 1 drips into tank 2.
Since water leaks from tank 1 into tank 2 at a rate of y gallons per minute, we get:
Time to fill tank 2 \(= \frac{capacity}{rate} = \frac{2x - 2y}{y}\)

Does tank 1 fill up before tank 2?
In other words:
Is the time fill tank 1 less than the time to fill tank 2?
Original question stem:
Is \(\frac{z}{x-y}< \frac{2x-2y}{y}\)?

Simplifying the question stem, we get:
\(zy < (2x-2y)(x-y)\)
\(zy < 2x^2-4xy+2y^2\)

Question stem, rephrased:
Is \(zy < 2x^2-4xy+2y^2\)?

Statement 1: \(zy < 2x^2-4xy+2y^2\)
The answer to the rephrased question stem is YES.
SUFFICIENT.

Statement 2:
Since the capacity of tank 2 = 2x-2y and the capacity of tank 1 = z, we get:
\(2x-2y < \frac{1}{2}z\)
\(4x-4y < z\)
\(4 > \frac{z}{x-y}\)
\(\frac{z}{x-y} < 4\)
No way to answer the original question stem or the rephrased question stem.
INSUFFICIENT.


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Reserve tank 1 is capable of holding z gallons of water. Water is pump   [#permalink] 29 Aug 2019, 12:17
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