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

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"

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.

Another "lone wolf". Here the complex equation is A

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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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

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.

Nice problem. Got to the answer but took more than 3 mins... Problem statement itself took very long to read and understand

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

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.


_________________

Please find the solution as depicted in figure

Answer: Option A

Capacity of tank 1 = \(z\) gallons ... (i)
Effective filling rate of tank 1 = \((x - y)\) gallons/min

=> Time taken to fill tank 1 = \(\frac{z}{(x-y)}\) minutes ... (ii)

Volume of water in tank 1 after 1 minute = \((x - y)\) gallons

=> Volume of tank 2 = \(2(x - y)\) gallons ... (iii)

Filling rate of tank 2 = \(y\) gallons/min

=> Time taken to fill tank 2 = \(\frac{2(x - y)}{y}\) minutes ... (iv)

We need to verify if: tank 1 fills up before tank 2

i.e. if \(\frac{z}{(x-y)} < \frac{2(x - y)}{y}\)

i.e. if \(yz < 2(x - y)^2\)

i.e. if \(yz < 2x^2 - 4xy + 2y^2\) ... (v)

From statement 1: The exact condition in (iii) above is mentioned in this statement --- Sufficient

From statement 2: Capacity of tank 2 is less than one half that of tank 1

=> \(2(x - y) < z/2\)

=> \(z > 4(x - y)\)

However, this is not enough to verify condition (v) above - Insufficient

Answer A
_________________

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?

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

Bunuel - Please can you help suggest if is there any method in which we can approach to solve in 2 mins ??

Thanks in advance

What is the difficulty level of this question.

Dear GMATGuruNY

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

Do we have any other approach to this question
GMATinsight
VeritasKarishma

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

Statement 2 - Capacity of tank 2 is less than one half that of tank 1

2(x−y)<z/2, over here why are we saying Capacity of tank 2 is 2(x-y) as that's not the overall capacity but the capacity after 1 min. How do we infer that total capacity of tank 2 will follow the same equation as after 1 min