Data Sufficiency

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 ( Where x>y) . The water that leaks out of tank 1 drips into tank 2, which also starts out empty. If the total capacity of tank 2 is twice the number of gallons that remains in tank 1 after one 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.

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 (where x > y). The water that leaks out of tank 1 drips into tank 2, which also starts out empty. If the total capacity of tank 2 is twice the number of gallons of water actually existing in tank 1 after one minute, does tank 1 fill up before tank 2?

(1) zy < 2x2 - 4xy + 2y2

(2) The total capacity of tank 2 is less than one-half that of tank 1.

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.

The correct answer is A.

Could you elaborate on the 2nd statement logic please. Thanks !

aditiniyer wrote:Could you elaborate on the 2nd statement logic please. Thanks !

Case 1: x=10 gallons per minute, y=2 gallons per minute, z=56 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 56 gallons, the time to fill tank 1 at a rate of 8 gallons per minute:
56/8 = 7 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.

In this case, Tank 1 fills before Tank 2.

Case 2: x=10 gallons per minute, y=2 gallons per minute, z=1000 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 1000 gallons, the time to fill tank 1 at a rate of 8 gallons per minute:
1000/8 = 125 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.

In this case, Tank 2 fills before Tank 1.

Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.