A certain pipe carries water between two tanks such that the height of

Volume = Base Area * Height of vassel

Base Area of First Tank = 3* Base area of Second Tank
Understand that bigger the base will lead to lesser reduction in height in comparison to other vessel which has smaller base
i.e. Rate of decrease in height*3 = rate of Increase in height (Change in height is inversely proportional to Base areas)

3*3 = rate of Increase in height = 9

Answer: OPtion B

Try "smart" numbers. The logic of above answers is the same, but it was easier for me to see, might be for others.

If a problem doesn't forbid emptying contents of one container into another when different dimensions are at issue but NOT different units, almost always, it helps to assume that one empties into the other. (I started trying to account for -3 inches per second. Bad idea.)

So the volume of water in the first tank, with height of 3 is EMPTIED in 3 seconds.

1. Numbers chosen

Let base area of Tank 1 = 3
Let height of Tank 1 = 3

Base area of Tank 1 is three times that of Tank 2=>

Tank 2 base area= 3/1 = 1

BASE AREA of both is (π*\(r^2\)) - No need to calculate it in volume formula; its value is assigned

2. Volume of water in Tank 1

--> V =(π*\(r^2\))*h

Base area = 3

Height of water is 3

V=3*3

Volume of water equals 9

3. Find height of Tank 2, into which the water goes. Volume is 9. Same formula, but solve for height (to which water will rise)

9 = (π*\(r^2\))*h

Base area = 1,so height of water:
9=(1)(h), h=9

Answer B: 9 inches per second

Let us say that the area of the base of the first tank is 6 square inches. Now 6 x 3 = 18 cubic inches of water is flowing in the other tank. The area of the second tank is 2 square inches. If the volume is 18 cubic inches and area is 2 square inches then the height is 18/2 = 9 inches.

Answer: B

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