A Hydrogenator water gun has a cylindrical water tank, which is 30

Volume as filled by hose=60×8×pi

Volume of cylinder=pi×r^2×30

60×8×pi=pi×r^2×30
r=4 so diameter=8

Answer:

D. 8 cm

volume = 480 *pi

Pi*r^2*h = 480 pi
r^2 * 30 = 480

so r = 4
And diameter = 8

D is correct
Volume as filled by hose=60×8×pi

Volume of cylinder=pi×r^2×30

60×8×pi=pi×r^2×30
r=4 so diameter = 8

Height of the cylinder = h = 30 cm

Rate of filling the tank = R = \(\pi \frac{{cm}^3}{s}\)

Time it takes to fill the tank = t=8 minutes = 480 seconds

Thus, volume of the cylinder (with radius of the base of the base of the cylinder = r cm) = V = R*t = \(480\pi\) = \(\pi*r^2*h\)

Thus r = 4 cm and diameter = 2r =8 cm. D is the correct answer.

Height of cylinder tank=30 cm
Rate of filling = Pie cu cm per second
8 minutes=8*60=480 seconds
Volume of tank =pie r^2*h=pie*r^2*30=480 pie
radius=4 cm
diamter=8 cm
Answer D

MANHATTAN GMAT OFFICIAL SOLUTION:

In 8 minutes, or 480 seconds, \(480\pi\) cm3 of water flows into the tank. Therefore, the volume of the tank is \(480\pi\). You are given a height of 30, so you can solve for the radius:

\(V = \pi r^2 * h\)

\(480\pi = 30\pi r^2\)

\(r^2 = 16\)

\(r= 4\)

Therefore, the diameter of the tanks base is 8 cm.

Answer: D.

A hydrogenator water gun has a cylinderical water tank ; which is 30 cm long .Using a hose, jack fills the tank with π cubic cm of water every scond. If it takes 8 minutes to fill the tank completely with water => the diameter of the tank is
[A] 4
[B] 8
[C] 12
[D] 16
[E] 20

Merging topics. Please refer to the discussion above.

\(H = 30\,{\rm{cm}}\,\,\,\,\,\,\,{\rm{;}}\,\,\,\,\,\,\,{{\pi \,\,{\rm{c}}{{\rm{m}}^3}} \over {\,1\,\,{\rm{second}}\,}}\,\,\,{\rm{filling}}\,\,{\rm{rate}}\)

\({\rm{?}}\,\,{\rm{ = }}\,\,D\,\,\,\left[ {{\rm{cm}}} \right]\)

Let´s use UNITS CONTROL, one of the most powerful tools of our method:

\({\rm{8}}\,{\rm{minutes}}\,\, \cdot \left( {{{\,60\,\,{\rm{seconds}}\,} \over {1\,\,{\rm{minute}}}}} \right)\,\,\,\left( {{{\pi \,\,{\rm{c}}{{\rm{m}}^3}} \over {\,1\,\,{\rm{second}}\,}}} \right)\,\,\, = \,\,\,\pi {\left( {{D \over 2}} \right)^2}H\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\, = {V_{{\rm{cylinder}}}}\,\,\left[ {{\rm{c}}{{\rm{m}}^3}} \right]\,\,\,} \right]\)

\(8 \cdot 60 = {\left( {{D \over 2}} \right)^2} \cdot 30\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\left( {{D \over 2}} \right)^2} = 16\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{D\, > \,0} \,\,\,\,\,\,\,\,{D \over 2} = 4\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = D = 8\,\,\,\,\,\,\,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

We can let r = the radius of the circular base of the gun's water tank, so the volume of the tank is:

V = πr^2 x 30

V = 30πr^2

Since the tank is filled with π cubic centimeters of water every second, and it takes 8 minutes, or 480 seconds, to fill the tank, the volume of the tank is also:

V = 480π

Therefore, we have:

30πr^2 = 480π

r^2 = 16

r = 4

Therefore, the diameter of the circular base of the gun's water tank is 2 x 4 = 8 cm.

Answer: D

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