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Marty has a right circular cylindrical pool of diameter 12 f
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25 Nov 2020, 09:31
25 Nov 2020, 09:31
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Question Stats:
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Marty has a right circular cylindrical pool of diameter 12 feet and his neighbor, Rusty, has a right circular cylindrical pool of diameter 18 feet. If the depths of the pools are equal, then the volume of water in Rusty’s pool is how many times that in Marty’s pool?
A. 1.5
B. 2.25
C. 2.5
D. 4
E. 4.25
A. 1.5
B. 2.25
C. 2.5
D. 4
E. 4.25
Re: Marty has a right circular cylindrical pool of diameter 12 f
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31 Dec 2020, 20:13
31 Dec 2020, 20:13
1
Depth of pools is same.
Radius of marty's pool R1 = 6ft
Volume of marty's pool Vm = πr^2h = π * 36 * h
h = \(\frac{Vm}{36*π}\)
Now,
Radius of Rusty's pool R2 = 9ft
Volume of Rusty's pool Vr = πr^2h = π * 81 * h
h = \(\frac{Vr}{81*π}\)
Compare h
\(\frac{Vm}{36*π}\) = \(\frac{Vr}{81*π}\)
Vr = \(\frac{81*Vm}{36}\)
Vr = 2.25 Vm
Answer B
Radius of marty's pool R1 = 6ft
Volume of marty's pool Vm = πr^2h = π * 36 * h
h = \(\frac{Vm}{36*π}\)
Now,
Radius of Rusty's pool R2 = 9ft
Volume of Rusty's pool Vr = πr^2h = π * 81 * h
h = \(\frac{Vr}{81*π}\)
Compare h
\(\frac{Vm}{36*π}\) = \(\frac{Vr}{81*π}\)
Vr = \(\frac{81*Vm}{36}\)
Vr = 2.25 Vm
Answer B
Marty has a right circular cylindrical pool of diameter 12 f
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26 Jan 2021, 07:57
26 Jan 2021, 07:57
1
Try to reduce the calculation and save time whenever possible.
We are asked to find out by how many times the water in the larger cylindrical pool is more than the smaller one.
The volume in a right cylindrical pool can be calculated with the formula \(πr^2*h\)
Since π and height(h) are common in the calculation of volume in both the pool we can cancel them out.
In effect the multiple by which the water in the larger pool is more than the smaller pool can be calculated as:
\(\frac{R^2}{r^2}\) where R=radius of the larger pool and \(r\) = radius of the smaller pool.
radius = \(\frac{1}{2}\) * diameter.
Hence the multiple is \(9^2/6^2\) = \(\frac{81}{36}\) = 2.25
Answer is B
We are asked to find out by how many times the water in the larger cylindrical pool is more than the smaller one.
The volume in a right cylindrical pool can be calculated with the formula \(πr^2*h\)
Since π and height(h) are common in the calculation of volume in both the pool we can cancel them out.
In effect the multiple by which the water in the larger pool is more than the smaller pool can be calculated as:
\(\frac{R^2}{r^2}\) where R=radius of the larger pool and \(r\) = radius of the smaller pool.
radius = \(\frac{1}{2}\) * diameter.
Hence the multiple is \(9^2/6^2\) = \(\frac{81}{36}\) = 2.25
Answer is B
