The radius of a sphere was increased by p percent. If the volume of th

Question

The radius of a sphere was increased by p percent. If the volume of the sphere was increased by 33.1%, what is the value of p?

(A) 5
(B) 10
(C) 11
(D) 12
(E) 15

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Anurocks1233 · Answer

One can solve it using compound interest formula as - in volume r is cubed , so think radius is 100 and with 3 interval of increase it should be 133.33 
Start with B
100 increased by 10% three times
100 * 1.1 = 110
110 same = 110+11 = 121
121 +12.1 = 133.1
Hence this the answer

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gmatchile · Answer

Remember. An increase is a positive change between a final amount and an initial amount.

An increase of 20% is:

Final amount - Initial amount = 20% of the initial amount
Final amount - initial amount = (20/100) x (Initial amount)

Thus, if the volume of a sphere was increased by 33.1%, then:

Final Volume - Initial Volume = 33.1% of Initial Volume

This is:

Final volume - initial volume = (33.1/100) x (initial volume)

Final volume = initial volume + (33.1/100) x (initial volume)

Final volume = (100/100) x (initial volume) + (33.1/100) x (initial volume)

Final volume = (133.1/100) x (initial volume)

4/3Pi(R)exp3 = (133.1/100) x 4/3 Pi(r)exp3

Then (R)exp3 = (133.1/100)x(r)exp3

Applying cube root to both sides, we have:

R = (133.1/100)exp(1/3) x r

R = (1331/1000) exp(1/3) x r

R = (11/10) x r

R - r = (11/10) xr -r
R -r = (11/10)xr - (10/10)xr
R-r = (1/10)xr
R-r = (10/100)xr

Then the radius increased by 10%

Answer B

hadimadi · Answer

Hi,

V_after=(4/3)*pi*(r*(1+p))^3
V_before=(4/3)*pi*r^3

V_after/V_before=1.33
 / =1.33
((r*(1+p))^3)/r^3=1.33
(r^3*(1+p)^3)/r^3=1.33
(1+p)^3=1.33
p^3+3p^2+3p+1=1.33
p^3+3p^2+3p=0.33
p(p^2+3p+3)=0.33

(A) 0.05*(0.05^2+3*0.05+3)=0.05*(0.0025+0.15+3)=0.05*(3.1525), so not (A)
(B) 0.1*(0.1^2+3*0.1+3)=0.1*(0.01+0.3+3)=0.1*(3.31)=0.331 -> (B)

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The radius of a sphere was increased by p percent. If the volume of th

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