A spherical balloon originally had a radius of x centimeters. The ball

Question

A spherical balloon originally had a radius of  x  centimeters. The balloon was partially deflated by letting out some air so that the volume of the balloon decreased by  a\% . If the balloon remained spherical while the air was being let out, which of the following expressions in  x  and  a  gives the radius, in centimeters, of the partially deflated balloon?

(Note: The volume of a sphere with radius r is  \frac{4}{3}πr^3 .)

A.  x(100 - a)^3

B.  x(1 - \frac{a}{100})^3

C.  x(\sqrt {100 - a})

D.  x(\sqrt {1 - \frac{a}{100}})

E.  x(1 - \sqrt {\frac{a}{100}})

JeffTargetTestPrep · Accepted Answer

The original volume of the balloon is:
(4/3)(pi)(x^3)

Since an a% decrease corresponds to a factor of (1 - a/100), the volume of partially deflated balloon is:
(4/3)(pi)(x^3)(1 - a/100)

If r is the radius of the partially deflated balloon, we have:

(4/3)(pi)(r^3) = (4/3)(pi)(x^3)(1 - a/100)

r^3 = (x^3)(1 - a/100)

r = (x)(root (1 -a/100))

Answer: D

lg800 · Answer

Any suggestions on ways to do this by plugging in?

DanTheGMATMan · Answer

You could just choose values for x and a but because of the complexity of the formula for volume here, I don't think it would make it much easier

DanTheGMATMan · Answer

A possible solution:

https://www.youtube.com/watch?v=yNeCzIJOEkU

liberoaut · Answer

Feel like I missed something, is there any reason why I can't simplify the cube root of 1-a/100 to 1 - cuberoot(a/100), therefore making E the answer?

Bunuel · Answer

In general,  \sqrt {1 - \frac{a}{100}}  is not equal to  1 - \sqrt {\frac{a}{100}} .

Usernamevisible · Answer

Bunuel  - given geometry is not there, can this question come in exam ?

Bunuel · Answer

This is GMAT Prep Question: https://gmatclub.com/forum/download/file.php?id=88182 It does not test any specific geometry knowledge, and the formula is provided, so questions like this can still appear on the exam. GMAT-Club-Forum-kzhumagh.png

Adit_ · Answer

4/3pix^3  = 4/3pi*y^3
y value is what we need.

Cancelling 4/3pi:
x^3  = y^3
Taking cube root:

y = x* ^1/3

Answer:  Option D

tanvi0915 · Answer

Using Percent formula along with volume
(4/3pix^3 - 4/3pir^3) / 4/3pix^3 = a/100
4/3pi(x^3 - r^3)/ 4/3pix^3 = a/100
(x^3 - r^3) = ax^3/100
x^3 - ax^3/100 = r^3
x^3 ( 1 - a/100) = r^3
Taking cube root:
r = x (1 - a/100 ) ^1/3

MakB · Answer

Why did we made volume of fully inflated balloon equal with deflated balloon ?
Wont it be less than the original?
Didnt get this one

Bunuel · Answer

We are not making the deflated volume equal to the original volume.

The right side,  \frac{4}{3}\pi x^3(1-\frac{a}{100}) , is already the  reduced  volume after the a% decrease. The left side,  \frac{4}{3}\pi r^3 , is the volume of the deflated balloon using its  new  radius  r .

So we are simply saying:  new volume written using r = new volume written as a reduced part of the original volume :

\frac{4}{3}\pi r^3=\frac{4}{3}\pi x^3(1-\frac{a}{100})

A spherical balloon originally had a radius of x centimeters. The ball

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A spherical balloon originally had a radius of x centimeters. The ball

Prep Toolkit

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