The area of circle O is added to its diameter. . .

[Vincen]

The area of circle O is added to its diameter. If the circumference of circle O is then subtracted from this total, the result is 4. What is the radius of circle O?

A) -2/pi
B) 2
C) 3
D) 4
E) 5

The OA is B.

What are the calculus needed to solve this PS question?

[Brent@GMATPrepNow]

The area of circle O is added to its diameter. If the circumference of circle O is then subtracted from this total, the result is 4. What is the radius of circle O?

A) -2/Ï€
B) 2
C) 3
D) 4
E) 5

Let r = radius of circle
Area of circle = πr²
Diameter = 2r
Circumference of circle = 2rπ

The area of circle O is added to its diameter...
We get: πr² + 2r

...If the circumference of circle O is then subtracted from this total, the result is 4
We get: πr² + 2r - 2rπ = 4
From here, we CAN just test the answer choices.
First, we can SKIP answer choice A, since the radius cannot have a negative value.

What about B (2)
Replace r with (2) to get: π(2)² + 2(2) - 2(2)π = 4
Simplify: 4Ï€ + 4 - 4Ï€ = 4
Simplify again: 4 = 4
PERFECT!

Answer: B

Cheers,
Brent

[Brent@GMATPrepNow]

The area of circle O is added to its diameter. If the circumference of circle O is then subtracted from this total, the result is 4. What is the radius of circle O?

A) -2/pi
B) 2
C) 3
D) 4
E) 5

Another approach:

Once we get the equation πr² + 2r - 2rπ = 4, we can also solve it algebraically.
To do so, we're going to factor the expression in parts
Here's what I mean...

We have: πr² + 2r - 2rπ = 4
Subtract 4 from both sides to get: πr² + 2r - 2rπ - 4 = 0
Rearrange terms to get: πr² - 2rπ + 2r - 4 = 0
Factor as follows: πr(r - 2) + 2(r - 2) = 0
Notice that we have (x-2) in common in both parts.
So, we can combine the parts to get: (Ï€r + 2)(r - 2) = 0
This means that EITHER πr + 2 = 0 OR r - 2 = 0
case a: πr + 2 = 0
This means: πr = -2
Since π and r are both POSITIVE, this equation has NO SOLUTION

case b: r - 2 = 0
This tells us that r = 2

Answer: B

Cheers,
Brent

[[email protected]]

Hi Vincen,

This question can be solved rather handily by TESTing THE ANSWER (one of the approaches that Brent showed), so I won't rehash any of that here. Instead, I want to point out a Geometry relationship that you might be able to use on Test Day.

In this prompt, we needed the "Pi"s to cancel out, since we were left with the number 4.

We were dealing with (π)(r²) and 2(π)(r) and the only way for those terms to cancel out is if r=2...

(π)(2²) and 2(π)(2) =
4Ï€ and 4Ï€

The geometry pattern is in reference to the radius of the circle. When comparing the area of a circle with the circumference of that same circle, the two values will be equal ONLY when the radius is 2.

If the radius is LESS than 2, then the circumference will be LARGER than the area.
If the radius is GREATER than 2, then the area will be LARGER than the circumference.

This math relationship might be useful on certain DS prompts and on PS questions in which you might need to approximate a value based on the radius (or vice versa).

GMAT assassins aren't born, they're made,
Rich

[Scott@TargetTestPrep]

The area of circle O is added to its diameter. If the circumference of circle O is then subtracted from this total, the result is 4. What is the radius of circle O?

A) -2/pi
B) 2
C) 3
D) 4
E) 5

We can let r = radius of the circle and create the following equation:

Ï€r^2 + 2r - 2Ï€r = 4

Ï€r^2 + 2r - 2Ï€r - 4 = 0

r(Ï€r + 2) - 2(Ï€r + 2) = 0

(r - 2)(Ï€r + 2) = 0

r = 2

or

Ï€r + 2 = 0

Ï€r = -2

r = -2/Ï€

Since r can't be negative, r must be 2.

Answer: B