The area of circle O is added to its diameter. If the circumference

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: 4π + 4 - 4π = 4
Simplify again: 4 = 4
PERFECT!

Answer:

ALTERNATE SOLUTION
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: (πr + 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

Area + Diameter - Circumference = 4

i.e. pi*r^2 + 2r -2*pi*r = 4

i.e. pi*r^2 + r * (2-2*pi) -4 = 0

Find roots using

[ -b +- root (b^2 - 4ac)] / 2a

The root comes out to be root (4*(pi+1)^2) = 2*(pi+1)

Solving,

b = 2 - 2*pi
a = pi
c = -4

So, the roots become

(-2 + 2*pi + 2*pi + 2)/(2*pi) and (-2 + 2*pi - 2*pi -2)/(2*pi)

Simplifying gives

r = 2 and r = -2/pi

Since radius cannot be negative,

r = 2

Answer is (B)

pi*r^2 + 2r -2*pi*r = 4

Simplifying the equation: pi*r(r-2)+2r=4

Without much algebraic: We can Test the answers quickly, then 2 is the only possible answer that will eliminate pi from equation.
Answer is B

answer is 2.

2r+pi*r2-2 pi r=4

KAPLAN OFFICIAL SOLUTION:

The key to solving this problem within the two-minute time frame on the GMAT is realizing what it is really testing. As is the case with many GMAT problems, this is not the type of question it seems to be at first. Many students, seeing information about a circle, start drawing a picture. If this were really a geometry problem, that would be the correct first step. However, this is actually an algebra problem in disguise.

The correct first step is to translate the information in the problem into an equation. ‘The area of a circle is added to its diameter’ becomes \(\pi r^2 + d\). Since we know that the diameter is twice as long as the radius, we can rewrite d as 2r, making the expression \(\pi r^2+ 2r\). Next, we are told that the circumference of the circle is subtracted from this total, making the expression \(\pi r^2 + 2r – 2\pi r\). Finally, we know that the result is 4. So, the entire equation is \(\pi r^2+ 2r – 2 \pi r = 4\). As the problem asks for the radius of the circle, all we need to do now is solve for r, which can be done in the following manner:

\(\pi r^2 + 2r – 2\pi r = 4\)

\(\pi r^2+ 2r – 2 \pi r – 4 = 0\)

At this point, be sure to note that you have a quadratic, which can be factored to:

\((\pi r + 2)(r – 2) = 0\)

As is the case with most quadratics, this equation has two solutions. Either \(\pi r + 2 = 0\), in which case \(r = -\frac{2}{\pi}\), or \(r-2 = 0\), in which case \(r = 2\).

Looking at the answer choices, you will notice that both of these are listed as options. But, because this problem is referring to a circle’s radius, which can only have a positive value, r must be positive and thus must equal 2.

Answer: B.
_________________

Its quicker to just plug in the values.

i have a doubt in this question? is it a GMAT standard question?

how can we add area (unit^2) to diameter (unit) and circumference (Unit).

The quickest way to do this is to simply plug in.

For future reference, if you can't get something going by 1 minute in, try to plug the answers in.

Hi All,

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) =
4π and 4π

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

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

πr^2 + 2r - 2πr = 4

πr^2 + 2r - 2πr - 4 = 0

r(πr + 2) - 2(πr + 2) = 0

(r - 2)(πr + 2) = 0

r = 2

or

πr + 2 = 0

πr = -2

r = -2/π

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

Answer: B
_________________

tHIS IS HOW I SOLVED

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

OR r² + 2r - 2r = 4
oR r² = 4 or r =2 ans B

With the exception of the (-)negative solution, each answer for the radius is a whole number.

Once you translate the question stem into algebra, you have two terms with (pi).

In order for the answer to be a Whole Number such as 2, 3, 4, or 5, these two terms with (pi) MUST CANCEL each other out.

(pi) * (r)^2 = (2) (pi) (r)

(r)^2 = 2(r)

(r)^2 - 2r = 0

(r) (r - 2) = 0

Since the circle actually exists, r must be a positive length

r = 2

Finally, to ensure that our assumption is valid, we can always test the answer choice.

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Alternative approach from a units standpoint.

An area with dimension L^2 where L is the dimension of length has both added and subtracted from it the dimension of length, so only the lengths can be combined to maintain integrity of the dimensions.

Yet the result is 4, with dimension unstated.

The only dimension it could be is L^2, area, since no area was subtracted.

So the length dimension must be 2, the radius.

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