# Tricky circular area math problem from the GRE

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1. Aug 19, 2013

### mgier001

Hey guys, I was wondering if anyone could help me out with this problem from the GRE practice exam:

I honestly don't know how to go about solving it. I'm guessing that the answer is either B or D, however I know that in this GRE exam you can never guess... My friends think the answer is B but I have not been able to come up with any logical reasoning behind any of the answer choices.
This problem is tricky so if anyone can help me through it I would greatly appreciate it.
Thanks,
-Matt

2. Aug 19, 2013

### mgier001

Is this thread misposted? I'm not sure if this problem would fall under a homework/coursework category since it technically is not homework or coursework... :\

3. Aug 19, 2013

### Mandelbroth

The area of each circle is $x$. The intersection of two circles has area $y$. The intersection of all three has area $z$. Quantity A is given by $y+(x-2y-z)$. What does this tell you?

Technically, I think this goes under homework.

4. Aug 19, 2013

### mgier001

Well the area of each circle x=40, and intersection y=15. However, z is not given. I would guess z to be 7.5, but I don't want to simply guess an answer.

Solving with 7.5 yields: 15+(40-(2x15)-7.5) = 17.5

5. Aug 19, 2013

### Mandelbroth

All you need to know is that $z$ is nonnegative. Other than that, it doesn't really matter what $z$ is. Try it. Algebra is fun. :tongue:

6. Aug 19, 2013

### mgier001

Do you mind explaining the equation? I would have subtracted 3y and added 2z.
x-3y+2z

7. Aug 19, 2013

### lurflurf

we have 7 regions
3A
3B
1C
each circle is made up of 1A+2B+C
so 40=a+2b+c
each intersection of two circles is made up of B+C
so 15=b+c
so 25=a+b

8. Aug 19, 2013

### Redbelly98

Staff Emeritus
Doing all that algebra is wasting valuable test-taking time. Take the smaller shaded region, and instead shade in one of the other two regions that have the same shape. Look at the overall shaded region that results.

p.s. Mod's note: moving this to homework area.

9. Aug 20, 2013

### reenmachine

Put the small shaded area in the circle that has the large shaded area , it leaves you with the intersection of two circles which is 15.

So the shaded area is 40-15 = 25.

Not sure if it's suppose to be more complicated , but the statement that all the intersections of two circles are identical intuitively suggest that each of the three small areas surrounding the triple intersection are of the same size.Do you think you had to prove that?

10. Aug 20, 2013

### skiller

I can't see where you get $y+(x-2y-z)$ from. Using your labelling:

the larger shaded area = $x - 2y + z$

and the smaller shaded area = $y - z$

so Quantity A = $(x - 2y + z) + (y - z) = x - y$

As a couple of of the posters have already stated, it is intuitively obvious that it is $x - y$ if you "move" the smaller shaded area into the other circle.

11. Aug 21, 2013

### BrettJimison

I agree with Red Belly, no algebra is needed. This is a logic question, no maths involved.

12. Aug 22, 2013

### Mandelbroth

I misinterpreted their meaning of intersection of 2 circles as "the intersection of 2 and only 2 circles." I apologize.

Regardless, B is correct.