What is the greatest possible number of points at which 11 circles wit

Question

Math Revolution and GMAT Club Contest Starts!

QUESTION #14:

What is the greatest possible number of points at which 11 circles with different radii intersected one another?

A. 45
B. 60
C. 85
D. 90
E. 110

Check conditions below:

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rockthegmat007 · Accepted Answer

Any Circle intersects another circle at maximum two intersection points.

With two circles : greatest number of intersection points = 2 
With third circle introduced to the first two (already intersecting at two points), the third circle can intersect both the circles at two different intersection points, introducing  4 more   intersection points ( third circle cuts each of the first two at two different poitns) . in addition to the two intersection points of first two.
i.e., with three circles , max # of intersection points = 2 + 2*2 = 6
with 4th circle introduced, it cuts the first three circles at two intersection points each i.e, 2*3 additional intersection points introduced now.
with 4 circles , max # of intersection points = (2 + 2*2)  (intersection points for 3 circles)  + 2*3  (additional intersection points introduced by 4th circle)   = 2 + 2*2 + 2*3 
Similarily , with 5th circle . total = 2 + 2*2 + 2*3 + 2*4
we see a series emerging here 
greatest number of of intersection points for m circles = 2 + 2*2 + 2*3 + .....+ 2*(m-1)

Applying the series extrapolation for 11 circles, greatest number of intersection points = 2 + 2*2 + 2*3 + .....+ 2*10

= 2 * (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 2 * (110)/2 = 110  ( calculation for the sum shown below) 
-- Calculation for the sum.
since the sum in the paranthesis is a sum of consecutive integers , we can calculate the average by (last + first) / 2 
Sum in paranthesis = average * # of terms = 10*((1 + 10)/2) = 110/2 
--

Correct Answer : E

dayasantoshi · Answer

The first 2 circles intersect at 2 points.
3rd circle can intersect the first two circles at 4 points. refer to the figure.

i.e Circle 1 = no intersection
 Circle 2 = 2 points
 Circle 3 = 2+4 points
 Circle 4 = 2+4+8 points
 .
 .
 .
 Circle 11= 2+4+8+10+12+14+16+18+20 = n(n+1)= 10*11=110 points.
 Implies N circles can be drawn to intersect N(N-1) points . In this case, 11 circles can be drawn that can intersect at 11*10=110 points

Ans E

Bunuel · Answer

Similar questions: https://gmatclub.com/forum/which-of-the ... 92274.html (triangle) https://gmatclub.com/forum/which-of-the ... 38952.html (parallelogram) https://gmatclub.com/forum/which-of-the ... 14013.html (square) https://gmatclub.com/forum/what-is-the- ... 97855.html (10 circles) https://gmatclub.com/forum/gmatbusters- ... 16475.html (6 circles) https://gmatclub.com/forum/what-is-the- ... 10362.html (11 circles) https://gmatclub.com/forum/which-of-the ... 80597.html (two parallel lines) Hope it helps.

riteshgmat · Answer

Answer is E i.e 110

Two circles can intersect at maximum 2 points.

Three circles can intersect at maximum 6 points..

Because we select 2 circles from 3 and each two circles can intersect at maximum 2 points

2 c^3_2  ... Three circles can intersect at 6 points

So 11 circles can intersect at maximum

2c^{11}_2  ... 2 * 11 * 5 = 110

appsy01 · Answer

1st Circle doesn't intersect itself - 2*(1-1)=0 points
2nd Circle intersects 1st at 2*(2-1)=2 Points
3rd Circle intersect 2nd and 3rd at 2*(3-1)=4 Points
4th Circle intersects other Circles at 2*(4-1)=6 points
...
11th circle intersects other circles at 2*(11-1)=20 points

Therefore total no of points = 2 + 4 + 6 + ... + 20
 Sum = 10/2 * (20+2) = 5*22 =110

Answer is E

kingjamesrules · Answer

Answer is E -> 110.

2 circle intersects at 2 points, 3 circles intersect at 4 points, 4 circles intersect at 6 points, 5 circles at 8 points, etc., Hence max # of points will be the result of addition of these points - as you can draw a circle anyway you want to ensure there are distinct points with each of these circle's intersection.

S = 2+4+6+... +20 (n=10) as there are 11 circles 
S = (a1+an)/2*n = (2+20)/2*10 = 110

Kurtosis · Answer

Given: Number of circles with different radii = 11

We can draw a few circles to obtain the below results

For 2 circles, max number of points of intersection = 2 = 2 * 1
For 3 circles, max number of points of intersection = 6 = 3 * 2
For 4 circles, max number of points of intersection = 12 = 4 * 3
We can observe a pattern here.

For n circles with different radii, max number of points of intersection can be deduced from the above results as n * (n-1).

Therefore, for 11 circles with different radii, max number of points of intersection = 11 * 10 = 110

Answer: E

abhib3388 · Answer

The answer is E
By drawing circles and deducing
2 circles intersect at 2 points(1x2)
3 circles intersect at 6 points(2x3)
4 circles intersect at 12 points(3x4)
11 circles intersect at 10x11=110 points

sri1sastra · Answer

2 circles can intersect at the most 2 points. ---> which is 2P2
3 circles can intersect at the most 6 points. ----> which is 3P2

Therefore it is similar to number of arrangements possible for 11 objects taken two at a time.
Therefore 11 circles with different radii will intersect at 11P2 = 11!/9! = 11*10= 110 points. Answer choice E.

MK1480 · Answer

1 circles have 0 intersection point ( 1 x 0)
2 circles have 2 intersection points (2 x 1)
3 circles have 6 intersection points (3 x 2)
4 circles have 12 intersection points (4 x 3)
....
....
a pattern develops!
So, 11 circles have (11 x 10) = 110 intersection points

However, an expression exists as well ie; C x (C-1) = I, where C is the number of circles and I the number of intersections.

Bunuel · Answer

MATH REVOLUTION OFFICIAL SOLUTION:

The greatest possible number of points of intersection between two circles is 2. The greatest possible number of points of intersection of three circles is 2+2*2 (The greatest possible number of points of intersection between two circles + The greatest possible number of points of intersection that one circle can have with other two circles). The greatest possible number of points of intersection of four circles is 2+2*2+2*3 (The greatest possible number of points of intersection of three circles + The greatest possible number of points of intersection that one circle can have with other two circles).

So, the greatest possible number of points of intersection of 11 circles is, 2+2*2+2*3+…..2*10=2(1+2+…..+10)=2 =110.  The correct answer is E.

Note: 1+2+…..+n=n(n+1)/2

Louis14 · Answer

Math Revolution and GMAT Club Contest Starts!

QUESTION #14:

What is the greatest possible number of points at which 11 circles with different radii intersected one another?
 
A. 45 
B. 60 
C. 85 
D. 90 
E. 110 
quote]

IanStewart Sir, can we expect such a question on the GMAT? I've never seen anything like it before in my long, arduous preparation. If possible, could you please explain the concept in simpler terms? Thanks a lot.

IanStewart · Answer

I've seen several thousand official questions, and I've never seen a question similar to this one. So it's very unlikely you'd encounter something like this on a real test. But I suppose it's not impossible. If a question doesn't require any specialized math knowledge and can be solved within 2 minutes, then it's within the scope of the GMAT, and that's true of this problem.

If you notice that two distinct circles intersect at 2 points at most, then to count the number of points where 11 circles can meet, we just need to count how many pairs of circles we have, which is 11C2 = (11)(10)/2! = 55, and multiply by 2, because each pair of circles meets in 2 places, not 1, so the answer is 2*55 = 110.

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