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# Math Revolution and GMAT Club Contest! What is the greatest possible

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Math Expert
Joined: 02 Sep 2009
Posts: 57244
Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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19 Dec 2015, 08:43
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35% (medium)

Question Stats:

66% (01:46) correct 34% (01:53) wrong based on 345 sessions

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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:

Math Revolution and GMAT Club Contest

The Contest Starts November 28th in Quant Forum

We are happy to announce a Math Revolution and GMAT Club Contest

For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday).

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize:

PS + DS course with 502 videos that is worth $299! All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process. Thank you! _________________ ##### Most Helpful Community Reply Intern Joined: 21 Jan 2013 Posts: 39 Concentration: General Management, Leadership Schools: LBS GPA: 3.82 WE: Engineering (Computer Software) Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 20 Dec 2015, 04:30 8 3 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 _________________ -- Consider +1 Kudos if you find my post useful ##### General Discussion Intern Joined: 30 Sep 2013 Posts: 20 Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 19 Dec 2015, 10:43 3 3 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 Intern Joined: 12 Jul 2015 Posts: 25 Location: United States Concentration: Strategy, General Management WE: General Management (Pharmaceuticals and Biotech) Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 19 Dec 2015, 12:05 5 3 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 Attachments File comment: circles circles.png [ 27.71 KiB | Viewed 5638 times ] Intern Joined: 01 Nov 2015 Posts: 36 Location: India Concentration: Marketing, Entrepreneurship WE: Engineering (Computer Software) Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 20 Dec 2015, 20:46 2 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 Intern Joined: 09 Jul 2015 Posts: 49 Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 21 Dec 2015, 17:54 2 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 _________________ Please kudos if you find this post helpful. I am trying to unlock the tests Marshall & McDonough Moderator Joined: 13 Apr 2015 Posts: 1680 Location: India Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 21 Dec 2015, 20:43 5 2 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 Intern Joined: 06 Jun 2015 Posts: 15 Location: India Concentration: Marketing, Technology WE: Engineering (Energy and Utilities) Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 21 Dec 2015, 21:31 1 ANSWER IS 110 SINCE EACH CIRCLE CUTS ANOTHER ONE AT 2 POINTS, AND THERE ARE 11 CIRCLES SO, SIMILAR TO ARRANGEMENTS. 20+18+16+14+12+10+8+6+4+2 = 110 Intern Joined: 21 Nov 2014 Posts: 31 Location: Viet Nam GMAT 1: 760 Q50 V44 Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 22 Dec 2015, 02:38 1 Between two circles with different radii, there will be at most 2 intersected points. And we have C(11,2) = 55 ways of selecting 2 circles from 11 circles. The maximum number of points is: 55*2 = 110. Answer E. _________________ GMAT Group for Vietnamese: https://www.facebook.com/groups/644070009087525/ Senior Manager Joined: 23 Sep 2015 Posts: 371 Location: France GMAT 1: 690 Q47 V38 GMAT 2: 700 Q48 V38 WE: Real Estate (Mutual Funds and Brokerage) Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 23 Dec 2015, 07:40 1 Drawing the circles we see that 2 circles will intersect at 2 points, 3 circles at 6 points, 4 circles at 12... 5 at 20 and so on 2 x 1 =2 3 x 2 = 6 4 x 3 =12 5 x 4 = 20 The pattern is n x (n-1), so 11 circles will intersect at 11 x 10 different points. Answer E _________________ Intern Joined: 17 Aug 2014 Posts: 10 Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 24 Dec 2015, 10:21 2 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 Intern Joined: 05 Jun 2013 Posts: 35 Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 25 Dec 2015, 05:52 1 1 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. Retired Moderator Joined: 22 Jun 2014 Posts: 1096 Location: India Concentration: General Management, Technology GMAT 1: 540 Q45 V20 GPA: 2.49 WE: Information Technology (Computer Software) Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 25 Dec 2015, 16:01 1 we can draw circles on paper and find the pattern: 2 circles intersect at 2 points 3 circles intersect at 6 points 4 circles intersect at 12 points sequence for intersection points is 2, 6, 12, 20 ..... the difference between the two terms is 2 more than that of the previous two terms. For such a sequence nth term can be written as - n^2 + n 2 circles - 1st case - n = 1 then nth term will be = 1^2 + 1 = 2 3 circles - 2nd case - n = 2 then nth term will be = 2^2 + 2 = 6 4 circles - 3rd case - n = 3 then nth term will be = 3^2 + 3 = 12 . . . 11 circles - 10th case - n = 10 then nth term will be = 10^2 + 10 = 110 Option E is the correct answer. _________________ Intern Joined: 04 Sep 2015 Posts: 35 Location: Germany Concentration: Operations, Finance WE: Project Management (Aerospace and Defense) Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 27 Dec 2015, 03:14 1 1 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. Math Expert Joined: 02 Sep 2009 Posts: 57244 Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 27 Dec 2015, 12:17 Bunuel wrote: 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: Math Revolution and GMAT Club Contest The Contest Starts November 28th in Quant Forum We are happy to announce a Math Revolution and GMAT Club Contest For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday). To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific. Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests. PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize: PS + DS course with 502 videos that is worth$299!

All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time.

NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!

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[10(10+1)/2]=110. The correct answer is E.

Note: 1+2+…..+n=n(n+1)/2
_________________
Manager
Joined: 11 Oct 2016
Posts: 76
Location: India
GMAT 1: 610 Q47 V28
Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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21 May 2017, 06:50
Thanks Bunuel

For maximum number of points
2+2*2+2*3....

for minimum number of points , equation will be
1+2+3+4....

Please correct me if i am wrong .

Thanks
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Intern
Joined: 01 Mar 2017
Posts: 14
Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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07 Aug 2018, 21:08
How would the problem change if the radii were equal?

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Senior Manager
Joined: 17 Mar 2014
Posts: 437
Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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13 Aug 2018, 21:24
Juliaz wrote:
How would the problem change if the radii were equal?

Posted from my mobile device

math Revolution, could you answer above question
Re: Math Revolution and GMAT Club Contest! What is the greatest possible   [#permalink] 13 Aug 2018, 21:24
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