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

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

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New post 19 Dec 2015, 07:43
2
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A
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D
E

Difficulty:

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Question Stats:

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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!


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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 20 Dec 2015, 03:30
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
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 19 Dec 2015, 09:43
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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
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 19 Dec 2015, 11:05
5
1
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
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 20 Dec 2015, 19:46
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
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 21 Dec 2015, 16:54
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
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 21 Dec 2015, 19:43
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
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 21 Dec 2015, 20:31
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
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 22 Dec 2015, 01:38
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.
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 23 Dec 2015, 06:40
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
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 24 Dec 2015, 09:21
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
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 25 Dec 2015, 04:52
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.
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 25 Dec 2015, 15:01
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.

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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 27 Dec 2015, 02:14
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.
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 27 Dec 2015, 11:17
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
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 21 May 2017, 05:50
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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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 07 Aug 2018, 20:08
How would the problem change if the radii were equal?

Posted from my mobile device
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible  [#permalink]

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New post 13 Aug 2018, 20:24
Juliaz wrote:
How would the problem change if the radii were equal?

Posted from my mobile device


math Revolution, could you answer above question
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Re: Math Revolution and GMAT Club Contest! What is the greatest possible &nbs [#permalink] 13 Aug 2018, 20:24
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