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A elderly man, in order to practise manual dexterity, draws

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A elderly man, in order to practise manual dexterity, draws [#permalink] New post 25 Jul 2006, 00:57
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A
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C
D
E

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A elderly man, in order to practise manual dexterity, draws 50 triangles a day by connecting three of the solid points on the grid made up of 16 squares shown in the attachment. He starts this activity on January 1 and is careful not to skip a day. On which day will he be forced to retrace one of his previously drawn triangles?



(A) February 12 (B) February 13 (C) February 14 (D) February 15 (E) None of these
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 [#permalink] New post 25 Jul 2006, 03:58
Think it is like that
Total number of selecting 3 poins out of 25 is 2300 triplets. Now remove 5C3*10 for the colinear triplets in vertical and horizontal lines of the grid. Remove 5C3*2 for the 2 diagonals 4C3*4 for the lines parallel to the diagonals, and finally 3C3 for the outer parallel lines to the diagonals. In total we have: 2300-( 100+20+16+4)=>2160 triangles. divide by 20 to find in how many days the different triangles will be depleted.
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 [#permalink] New post 25 Jul 2006, 04:48
Answer E

As shown in the figure there are (5*5)25 distinct points in the figure.
To draw each triangle you need 3 distinct points.
He cannot retrace a line already drawn. It means that he cannot select a combination of points for a line which has already been selected.

Now , you as look at this problen as follows
How many ways you can select a combination of 3 points from 25 points ?
Using the formula for Combination
25!
--------- = 2300 , So he can draw 2300 triangles without retracing.
3!(25-3)!

Old man draws 50 triangles a day , so he will need 46 days.
January has 31 days , So he would finish drawing all possible triangles by Feb 15th. On the 16th he would have to retrace . Hence E
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 [#permalink] New post 25 Jul 2006, 23:36
In each square there are 4 points. YOu can join three points to make 3 triangles.

hence with 16 squares you can make 16*3 = 48 triangles.

Hence in the 49th day he will be forced to retrace one of the triangles.

49th day is not one of the options. Hence E
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 [#permalink] New post 25 Jul 2006, 23:49
SInce he is drawing 50 triangles a day, and can draw 2160 triangles then in January he will have drawn 50*31=1550 triangles. He has 12 more days to draw different triangles, and on 13 of February he will start redrawing his triangles.
So the ans is B
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 [#permalink] New post 26 Jul 2006, 00:35
Still no definite clear answer&explanation? Please dont leave questions unaswered:(
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 [#permalink] New post 26 Jul 2006, 02:32
I thank you for your patience. I want somebody to have the pleasure of solving it. BG is close...
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 [#permalink] New post 26 Jul 2006, 11:52
Total triangles possible = 25C3 - 10 * 5C3 = 2200

Total days = 2200/50 = 44
He will finish on Feb 13
On Feb 14 he is forced to redraw.

Answer = C
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 [#permalink] New post 26 Jul 2006, 13:55
25 points. 3 needed to make a triangle

Ways of choosing 3 pts = 25c3
invalid cases (3 pts on same line) = 10 x 5c3

triangles = 25c3 - 10x5c3 = 2200

days these will last = 2200/50 = 44

Jan has 31 days so 13 left in feb.

on 14th feb, he will reuse. C?
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 [#permalink] New post 26 Jul 2006, 14:11
BG is closer, but you're on the right track!
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 [#permalink] New post 26 Jul 2006, 14:28
Ok i figured some more stuff....

25C3 = 2300

Number of straight lines with 5 points = 12 (gives 12x5c3 = 120)
number of straight lines with 4 points = 4 (gives 4x4c3 = 16)
'' 3 '' = 16 (give 16x3c3 = 16)

So subtract 152 from 2300 and we get 2148

2148/50 = 42 days => 11 days in feb so feb 12th he will restart? A?
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 [#permalink] New post 26 Jul 2006, 14:42
ps_dahiya wrote:
Futuristic wrote:
'' 3 '' = 16


where are 16???


One is from top centre to bottom left. Did you like this question?
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 [#permalink] New post 26 Jul 2006, 14:45
The 16 lines with 3 points is the tricky part. Since I don't have the time to draw it out consider the grid to be extending from 0,0 into the 1st quadrant of the xy plane only. Now the following sets of co-ordinates will define lines having exactly 3 points in the grid.

1. (0,0), (2,4)
2. (1,0), (3,4)
3. (2,0), (4,4)

Rotate the grid by 90 degrees and you'll get 3 more unique lines in the grid for a total of 12.

For the remaining 4 lines, consider the following co-ordinates

1. (0,2),(2,4)
2. (2,0),(4,2)

In this case, you will find there are 4 unique lines.

Hope this clarifies it. kevin this was a killer. i hope you don't ever get to set a gmat paper!
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 [#permalink] New post 26 Jul 2006, 14:47
Thanks to everybody who tried this. Shall I make it easier for my book?
  [#permalink] 26 Jul 2006, 14:47
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