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

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

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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

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.

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

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

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!