Suppose the number of squares with sides of length 4 units

Question

Suppose the number of squares with sides of length 4 units that can be formed by joining 4 points in the grid above is n4, while the number of squares with sides of length 5 units that can be formed by joining 4 points in the grid above is n5. Then n4/(n4+n5) is closest to =

(A) 0.51 (B) 0.61 (C) 0.64 (D) 0.75 (E) 0.96

cbrf3 · Answer

ans is "A"
Took the number of points to be 20 as it is divisible by 4 and 5
so the number would be 5/(4+5) = 5/45 = 0.51 approx

ps_dahiya · Answer

I am getting B.

n4 = 36
n5 = 25

n4/(n4+n5) = 36/61 = approx 0.6

kevincan · Answer

Just because I make the questions up doesn't mean I know the answers, so explain away. Many times people (esp. you) have come up with great explanations

ps_dahiya · Answer

Here is my explanation:
n4:
starting from top left to bottom right there are 36 points that can be the left edge of the 4 unit squares.

n5:
starting from top left to bottom right there are 25 points that can be the left edge of the 5 unit squares.

We should not count the diagonal squares because the side of those squares will not be 4 or 5 units but will be 4*SQRT(2) units and 5* SQRT(2) units.

cbrf3 · Answer

I took the lease common multiple of 4 and 5 which is 20 to be the number od points so with that...we can make 5 4 unit squares or 4 5 unit squares...whjch gives the fraction as 5/(5+4) which is 0.51....
Does this makes sense?

bondguy · Answer

if x is the length of the grid , then total area is x^2 square units
area of square of 4 units is 16 square units
no. of squares of 4 units = x^2/16 = n4
similiarly
no. of squares of 5 units = x^2/25 = n5

n4/(n4 + n5) = 
so n4/(n4 + n5) = (x^2/16) / (x^2/16 + x^2/25)
= (1/16)/(1/16 + 1/25)
= (1/16)/(25 + 16)/16*25
= 25/(25+16)
= 25/41
= 0.61

natehill · Answer

I get the same as bond guy, but by thinking about it differently

There are 5 positions vertically and 5 positions horizontally for a 4x4 square.
n4 = 25
The 5x5 square can take 4 unique positions vertically and horizontally.
n5 = 16
n4/(n4+n5) = 25/41 = ~0.61

kevincan · Answer

Must squares have sides that are vertical and horizontal?

freetheking · Answer

If so, 

n4 = 6*6 
n5 = 5*5 + ways of form 8*6 and 6*8 rectangles. 
= 5*5+8*2 

thus, 36/(25+16+36) = 0.47.......closest to A

Zooroopa · Answer

I suppose we need to join 5 points to get a length of 5 units.

If we consider non-vertical and non-horizontal sides, as explained by ps_dahiya we don't get 4 or 5 units but would get 4√2 or 5√2 units (by joining 4 or 5 points diagonally).

My result is close to one of the choices but isn't matching exactly.

I set the bottommost point as (0,0) and got a grid ranging from (0,0) to (8,0) and (0,0) to (0,8) with squares in every integral coordinates.

N4 = {set of squares starting from (0,8) and going left and going down} + {set of squares starting from (1,7) and going left and going down}
+ {set of squares starting from (2,6) and going left and going down}
+ {set of squares starting from (3,5) and going left and going down}
= (5+5) + (4+4) + (3+3) + (2+2) + (1+1) = 30

Similarly N5 = 16

So the answer that I'm getting is 30/46 = 0.65. This is close to C but not exactly so.

Am I missing some square units?

Zooroopa · Answer

Holy Cow! I think I am off!

Kevin - the points need not be consecutive. Is this true?

kevincan · Answer

I don't agree with this

kevincan · Answer

Clue:What can be the ratio of the sides of a right trinagle with hypotenuse 5?

freetheking · Answer

Did I do something wrong???
ways of form 8*6 and 6*8 rectangles. => 3:4:5..

kevincan · Answer

See attached

Suppose the number of squares with sides of length 4 units

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards