insects_geometry

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gmatmachoman: Legendary Member; Posts: 2326; Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members; Thanked: 173 times; GMAT Score:710

by gmatmachoman » Sun May 30, 2010 2:06 am

An insect is crawling to climb a 20 metre long pipe of diameter 3.5 metres. If the insect take 4
circular rounds to reach the top of the building, then what will be total distance (in nearer
metres) traveled by insect to reach the top?

(1) 47 m (2) 48 m (3) 49 m (4) 50 m

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Source: Beat The GMAT — Problem Solving | Sun May 30, 2010 2:06 am

clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Sun May 30, 2010 2:57 am

i got 3,5*5*pi=52,5 that is close to E
is it right answer

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gmatmachoman: Legendary Member; Posts: 2326; Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members; Thanked: 173 times; GMAT Score:710

by gmatmachoman » Sun May 30, 2010 4:32 am

clock60 wrote:i got 3,5*5*pi=52,5 that is close to E
is it right answer

I am not able to get what u keyed in!

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liferocks: Legendary Member; Posts: 576; Joined: Sat Mar 13, 2010 8:31 pmFollowed by: 1 member; Thanked: 97 times

by liferocks » Sun May 30, 2010 6:07 am

path of traverse of the ant can be divided into 8 equal line whose length equal to the length of hypotenuse of a right angled triangle with onr side 3.5 and another 20/4=5
so length is 8*V(5^2+49/4)=8V(149/4)=4V149
Now V149=12.2 approximately
hence length of path=4*12.2=48.8 or 49 approximately

Ans option D
What is OA?

"If you don't know where you are going, any road will get you there."
Lewis Carroll

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ronsom: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Fri Apr 16, 2010 12:01 pm; Thanked: 2 times

by ronsom » Mon May 31, 2010 2:12 am

I'm getting something around 54 cms!
Please see the attached diagram (not drawn to scale)
The 4 circular rounds have been divided into 8 semi-circles.
The diameter of a an approximate semi-circle (AB) has been calculated using pythagoras theorem where the diameter of the cylinder and the length of side (20/8 = 2.5cms) form the 2 perpendicular arms. The periphery of the semi-circle comes to 6.75.
For 8 semi-circles its 6.75 * 8 = 54!
Nearest is 50.. but with the options being 47,48, 49 and 50.. its hard to say which one is right. A rounding error could lead to either of these..

What's the source of this question?

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kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » Mon May 31, 2010 5:51 am

The ant is tracing a 3 D spiral or Helix akin to found in a DNA.

When you wind a rt angle triangle along the pipe the hypotenuse traces the spiral in this case the path of the ant.
The perpendicular of the rt angle triangle is 20/4 = 5, cos the ant gains 20 mts in 4 circles.

The base is TT D = 22/7*7/2 = 11.
Hypotenuse is 11Â²+5Â² = 121+25= âˆš146.
4 rounds, means four circles so 4*âˆš146
But, âˆš 144 = 12 so the answer is a little more than 4*12 = 48.

IMO 48

@ gmatmachoman terrific problem I rate ur recent postings at par with sanju09

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nikhilkatira: Master | Next Rank: 500 Posts; Posts: 318; Joined: Mon Jul 13, 2009 3:55 am; Thanked: 12 times

by nikhilkatira » Mon May 31, 2010 6:04 am

Is this really a gmat type question ?

Source pls ?

Best,
Nikhil H. Katira

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kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » Mon May 31, 2010 6:44 am

nikhilkatira wrote:Is this really a gmat type question ?

Source pls ?

If it is, then like the ant I will complete 4 problems in 20 mins. Spiralling downwards rapidly from the median score which I think is 500.
Or on a +ve note you must be maxing your quant, so CAT is dishing out such a twister.

But why not, practise at high altitude. Over to u gmatmachoman.

Frankly, the concept of 3D spiral is beyond GMAT.

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jube: Master | Next Rank: 500 Posts; Posts: 186; Joined: Fri May 28, 2010 1:05 am; Thanked: 11 times

by jube » Mon May 31, 2010 8:18 am

liferocks wrote:path of traverse of the ant can be divided into 8 equal line whose length equal to the length of hypotenuse of a right angled triangle with onr side 3.5 and another 20/4=5

hey, can you explain how you get this? From the way I'm visualizing it, the ant is tracing a spiral path around the cylinder. But I'm not sure what to do next. Is this the right approach?

thanks!

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