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insects_geometry

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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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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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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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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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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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 11²+5² = 121+25= √146.
4 rounds, means four circles so 4*√146
But, √ 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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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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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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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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