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
insects_geometry
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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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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?
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
Lewis Carroll
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?
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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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
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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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.nikhilkatira wrote:Is this really a gmat type question ?
Source pls ?
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.
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?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
thanks!