8 circles of diameter 1 cm are kept in a row, each circle touching its

Question

8 circles of diameter 1 cm are kept in a row, each circle touching its neighbours. How many paths of length 4π are possible to go from A(0,0) to B(8,0) if you are not allowed to traverse back?

A. 128
B. 512
C. 256
D. 240
E. 300

GMATinsight · Answer

There are two ways to reach from start to first tangential point

There are again 2 ways to reach from first tangential point to second tangential point of circles

There are always 2 ways to reach from one tangential to other in case of other circles as well

hence total ways to reach from start to end point = 2^8 = 256

Answer: option C

umasarath52 · Answer

Circumference of each circle is => 2*pi*r => 2*pi*1/2 => pi
if we divide each circle into two semi circles, we will have 16 semi circles with each circumference as pi/2.

Every circle has two choice of paths, upper semi circle or lower semi circle.

From each circle, there are 2 ways to select one semi circle. SInce there are 8 semi circles, the number of ways to select is 2^8 => 256

Answer C

EMPOWERgmatRichC · Answer

Hi All,

This question is poorly written, so even though the basic concepts (geometry and permutations) will show up on Test Day, there are far more realistic practice resources that you could be using to learn those concepts.

Among the non GMAT-like details, the description of the circles should be "tangent to one another along a straight line" and the prompt refers to graphing co-ordinates, but doesn't establish that we're actually dealing with the XY co-ordinate plane.

In real basic terms, each of the 8 circles offers 2 options to get 'around it', which means that there are 2^8 ways to get from the 'beginning' to the 'end.' GMAT question writers are far more rigorous and detail-oriented about how they craft their questions and you should be sure that you're working with resources that train you to face the terminology and concepts that will appear on Test Day.

GMAT assassins aren't born, they're made,
Rich

Luca1111111111111 · Answer

Hello, 
In what way does the length of 4pi influence our calculations? 
The existence of 2 ways per circle ^8 seams too easy as an answer and I am wondering how the answer would change if the length of 4pi would be changed.
Thank you

EMPOWERgmatRichC · Answer

Hi Luca1111111111111,

The prompt tells us that we are "not allowed to traverse back", which implies that we cannot "trace our steps" when traveling along the circles. In simple terms, you cannot 'go backwards' or go around a particular circle multiple times. By telling us this, the length of the circles actually ends up having NOTHING to do with the answer to the question - we can only travel on 1/2 of each circle and then we have to transfer to the next circle. We're going from point A to point B, so there are 2^8 possible paths to take.

As it stands, since each circle has a diameter of 1 cm, each has a circumference of π cm. When you travel around 8 "halves" (which measure (1/2)π cm each), you'll travel 4π cm of total distance. If you change the total distance traveled, then you also have to change other aspects of the question to account for that new distance.

GMAT assassins aren't born, they're made,
Rich

Luca1111111111111 · Answer

Thank you very much - great explanation! 
May I ask another question (hopefully not too stupid): The formula to calculate the a circle's circumference is 2πr. Isn't the circumference of the circle 2π then and of the halves just π?

EMPOWERgmatRichC · Answer

Hi Luca1111111111111,

We're told that each circle has a DIAMETER of 1 cm, which means that each RADIUS is 1/2 cm. You are correct that the formula for a circle's circumference = 2π(radius). Here, that would be 2π (1/2) = π... so each half-circumference would be (1/2)π.

GMAT assassins aren't born, they're made,
Rich

philipssonicare · Answer

Doesn't the length of 4π limit the start to the first 5 circles?

From the first circle there are 2^4 options
From the second circle there are 2^4 options
Up to the fifth circle having 2^4 options (the circles thereafter can't have lines of 4π length without going backwards)

5·2^4=90

How is the above thinking wrong?

Kinshook · Answer

Given:  8 circles of diameter 1 cm are kept in a row, each circle touching its neighbours.

Asked:  How many paths of length 4π are possible to go from A(0,0) to B(8,0) if you are not allowed to traverse back?

There are 8 circles touching one another with circumference  = \pi 
There are 2 options for each circle = upper semi-circle or lower semi-circle of lengths  = \pi/2 
So total ways (options) for path of length  ( 8 * \pi/2 = 4 \pi) = 2^8 = 256

IMO C

Kinshook · Answer

There are 8 semi-circles of length (\pi/2) making a total path length of  8 * \pi/2 = 4 \pi

So there is no other option than to choose either a upper semi-circle or a lower semi-circle for each circle in the path without going backwards.
Total ways =  2^8 = 256

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