# Manhattan GMAT Challenge Problem of the Week – 19 Sept 2011:

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## Question

The 8 spokes of a custom circular bicycle wheel radiate from the central axle of the wheel and are arranged such that the sectors formed by adjacent spokes all have different central angles, which constitute an arithmetic series of numbers (that is, the difference between any angle and the next largest angle is constant). If the largest sector so formed has a central angle of 80°, what fraction of the wheel’s area is represented by the smallest sector?

A. 1/72

B. 1/36

C. 1/18

D. 1/12

E. 1/9

## Answer

Each spoke is a radius of the circular wheel. If there are 8 spokes, then the circle is broken up into 8 sectors. The central angles of these sectors are 8 different numbers: call them *a, b, c, d, e, f, g,* and *h,* with *a* as the smallest number and *h* as the biggest number.

All of these angles add up to 360°, the total central angle in a circle.

Since there are 8 angles, the average of all the angles must be 360/8 = 45°.

Since the angle measures are evenly spaced as a series, the average of all the angles must also be the average of the smallest & the largest angles. That is, (*a* + *h*)/2 = 45.

Finally, since *h* = 80, we can figure out *a*, which equals 10.

A 10° angle is 10/360 = 1/36 of the circle. The sector with this angle occupies just 1/36 of the wheel’s area.

**The correct answer is B.**

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## 9 comments

tarun on September 23rd, 2011 at 9:29 am

Hi

Good ques. I knew the fact that the average of an evenly spaced set is equal to the avg of the first and last term but this thing didnt strike me when i was doing the ques. is there any other method of solving this ques if such things dont strike the student while attempting the exam?

thanks tarun

Mohan on September 23rd, 2011 at 2:14 pm

Yep, fortunately there is an easier way out:

Let us assume the smallest angle is 'y'. Given that all the angles are in arithmetic progression, the angles can be represented as y, y+x, y+2x, y+3x, y+4x, y+5x, y+6x, y+7x - where x is the interval between any successive pairs.

It is given that largest angle is 80. This translates as => y + 7x = 80

We also know that sum of all angles is 360 (circle!) => 8y + 28x = 360

Multiplying the first equation with '4' and working these two equations we get 4y = 40 => y = 10

Smallest angle is 10 so the fraction of circle is 10/360 => 1/36 (Choice: B)

Cheers!

tarun on September 24th, 2011 at 12:01 am

awesome mohan...yep pretty basic method to solve it...just didnt strike....thanks a lot

rgds

tarun

punit on September 25th, 2011 at 11:53 pm

Since it is given that the difference between any angle and the next largest angle is constant , so the the angles are in A.P (Arthimetic Progression) . We know that the sum of angles is 360.So,using the formula of n terms of A.P i.e

S(n)=n/2(first term+last term) n=8

for the question here first term=smallest angle and

last term=largest angle

360 = 8/2(first term+80)

first term=10

i.e smallest angle is 10 degree

which is 1/36 of the wheel area .Hence the correct answer is B.

Simon on September 26th, 2011 at 3:16 pm

Last term K (8) + X = 80

K(1) + X…. +…….K(8) + X = 360

So we have two equations 8K + X = 80

36K + 8X = 360

Solving two equations we get X=0 and K =10

Plug in K (1) + X = 10 (1) +0 = 10

Smallest angle is 10.

The official answer above is probably the short version but this method can be used if we can’t think of anything else.

Abhinav Saxena on September 27th, 2011 at 6:39 am

8 even numbers in AP series are: a-4d, a-3d, a-2d, a-d, a+d, a+2d, a+3d, a+4d

hence

(a-4d) + (a-3d) + (a-2d) + (a-d) + (a+d) + (a+2d) + (a+3d) + (a+4d) = 360

8a = 360

a = 360 / 8 = 45

largest sector

a+4d = 80

45+4d = 80

d = 8.75

hence, smallest angle which is a-4d = 45-35 = 10 degree.

And fraction of the wheels area is 10/360 = 1/36

Hence B is the correct answer.

snk on September 27th, 2011 at 8:50 am

how do you know that largest sector = 80 ??

tarun on September 27th, 2011 at 8:53 am

hi snk

its given in the question

cheers

snk on September 27th, 2011 at 9:28 am

yes it is..thanks..