Question 1205417
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A roll of tape is on a spool with diameter 3.6 cm. The diameter of the spool plus the diameter 
of the tape is 5.4 cm. The tape is 25.4 m long. The number of layers on the tape is closest to
a) 160
b) 180
c) 215
d) 230
e) 1773
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<pre>
The minimum diameter (the diameter of the first layer) is d = 3.6 cm; 
the circumference (the length) of the first layer is  {{{pi*d}}} = 3.14159*3.6 = 11.309724 cm.


The maximum diameter (the diameter of the last layer) is D = 5.4 cm; 
the circumference (the length) of the last layer is  {{{pi*D}}} = 3.14159*5.4 = 16.964586 cm.


The lengths of consecutive layers make an arithmetic progression with the common  difference  {{{2*pi*delta}}},
where  {{{delta}}}  is the thickness of the tape. Therefore, the number of layers n can be found from 
the formula of the sum of an arithmetic progression


    2540 = {{{((first_term+last_term)/2)*n}}} = {{{((11.309724+16.964586)/2)*n}}} = 14.137155*n,


    n = {{{2540/14.137155)}}} = 179.67  (approximately).


<U>ANSWER</U>.  The number of layers on the tape (approximately 180), is closest to option b).
</pre>

Solved.