Claudio wants to be well prepared for a long marathon that will occur

Question

GMATH  practice exercise (Quant Class 17)

Claudio wants to be well prepared for a long marathon that will occur in the near future. His running coach has decided that the training sessions must be done once every two days, starting with a 15-kilometer distance route, and adding exactly 500 meters (=half a kilometer) to the route at every new training session. Claudio´s coach believes at least 1800 kilometers must be run, during the whole training period, before someone is considered ready for this challenge. According to these assumptions, what is the minimum number of training sessions that Claudio needs for this long marathon preparation?

(A) 59
(B) 60
(C) 61
(D) 62
(E) 63

fskilnik · Answer

?\,\,\,:\,\,\,\min \,N\,\,{\rm{for}}\,\,{\rm{sum}}\,\,\, \ge \,\,\,1800\,

\left. \matrix{
 {a_1} = 15 \hfill \cr 
 {a_2} = 15 + 1 \cdot {1 \over 2} \hfill \cr 
 {a_3} = 15 + 2 \cdot {1 \over 2} \hfill \cr 
 \vdots \hfill \cr 
 {a_N} = 15 + \left( {N - 1} \right) \cdot {1 \over 2}\,\,\, \hfill \cr} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{Arith}}{\rm{. Seq}}{\rm{.}}} \,\,\,\,\,1800\,\, \le \,\,\,\,N \cdot {1 \over 2}\left  = {{N\left( {N + 59} \right)} \over 4}

{\rm{Trying}}\,\,\left( B \right)\,\,N = 60\,\,\,\,\, \Rightarrow \,\,\,\,\,{{N\left( {N + 59} \right)} \over 4} = {{{{60}^2} + 60\left( {60 - 1} \right)} \over 4} = {{2 \cdot 3600 - 60} \over 4} = 1800 - 15\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\left( C \right)

\left( * \right)\,\,N = 61\,\,\,\,\, \Rightarrow \,\,\,\,{{N\left( {N + 59} \right)} \over 4}\,\, = \,\,\,\underbrace {1800 - 15}_{{\rm{sum}}\,\,{\rm{for}}\,\,N\,\, = \,\,60}\,\,\, + \,\,\underbrace {15 + 60 \cdot {1 \over 2}}_{{a_{61}}}\,\,\, = 1830\,\, > \,\,1800

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

willykuo · Answer

suggest try (C) first as it's easier to calculate (61-1 = 60, i mean this part)

FerociousStorm · Answer

I found it easier to take an option and do backward solving.

so lets take 60, if Claudio ran 60 marathons then he increased the distance 59 times by 0.5 miles with each consecutive run.

so the distance covered during the final marathon would be 15 + 59*0.5 = 44.5 kmph

now average speed for the whole run -> (15 + 44.5)/2 = 59.5/2

average speed * total days = distance

59.5/2 * 60 = 1785

since Claudio needs to run 1800 kms, he must run one more day than 60. which is option C

willykuo · Answer

There is a calculation mistake here
15 + 59*0.5 = 44.75

should be:
15 + 59*0.5 = 44.5

suggest you take 61 first as 15 * 60*0.5 is much easier to calculate

FerociousStorm · Answer

Thanks, i corrected it. And point of taking 60 first was to show the additional step at last, which is adding or subtracting days.

Claudio wants to be well prepared for a long marathon that will occur

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