Latest MBA Aptitude Question SOLUTION: A monkey wants to climb up a pole of 50m height.He first climbs up 1m but he falls back by the same height.again he climbs up 2m but he falls back by 1m.Once again he climbs up 3m

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(#M40015609) MBA Logical Reasoning question time speed and distance, work, alligations & mixtures Keep an EYE

A monkey wants to climb up a pole of 50m height.He first climbs up 1m but he falls back by the same height.again he climbs up 2m but he falls back by 1m.Once again he climbs up 3m but he falls back by 1m.Again he climbs up 4m but he falls back by 1m.In this way he reaches at the top of the pole.if it is known that the monkey needs 10 sec for 1m in upward direction and 5 sec for 1m in downward direction,then the total time required by monkey to reach at the top of the pole.when he reaches at the top he wont slip back.the total time required by monkey to reach at the top of the pole is:-

1) 9 min. 10 sec
2) 10 mins.
3) 8 min. 20 sec.
4) none of these

Asked In MBA (11 years ago)

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K Vedant Laskar  (11 years ago)

The ans is 4.
It takes a total of 9min. 20 sec

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K Vedant Laskar  (11 years ago)

Sorry for the mistake.
The answer is 9min 10 sec
in 10 go the monkey takes a total of 55 steps - 10 slips = 45 position.
thus time taken = 550-50 = 500 sec.
plus 50 sec to get into the top. thus total 500+50 sec= 550 sec= 9min10sec. :)

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Pranay  (7 years ago)

the monkey climbs and falls, so after (n+1) falls,it is at the height of
(1−1)+(2−1)+(3−1)+…+(n+1−1)=1+2+3.......n=(n(n+1))/2
we must find max{n} such that
(n(n+1))/2≤50
from hit and trial, we conclude that n=9
so after 10 falls the monkey is at the height of 45m
It takes 10 sec to move 1m in upward direction
Also,It takes 5 sec to move 1m in downward direction
So total time taken=(45*10)+(10*5)=550 sec=9 min 10 sec
where (45*10) denotes time for upward movement and (10*5) denotes time for downward movement

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amit  (4 years ago)

option D
1-1/2-1/3-1/4-1/5-1/6-1/7-1/8-1/9-1/10-1/ =45meters SO, last jump 5meters ka lena h bs.............now time spent calculations
»(jo first position pr likhe h wo forward distance traveled h so 55*10=550 secs and jo 2nd postition pr negative wale h wo backward h so 10*5=50 secs total 600 and last jump forward me 5mtrs ka tha so 10*5=50 secs..............total time spent 650 secs

Note:
forward me 10 sec for every meter so 55meters forward distance is multiplied by 10 =550secs and backward 5sec for every mtr so bckwd distce of 10 mtrs is multiplied by 5=50secs

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