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Problem edited since i missed a line from the problem, A monkey wants to climb up a pole of 50 metre height. he first climbs up 1 m but he falls back by same height. Again he climbs up 2 m but he falls back by 1 m. Again he climbs up 3 m and falls back by 1 m. Again he climbs up 4 metre but falls back by 1 m. In this way he reaches top of the pole. If it is known that the monkey needs 10 secs for 1 metre in upward direction and 5 secs for 1 metre in downward direction. Then find the total time required by monkey to reach at the top of the pole. Note that, when he will reach at the top he will not be slipped back hence find the total time required by this monkey to reach on the top of the pole.

I have solved this question and i have getting over 10 minutes as answer. But book's answer is 9 minutes 10 seconds. I am not getting why the answer differs.

solution: (1,1) ,(2,1) ,(3,1),(4,1)the pair (x,y) where x=climbing interval and y=falling interval now then..... 1st time =0m 2nd time =2-1=1m 3rd time =3-1=2m 4th time =4-1=3m effective distance covered in 1 cycle=(0+1+2+3)=6m total distance covered =climbing distance + falling distance...... (1) for each cycle..... climmbing distance =(1+2+3+4)=10 but ...... falling distance = (1+1+1+1)=4

total time for each cycle=time(climbing dist)+time(falling dist) =10*10+4*5=120sec this process repeats for 8 interval
hence 120*8=960 sec after this i am not able to solve the problem

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Perhaps you could show us your work, and we could suggest where your error is. –  Matthew Conroy Jan 20 '11 at 6:51
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You probably accounted extra time for the monkey to slip at the top. –  PEV Jan 20 '11 at 6:55
    
So the steps repeat as 1,2,3,4,1,2,3,4,...? Or are they 1,2,3,4,5,6,7,... –  Aryabhata Jan 20 '11 at 9:37
    
Is the progression 1, 2, 3, 4, ...? Or 1, 2, 4, ... as originally stated? –  Isaac Jan 20 '11 at 15:52
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6 Answers

If $n$ is the number of times (and hence a total of $n$ meters) the monkey falls back, then when he reaches the top of the pole, the total distance he has moved upwards is $50+n$ meters (why? Hint: Instead of the monkey falling back, assume it was the pole that moved up by 1m).

Thus the time taken in seconds is $(50+n)\times 10 + n \times 5$.

Assuming the book answer is correct, we get

$$(50+n)\times 10 + n \times 5 = 550$$

i.e.

$$ 500 + 15n = 550$$

Thus

$$ n = 50/15 = 10/3$$

Thus the book answer you quote is wrong, irrespective of the steps of upward progression, if he falls back $1$m each time. The answer I get, assuming the upward progression as $1,2,4,8,16,32$ is less than $10$ minutes.

Are you sure you read the book answer correctly? Is it possible that you converted the answer given in seconds to minutes(and seconds) and might have made a mistake there?

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could u plz explain why u have not considered climbing interval mentioned as(1,2,3,4) as u have considered only falling interval(1m) ,that rising interval could have significant impact on the answer and again the answer is 9min and 10sec only –  user5918 Jan 20 '11 at 7:11
    
@user: First, the upward direction you gave was 1,2,4. Where is the 3? Second, if he always falls back by 1m and there is always enough space to do that, the upward steps don't really matter: the (why?) in the answer. –  Aryabhata Jan 20 '11 at 7:13
    
@user: I really suggest you try to understand this. Since you seem to be practising for CAT, you need quick methods. The method I suggest is quick: All you need to do is find out the number of times the monkey falls back. –  Aryabhata Jan 20 '11 at 19:59
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Distance travel (time taken) : 0(15 sec) + 1(25 sec) +2 (35sec)+ 3(45 sec)+ 4 (55 sec)+ 5 (65 sec)+ 6 (75sec) + 7 (85sec) + 8 (95sec) + 9 (105Sec)= 45 meter (600 sec)

for next 5 meter upward climbing will take 50 sec so total time is 650 sec= 10 min 50 secs

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I suspect that your work assumed that the monkey climbed 1, 2, 3, 4, 5, 6, ... meters (for which I get over 10 minutes), but the problem as you've stated it suggests that the monkey climbs 1, 2, 4, 8, 16, 32, ... meters or 1, 2, 4, 7, 11, ... meters or some other pattern—more information would be helpful.

In any case, the way that I'd do the problem is to start with a spreadsheet. Label the columns up-distance, up-time, down-distance, down-time, total-distance, total-time. In the up-distance column, fill in whichever progression of distances is intended. In the up-time column, multiply the up-distance column by 10. In the down-distance column, fill in all 1s. In the down-time column, multiply the down-distance column by 5. In the total-distance column, take the previous total-distance, add the up-distance, and subtract the down-distance. In the total-time column, take the previous total-time, add the up-time, and add the down-time.

Now, go down the spreadsheet and look for the last row where the distance is less than 50. Use this row to figure out whether or not the monkey already reached 50 meters or how much more time the monkey needs to reach 50 meters.

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Since the numbers are small, it's pretty easy just to work out by brute force.

Round 1, 15 seconds, 0m gained Round 2, 40 seconds, 1m gained Round 3, 85 seconds, 4m gained Round 4, 170 seconds, 11m gained Round 5, 335 seconds, 26m gained

He's now in striking distance. Add 24m, times 10 seconds gives 575 seconds, or 9:35s.

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In each step the monkey climbs and falls, so after $n+1$ steps he is at the heigth of

$$(1-1)+(2-1)+(3-1)+\ldots+(n+1-1)= 1+2+\ldots+n+(n+1)-(n+1)=\frac{n(n+1)}{2}$$

we must find $max\{n\}$ such that

$$\frac{n(n+1)}{2}\leq 50$$

from this we conclude that $n=9$, so after $10$ steps the monkey is at the heigth of $45m$

the time spent is then

$$10\times(1+2+\ldots+8+10)+5\times(1+1+\ldots+1)=10\times55+5\times 10=600$$

but to reach the top we need to add more $5\times10$ s so the total time is $650$ s and this corresponds to $10$ min $50$ sec.

If instead of adding the time spent in falls we had subtracted we would have $$10\times(1+2+\ldots+8+10)-5\times(1+1+\ldots+1)=10\times55-5\times 10=500$$ and the total time would be $550$ sec corresponding to $9$ min $10$ sec and is wrong.

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In fact, we can prove that no matter what the upward step progression (enough to fall back 1m each time, though), $9$ min $10$ sec cannot possibly be the right answer! See my answer. –  Aryabhata Jan 21 '11 at 7:15
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You started right approach, but there is tricky part in last jump:

First part for upward and second for downward

(1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (7,1) (8,1) (9,1) - Till here monkey climbed upward 45 M, time= 450sec fell downn 10 times= 50sec total time =500sec

In last only 5 required so if monkey climbed 10 m, then fall back 1 M. So last one will be (5,0)

Now total time taken 550 sec: 9 Minute and 10 second

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