reach the tree top

[sanju09]

A monkey is climbing a coconut tree, which is 35 feet tall. It climbs 3 feet in each jump, but slips down by 1½ feet after each jump it makes. When the monkey rests for 1 minute, it's able to make a jump 4 feet high. If the monkey takes an overall rest of 5 minutes during the operation, in how many minimum jumps will it reach the tree top?
(A) 18
(B) 19
(C) 20
(D) 24
(E) 25

[liferocks]

For jump of 3 feet high effective distance covered 3-1.5=1.5feet
and for jump of 4 feet high effective distance covered 4-1.5=2.5feet
now the monkey takes 5 minute breaks i.e total 5 4 feet high jumps.
it covered 5*2.5=12.5feet in 5 jumps
it leaves 35-12.5=22.5feet
this distance it will cover in 22.5/1.5=15 jumps
so total 15+5=20 jumps
now for the last jump it will not slip down
hence total jumps =20-1=19
ans option B

[Fiver]

Agree with B-19.

Because of the 5 4-feet jumps, the net distance covered is 12.5 and the remaining distance is 22.5.
Now the final jump must be a full and exact 3 feet jump, else the monkey would fall off the tree.
Reduce 3.5 from 22.5 = 19.5.
Divide 19.5 by 1.5 to get 13.

Hence the total number of jumps is [spoiler]5+13+1=19[/spoiler]

[akhpad]

Here, amount of slip is not mentioned when monkey jumped by 4 feet. How can I assume 1.5 feet.

Lets assume it is 1.5 feet for time being.

Monkey rests for 5 minutes and jumped 5 times of 4 feet.
Net jumped = 5 * (4 - 1.5) = 12.5 feet

Remaining = 35 - 12.5 = 22.5

Now,
15 * 1.5 = 22.5
13* 1.5 + 1*3 = 22.5
Number of jumped = 13+1 = 14

Total jumped = 14+5 = 19

I also got B

[sanju09]

Here, amount of slip is not mentioned when monkey jumped by 4 feet. How can I assume 1.5 feet.

Lets assume it is 1.5 feet for time being.

Monkey rests for 5 minutes and jumped 5 times of 4 feet.
Net jumped = 5 * (4 - 1.5) = 12.5 feet

Remaining = 35 - 12.5 = 22.5

Now,
15 * 1.5 = 22.5
13* 1.5 + 1*3 = 22.5
Number of jumped = 13+1 = 14

Total jumped = 14+5 = 19

I also got B

Please note the following in the stem

slips down by 1½ feet after each jump it makes.

[spoiler]each for?[/spoiler]

[singhpreet1]

For jump of 3 feet high effective distance covered 3-1.5=1.5feet
and for jump of 4 feet high effective distance covered 4-1.5=2.5feet
now the monkey takes 5 minute breaks i.e total 5 4 feet high jumps.
it covered 5*2.5=12.5feet in 5 jumps
it leaves 35-12.5=22.5feet
this distance it will cover in 22.5/1.5=15 jumps
so total 15+5=20 jumps
now for the last jump it will not slip down
hence total jumps =20-1=19
ans option B

could someone please explain why will he not fall down on the last jump all of a sudden because till now the monkey was losing 1.5 feet on every jump so isnt it assumed that he will slip down again, despite the question saying minimum jumps?

5 minutes break gives 5*2.5=12.5 gained
remaining distance to be covered: 35-12.5=22.5
15*1.5= 22.5, so shouldnt it be 20 total jumps rather than 19?

guys???????

would any GMAT expert please give a suitable explanation..would be highly appreciated.

Thanks,
Preet Singh

[amising6]

let say after 1st jump monkey will cover 3-1.5=1.5 feet net
now let us assume monkey start taking rest and next 5 jump he will take after taking rest of 1 mins after evry jump
in this he will be covering 4-1.5=2.5 feet net distance after every jump
so distance travelled in this 5 jump =2.5*5=12.5
till now number of jump =1+5=6 jump
distance covered 1.5+12=14 feet
now again he will take normal jump where net distance covered in each jump is 1.5 feet
so in next 13 jump he will cover 13*1.5=19.5 feet
so till now total distance travelled 14+19.5=33.5
total number of jump will be 6+13=19
now the moment he will take next jump since he covers 3 feet in each jump and distance remaining to be covered is 1.5 feet
he will reach top in 20 jump
cheers

[jube]

Rate of movement without 1 min rest = 3-1.5=1.5
Rate of movement with 1 min rest = 4-1.5 = 3.5

Jump 1 (1.5) - 1 min rest - Jump 2 (2.5) - 1 min rest - Jump 3 (2.5) - 1 min - Jump 4 (2.5) - 1 min - Jump 5 (2.5) - 1 min - (Jump 6) (2.5)

total ft covered till now = 16.5
total jumps till now = 6

distance left = 18.5
jumps needed = 18.5/15 = 13 (12 jumps will cover 18 ft so 12+1 for 19.5)

therefore total jumps = 13+6=19

[kvcpk]

This problem is easy to solve from behind.. I doubt if its GMAT problem..
This was one of those childhood puzzles..

35 feet.. last jump si the maximum distance jump (assume)
so monkey needs to cover 31 feet. Thats it and it has 4 minutes to take rest..

assume it used all its 4 minutes in the last jumps.. so

effective jump is 4 - 1.5 = 2.5 so 2.5 * 4 = 10

remaning distance to travel is 31-10 = 21

with normal jump effectively it covers 1.5 feet.

so it needs 14 jumps for 21 feet.

so total is 14 + 4 + 1 = 19

[amising6]

Rate of movement without 1 min rest = 3-1.5=1.5
Rate of movement with 1 min rest = 4-1.5 = 3.5

Jump 1 (1.5) - 1 min rest - Jump 2 (2.5) - 1 min rest - Jump 3 (2.5) - 1 min - Jump 4 (2.5) - 1 min - Jump 5 (2.5) - 1 min - (Jump 6) (2.5)

total ft covered till now = 16.5 (note dude till here distamce covered will be 14)
total jumps till now = 6

distance left = 18.5 (21)
jumps needed = 18.5/15 = 13 (12 jumps will cover 18 ft so 12+1 for 19.5) jump needed will be 21/1.5=14

therefore total jumps = 13+6=19

total jump =20

[jube]

Rate of movement without 1 min rest = 3-1.5=1.5
Rate of movement with 1 min rest = 4-1.5 = 3.5

Jump 1 (1.5) - 1 min rest - Jump 2 (2.5) - 1 min rest - Jump 3 (2.5) - 1 min - Jump 4 (2.5) - 1 min - Jump 5 (2.5) - 1 min - (Jump 6) (2.5)

total ft covered till now = 16.5 (note dude till here distamce covered will be 14)
total jumps till now = 6

distance left = 18.5 (21)
jumps needed = 18.5/15 = 13 (12 jumps will cover 18 ft so 12+1 for 19.5) jump needed will be 21/1.5=14

therefore total jumps = 13+6=19

total jump =20

ahh, you're right! but that means there's something wrong with what I was doing - not sure what that is. :\

[mj78ind]

May be I am overthinking this, the monkey can jump 4 ft after a minute's rest. Hence, first time it needs no rest and then it can jump 5 more times with 5 mins of rest. hence distance covered = 2.5*6 = 15 ft.

Now to cover 20 ft, last jump will cover 3 ft and all others will cover 1.5 hence 17 ft with 1.5 ft each = 17/1.5 = 12 jumps.

Thus total jumps = 6 + 1 + 12 = 19.

[singhpreet1]

May be I am overthinking this, the monkey can jump 4 ft after a minute's rest. Hence, first time it needs no rest and then it can jump 5 more times with 5 mins of rest. hence distance covered = 2.5*6 = 15 ft.

Now to cover 20 ft, last jump will cover 3 ft and all others will cover 1.5 hence 17 ft with 1.5 ft each = 17/1.5 = 12 jumps.

Thus total jumps = 6 + 1 + 12 = 19.

hahaa...i would love to agree with you mj78ind ,though we need to rely strictly on the data provided to us and im afraid nothing of the sorts is provided....so i would still go with 20 jumps as the final answer..unless someone is willing to logically prove me wrong with the specifically given data and no assumptions.

thanks.

Preet Singh

[amising6]

May be I am overthinking this, the monkey can jump 4 ft after a minute's rest. Hence, first time it needs no rest and then it can jump 5 more times with 5 mins of rest. hence distance covered = 2.5*6 = 15 ft.

Now to cover 20 ft, last jump will cover 3 ft and all others will cover 1.5 hence 17 ft with 1.5 ft each = 17/1.5 = 12 jumps.

Thus total jumps = 6 + 1 + 12 = 19.

hahaa...i would love to agree with you mj78ind ,though we need to rely strictly on the data provided to us and im afraid nothing of the sorts is provided....so i would still go with 20 jumps as the final answer..unless someone is willing to logically prove me wrong with the specifically given data and no assumptions.

thanks.

Preet Singh

i guess you are right see this explaination it does make sense

Thu Jun 17, 2010 6:39 am

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let say after 1st jump monkey will cover 3-1.5=1.5 feet net
now let us assume monkey start taking rest and next 5 jump he will take after taking rest of 1 mins after evry jump
in this he will be covering 4-1.5=2.5 feet net distance after every jump
so distance travelled in this 5 jump =2.5*5=12.5
till now number of jump =1+5=6 jump
distance covered 1.5+12=14 feet
now again he will take normal jump where net distance covered in each jump is 1.5 feet
so in next 13 jump he will cover 13*1.5=19.5 feet
so till now total distance travelled 14+19.5=33.5
total number of jump will be 6+13=19
now the moment he will take next jump since he covers 3 feet in each jump and distance remaining to be covered is 1.5 feet
he will reach top in 20 jump
cheers

[Asif]

I think the answer can be 19 or 20. I will not go for the methods as both the answers are logically right. its the way people see it differently. Monkey takes rest before the 1st jump will take you to answer 19 and the monkey takes rest after the first jump will lead to 20. But its not written or implied what is the actual thing.