Thread: Which frog is faster?

Dr. John is studying six frogs he found in the swamp. These frogs are identical to normal frogs except that the distances they jump are determined by a pattern that is unique to each frog. For example, Frog A jumps 2 feet every time, but Frog B jumps 2 feet on odd numbered jumps and 3 feet on even numbered jumps.

Dr. John writes this in his notes as Frog A (2) and Frog B (2,3). The other frogs are in Dr. John's notes as follows: Frog C (1,5,2), Frog D (2,3,2,4), Frog E (4,3,1,3,2) and Frog F (1,1,1,1,1,4).

The cage in which the frogs are kept is outside Dr. John's lab. One night the cage door pops open. Two hours later, the open door is discovered. If each frog leaves the cage when the door pops open, jumps once every minute and starts by jumping the first jump in its pattern as its first jump out of the cage, which frog will be the greatest distance from the cage when their escape is discovered?

I think the answer is Frog E.

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suresh

I think your answer is wrong.

Here is the answer for your question....

FrogA : 2feet * 120 = 240feet
FrogB : 5feet * 60 = 300feet ( 2+3 = 5 & 120/2 = 60)
FrogC : 8feet * 40 = 320feet ( 1+5+2 = 8 & 120/3 = 40)
FrogD : 11feet * 30 = 330feet ( 2+3+2+4 = 11 & 120/4 = 30)
FrogE : 13feet * 24 = 312feet ( 4+3+1+3+2 = 13 & 120/5 = 24)
FrogF : 9feet * 20 = 180feet ( 1+1+1+1+1+4 = 9 & 120/6 = 20)

So the answer is FrogD and it reach 330feet.

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suresh

the average speed of frogs are
A-2
B-2.5
C-8/3=2.66666667
D-11/4=2.75
E-12/5= 2.4
F-9/6=1.5

So the winner is D

Frog D is faster than the others.

120 jumps



a=2*120=240ft

b=2*60+3*60=300ft

c=1*40+5*40+2*40=320ft

d=2*30+3*30+2*30+4*30=330ft

e=4*24+3*24+1*24+3*24+2*24=240ft

f=1*20+1*20+1*20+1*20+1*20+4*20=180ft

so frog d will be in greater distance

frog D . . . ( 330 feet )

I think the answer is frog D.

hi,

hope the answer is D

Thanks
Sushma

i think the answer is Frog D..330 feet away from the cage...

its frog E

Hi,

Here is the answer:

1)Two hrs later it is found that the door is discovered

therefore 2 hrs=120 mins

2) All frogs are jumps once @ every minute

therefore

A = 2 * 120 = 240 feet
B = (2+3)/2 * 120= 300 feet Frog D (2,3,2,4), Frog E (4,3,1,3,2) and Frog F (1,1,1,1,1,4).
C = 8/3 *120 = 320 feet
D = 11/4 *120 = 330 feetE = 13/5 * 120= 312 feet
F = 9/6 *120 = 180 feet

So as per above calculation Frog D is covered greatest distance

Its Frog D.

Frog D (2,3,2,4) = 2+3+2+4 = 11 feets in 4 minutes so in two hours = 120/4 = 30 * 11= 330 (Max)