Answer / kukku

x=vel of d
y=vel oh h
t1=time taken by d
t2=time taken by h
we have 4t1=5t2..............(1)
also
3xt1=4yt2....................(2)
comparing 1 and 2
x/y=16/15

Answer / srikanth puli

5*3 = 15,
4*4 = 16...
dog speed = 16 units
hare speed = 15 units

Answer / monk

let 3 leap of dog = x = 4 leap of hair
let time taken by dog to take three leaps be 't'

dog takes 4 leap for every 5 leap of hair
so for every one leap of dog hair take 5/4 leaps

hence time taken by hair to take 3 leaps = (4/5)*t
time taken by hair to take 4 leaps =(4/5)*(1/3)*4*t=(16/15)t
now
speed of dog'Vd'=x/t
speed of hair'Vh'=x*15/16t

hence Vd:Vh = 16:15

Answer / jainachran

Ans is 16:15.
Given that 3 Leap of dog=4 Leap of hare (means they travel same distance)

3 Ld = 4 Lh


Initially they said that 4 leaps of dog are equal to 5 leaps of hare.
so, speed of Dog=(Distance of dog 4Ld)/time.
similarily,speed of hare = (distance of hare 5Lh)/time.

now compare speeds

Sd/Sh=[4Ld/t] / [5Lh/t]

we get Sd/Sh=4Ld/5Lh
where,Ld/Lh=4/3

Sd/Sh=16/15

Answer / shristi gour

Given
3dogs= 4cats
Or
Dog/ cat= 4/3
Let cat's 1leap= 3 metres
And dog's 1 leap= 4 metres
Then,
Ratio of speed of cat and dog. =3*5/4*4 = 15:16

Answer / pritesh moure

Let 1 leap of dog= x meter
1 leap of hare= y meter
So distance covered by 3 leaps of dog & 4 leaps of hare are equal
i.e. 3x=4y, x/y =4/3........(i)
Let time taken by 4 leaps of dog be 't' which is equal to time taken by 5 leaps of hare.
So time taken to cover 1 leap
By dog= t/4
By hare= t/5
Now Speed= (Distance covered in 1 leap)/ (Time taken to cover 1 leap)

Speed of dog(S1) = x/(t/4)=4x/t
Speed of hare(S2) = y/(t/5)=5y/t

So S1/S2 =(4x/t)/(5y/t)= 4x/5y=(4×4)/(5×3)= 16/15. (x/y=4/3 by (i) )

Ans: 16/15.

Answer / shani

dog 3 leaps = hare 4 leaps
taking lcm = 12, so both taking 12.
dog 1 unit =4
& hair 1 unit = 3.
so, dog taking 4 leaps =12+4 =>16.
hair taking 5 leaps =12+3 =>15.

ratio of speed =16/15.

Answer / surya

i get this question from rs agarwal in that ans is 16:15 but
here all of you tell different answers i am just confuse to
see these all

Answer / diffraction

let us assume



speed of dog D speed of hare H
leap of dog X leap of hare Y

now
a dog takes 4 leaps for every 5 leaps of a hare
let time T
in time T

dog =4X/D=T hare=5Y/H=T


3 leaps of a dog are equal to 4 leaps of the hare

3Y=4X

so

3Y/D=5Y/H=T

so D/H=3/5

Answer / vicky pep

according to the second data the dog ll finish .75 of leap
for every 1 leap of hare, so dog have to run faster than
hare to take 4 leap for every 5 leap.

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