Answer :
Answer:
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
Step-by-step explanation:
Answer:
72 leaps
Step-by-step explanation:
Given: A dog chasing a rabbit, rabbit runs 60 leaps before the dog. When the dog jumps 2 leaps rabbit can jumps 3 leaps. But rabbit have to jump 7 leaps to cover the distance that dog cover by 3 leaps.
To Find: number of leaps that the dog jumps when rabbit and dog meet.
Solution:
Let the units of leaps jumped by rabbit be r
Let the units of leaps jumped by dog be d
7 leaps of rabbit =3 leaps of dog
[tex]7\text{r}=3\text{d}[/tex]
[tex]1\text{r}=\frac{3}{7}\text{d}[/tex]
rabbit runs 60 leaps before the dog,
therefore
distance between dog and rabbit=60r=[tex]60\times\frac{3}{7}\text{d}[/tex]
When the dog jumps 2 leaps rabbit can jump 3 leaps
speed of rabbit([tex]\text{S}_{r}[/tex])=[tex]3\text{r}[/tex]=[tex]3\times\frac{3}{7}\text{d}[/tex]
[tex]\frac{9}{7}\text{d}[/tex]
speed of dog([tex]\text{S}_{d}[/tex]) = [tex]2\text{d}[/tex]
As dog and rabbit are running is same direction,
relative speed of dog and rabbit ([tex]\text{S}_{rel}[/tex])=speed of dog-speed of rabbit
([tex]\text{S}_{d}[/tex])-([tex]\text{S}_{r}[/tex])
[tex]2\text{d}-\frac{9}{7}\text{d}[/tex]
[tex]\frac{5}{7}\text{d}[/tex]
the distance to be covered by dog before catching rabbit
[tex]60\text{r}[/tex]
[tex]\frac{3}{7}\times60\text{d}[/tex]
Now,
[tex]\text{time}=\frac{\text{distance}}{\text{speed}}[/tex]
[tex]\text{time}=\frac{\frac{3}{7}\times60\text{d}}{\frac{5}{7}\text{d}}[/tex]
[tex]\text{time}=36\text{units}[/tex]
leaps jumped by dog in [tex]36\text{units}[/tex] of time
[tex]2\times36=72\text{leaps}[/tex]
Hence the leaps jumped by dog when dog and rabbit meet are 72