Question

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. How many leaps that the dog jumps when rabbit and dog meet?

Answer 1

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 2

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

$7\text{r}=3\text{d}$

$1\text{r}=\frac{3}{7}\text{d}$

rabbit runs 60 leaps before the dog,

therefore

distance between dog and rabbit=60r=$60\times\frac{3}{7}\text{d}$

When the dog jumps 2 leaps rabbit can jump 3 leaps

speed of rabbit($\text{S}_{r}$)=$3\text{r}$=$3\times\frac{3}{7}\text{d}$

                           $\frac{9}{7}\text{d}$

speed of dog($\text{S}_{d}$) = $2\text{d}$

As dog and rabbit are running is same direction,

relative speed of dog and rabbit ($\text{S}_{rel}$)=speed of dog-speed of rabbit

                                             ($\text{S}_{d}$)-($\text{S}_{r}$)

                                             $2\text{d}-\frac{9}{7}\text{d}$

                                             $\frac{5}{7}\text{d}$

the distance to be covered by dog before catching rabbit

                                             $60\text{r}$

                                             $\frac{3}{7}\times60\text{d}$

Now,

                                              $\text{time}=\frac{\text{distance}}{\text{speed}}$

                                              $\text{time}=\frac{\frac{3}{7}\times60\text{d}}{\frac{5}{7}\text{d}}$

$\text{time}=36\text{units}$

leaps jumped by dog in $36\text{units}$ of time

                                               $2\times36=72\text{leaps}$

Hence the leaps jumped by dog when dog and rabbit meet are 72