The Cat and The Mouse

After the cat jumps 2 times the mouse jumps 5. The jump of the cat are 3 times more large than the mouse jump. Therefore after the cat jumps 6 mouse's jumps, the mouse jumps 5.

We have this:

The cat jumps 2 cat's jumps or 6 mouse's jumps, and the distance of the cat and the mouse is 35 - 6 = 29, but the mouse increase 5 jumps, 29+5 = 34.

After the cat jumps 2 times the distance are reduced 1 mouse's jump. To reduce 35 the cat need to jump 35*2 = 70 cat's jumps.

Answer $\boxed{70}$

let y be the distance traveled by the mouse (based on the number of jumps it did). y=35+5x where x is how many times it jumps.

let z be the distance traveled by the cat (based on the number of jumps it did). z=2d, where d is stated as thrice x(the number of jump/s made by the mouse)

As stated in the problem, the problem, the two will eventually meet, which arrives to another equation.

y=z(after some jumps) 35+5x=2d (we may now replace d by 3x since that was given)

35+5x=2(3x) ==> x=35 to get the number of jumps by the cat, just substitute x in the equation z=2(3x)

You should get z=70 jumps :)