[Guest]

A man was being accused of having stolen jewelery and trying to run away with them, when he was caught by a policeman who overtook him. In cross examination the lawyer for the accused asked the police officer how he could catch up the accused who was already twenty seven steps ahead of him, when he started to run after him.

Officer: "Yes sir, he takes eight steps to every five of mine."

Lawyer: "How did you ever catch him then, if that was the case?"

Officer: "I have got a longer stride -- two steps of mine are equal to his five. So the number of steps I required were fewer than his."

And the lawyer was satisfied by the explanation.

Now, can you find out how many steps the officer needed to catch up with the thief?

[Guest]

If the thief's stride is 2 units, the officer's is 5 units.

The thief has a 54 unit head start.

If the thief takes 8 steps (16 units) per X amount of time, and the officer takes 5 (25 units), the thief's distance from the start at time T is 54+16(T/X), and the officers is 25(T/X).

54+16(T/X)=25(T/X) when they meet

T=6X

The officer took 30 steps.

[Guest]

Policeman took 30 steps.

[Guest]

30 steps

[bonanova]

OP does not say whose steps [officer's or thief's] measured the 27-step lead.

It's reasonable to assume it was the thief's.

If so,

Let N_o,t, S_o,t and f_o,t be the number of steps, length of a step and frequency of steps, respectively, of the officer and thief.

The distance from the officer to the point of capture is

[Eq. 1] N_oS_o = [27+N_t]S_t.

The time to reach the point of capture is

[Eq. 2] N_oS_o/f_o = N_tS_t/f_t

We're given

[Eqs. 3] S_o/S_t = 5/2; f_o/f_t = 5/8

Two equations in two unknowns.

N_t = 48; N_o = 30.