Footrace

Let –

Tusher’s speed be $T$ .

Avik’s speed be $A$ .

Kamal’s speed be $K$ .

$A=2T$

$K= \frac{T}{4}$

$A = 2T$

$K=\frac{T}{4}$

Because,

2T = T/4*(4 * 2)

2T = T/4*(8)

A = K*8

so, A = 8K

Given,

$A-K = 91 m$

$8K-K = 91 m$

$7K = 91 m$

$K= \frac{91}{7}m$

$\boxed{K = 13 m}$

Total distance = $A + K$

= $8K + K$

= 9K

= 9 * (13 m)

= 117 m (ans.)

Define the following variables:

The speed of Avik, Kamal and Tusher are $v_a$ , $v_k$ and $v_t$ respectively.

The distance Avik and Kamal ran are $d_a$ and $d_k$ respectively.

The time they ran is $t$ .

From the given information, we know that:

Avik runs twice as fast as Tusher: $v_a=2v_t \implies$ Eq.(1)

Kamal runs 4 times slower than Tusher: $v_k = \dfrac{1}{4}v_t \implies v_t=4v_k \implies$ Eq.(2)

Substitute Eq.(2) into Eq.(1), we have:

$v_a = 2(4v_k)\implies v_a=8v_k \implies$ Eq.(3)

Next, we know that after running some time, the distance between Avik and Kamal is $91$ m.

$\implies d_a - d_k = 91$

We know that

Distance = Speed $\times$ Time

Therefore,

$v_at-v_kt=91$

Substitute Eq.(3) in here:

$8v_kt-v_kt=91\\ 7v_kt=91\\ d_k = \dfrac{91}{7} = 13$

Calculate the final answer:

$d_a+d_k\\ =91+d_k+d_k\\ =91+2(13)\\ =\boxed{117}$