Amar, Akbar, Anthony

Consider the travelling time of Anthony and Akbar:

1. Anthony first traveled by car for $t_1$ hours and then walked for $t_2$ hours
2. Akbar first walked for $t_3$ hours and then traveled by for $t_2$ hours
3. Amar dropped Anthony at $t_1$ and turned back to picked up Akbar at $t_3$
4. Anthony and Akbar arrived at the destination at the same time

In equations $\begin{cases} 25t_1+5t_2 = 100 & ...(1) \\ 5t_3+25t_4 = 100 & ...(2) \\ 25t_1 - 25(t_3-t_1) = 5t_3 & ...(3) \\ t_1 + t_2 = t_3 + t_4 & ...(4) \end{cases}$

$\begin{aligned} (1)=(2): \quad 25t_1+5t_2 & = 5t_3+25t_4 \\ 5t_1+t_2 & = t_3+5t_4 & ...(5) \end{aligned}$

$\begin{aligned} (5)=(4): \quad 4t_1 & = 4t_4 \\ \implies t_1 & = t_4 & ...(6) \\ \implies t_2 & = t_3 & ...(7) \end{aligned}$

$\begin{aligned} (3): \quad 25t_1 - 25(t_3-t_1) = 5t_3 \\ 50t_1 & = 30t_3 \\ \implies 25t_1 & = 15t_2 & \small \blue{\text{Since }(7): t_2=t_3} \end{aligned}$

$\begin{aligned} (1): \quad 25t_1 + 5t_2 = 100 \\ 15t_2 + 5t_2 & = 100 \\ \implies t_2 & = 5 \\ \implies t_1 & = \frac 35 t_2 = 3 \end{aligned}$

Therefore $t_1 + t_2 = 3 + 5 = \boxed 8$ hours.

Let Amar and Anthony went together by car for $t_1$ hrs. Amar drove alone to pick up Akbar for $t_2$ hrs. and the total time of journey is $t$ hrs. Then we get

$25t_1-5t_1=25t_2+5t_2\implies t_2=\dfrac{2t_1}{3}$

$100-25t_1=5(t-t_1)\implies t=20-4t_1$

$100-5(t_1+t_2)=25(t-t_1-t_2)\implies t=4+\dfrac{4t_1}{3}$ .

So $20-4t_1=4+\dfrac{4t_1}{3}\implies t_1=3$ and hence $t=4+\dfrac{4\times 3}{3}=\boxed 8$ hrs.