$A$ can walk certain distance in $20$ days when he rests $6$ hours a day Problem : $A$ can walk certain distance in $20$ days when he rests $6$ hours a day. How long will he take to walk the distance twice as fats and resting twice as long each day ?

Solution :  let distance be $x$

time taken without rest  $=20 -5 =15$ days

speed =$ \frac {x}{15}$

new speed = $ \frac {2x}{15}$

distance =  $x$

time  taken without rest=  $ \frac {15x}{2x}$

=7.5

time taken with rest $=12 *7 =84$ hours

$3.5$days

total time taken $=7.5 +3.5=11$ days

But this answer is not correct.

Help me
When he rests $6$ hours a day, that would mean he would walk $24-6=18$ hours a day, or $\frac{18}{24}=\frac{3}{4}$ days everyday. He does this for $20$ days, so the total time he walks would be $20 \times \frac{3}{4} = 15$ days. 

If the distance he walks is $x$, his speed would be $\frac{x}{15}$.

Then his new speed is $\frac{x}{15}\times{2}=\frac{2x}{15}$

He walks for the same distance, so the product between the new speed and the new time would still be $x$: (remember: $\frac{\text{distance}}{\text{time}}=\text{speed}$)

$$
\frac{x}{t}=\frac{2x}{15} \\x \times 15 = 2x \times t \\ t = \frac{15x}{2x}=\frac{15}{2}
$$
This means it would take him $\frac{15}{2}$ days of nonstop walking to walk the same distance.

But he rests twice more than he did before, so he would walk $24-6\times2 = 12$ hours a day, or $\frac{12}{24}=\frac 1 2$ day everyday. That is, he would walk $\frac{1}{2}$ of some number of days, and at the end, he would have walked for $\frac{15}{2}$ days. Then:

$$
\text{Total days} \times \frac{1}{2} = \frac{15}{2} \\
\text{Total days} = 15 \text{ days}
$$

Hope this helps.
20 x 18 x 1(let speed is 1km/hr)  x 2(distance twice) = D(Number of days) x 12 x 1(speed)

D = 60 days