Little Red Riding Hood Riddle

Let $S$ be the distance, in km., between the two houses; $s_1$ be the distance of the first leg travelled by both ladies (where the girl walked and grandmother sat in the cart) with time taken $t_1$ in hours; $s_2$ be that of the second leg (where both ladies walked while the cart was catching up with the girl); and $s_3$ be that of the third leg (where the girl sat in the cart and the grandmother walked). We could then set up the equations as followed:

$s_1 = t_1 (7+14) = 21 t_1$ $s_2 = t_2 (7+16) = 13 t_2$ $s_3 = t_3 (14+6) = 20 t_3$

Now with insignificant time getting on and off the cart, we could visualize the whole $S$ from the point at the tree log fallen, where time $t_1$ had passed. From that spot, the grandmother walked home while the cart was heading to the little girl's house, and since they arrived at the same time, the time taken was also equal. This time is the total time subtracted by $t_1$ or $2-t_1$ , and we could set up the equation as:

$S = (2-t_1)(14+6) = 20(2-t_1)$

Therefore, we would attempt to calculate $t_1$ by revisiting the time the woodsman reversed his direction. From that point on, he would need to spend another $t_1$ time to reach their first meeting point. That would leave him $t_2 - t_1$ time to catch up with the little girl, where his advanced speed difference is $14-7 = 7$ km./hour, and she already made a distance of $7(2t_1) = 14t_1$ ahead of him. Therefore, $t_2 - t_1 = \dfrac{14t_1}{7} = 2t_1$ , and $t_2 = 3t_1$ .

From the last clue, the distance travelled by the ladies on foot equaled to that by the cart. We could then conclude that half of $S$ was travelled by the cart with the ladies on: $14t_1 + 14t_3 = 14(t_1 + t_3)$ . Since $2 = t_1 + t_2 + t_3$ , then $t_3 = 2 - t_1 - t_2 = 2-4t_1$ . Altogether, we will obtain another equation:

$\dfrac{S}{2} = 14(2-3t_1)$

Equivalently, we could equalize both equations above:

$S = 28(2-3t_1) = 20(2-t_1)$

Hence, $7(2-3t_1) = 5(2-t_1)$ . Then, $14-10 = t_1(21-5)$ ; $4 = 16t_1$ . Thus, $t_1 = \dfrac{1}{4}$ .

Substituting the value, we will obtain $S = 20(2-t_1) = 40 - 5 = \boxed{35}$ .

Let $A$ be the distance that the girl rode in the cart, let $B$ be the distance of that the girl walked, let $C$ be the distance that the grandmother rode in the cart, and let $D$ be the distance that the grandmother walked.

Since the girl walked $7$ km/h and the cart drove at $14$ km/h and the whole trip was $2$ hours, $\frac{B}{7} + \frac{A}{14} = 2$ .

Since the cart drove at $14$ km/h and the grandmother walked $7$ km/h and the whole trip was $2$ hours, $\frac{C}{14} + \frac{D}{7} = 2$ .

Since the cart drove over sections $C$ , then $C$ , then $B$ , and then finally over $A$ at $14$ km/h over $2$ hours, $14 \cdot 2 = A + B + 2C$ .

Since the total distance both ladies traveled by the cart equaled the distance traveled by on foot, $A + C = B + D$ .

These four equations solve to $A = 14$ , $B = 7$ , $C = 3.5$ , and $D = 10.5$ , so the houses were $A + B + C + D = \boxed{35}$ km apart.

Once home, Little Red Riding Hood realized that her grandmother must be hungry after all the walking she had to do, so she thoughtfully made some baked goods to bring over in a basket ...