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In $11$ hours $C$ walks $12.5 \text{ km}$ less than what $D$ does in $12$ hours; and in $5$ hours $D$ walks $3.25 \text{ km}$ less than what $C$ does in $7$ hours. How many $\text{km}$ does each walk per hour?

Attempted solution:

Let speed of $C$ and $D$ be $S_C$ and $S_D$ respectively. Let $D$ travel $x \text{ km}$ in $12$ hours. Let $C$ travel $x \text{ km}$ in $7$ hours.

Accordingly:

$$S_C=\frac{x-12.5}{11}$$

$$S_D=\frac{x}{12}$$

$$S_C=\frac{y}{7}$$

$$S_D=\frac{y-3.25}{5}$$

Fur variables, four unknowns, we should get the answer on solving.

On solving, I get $S_C= 7.07$ and $S_D=4.25$.

However the answer is $S_C= 3.5$ and $S_D=4.25$.

I am aware of other method of solving. But I want to know where I have gone wrong in formulating these equations. Please help.

2 Answers 2

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We know that speed is the ratio of distance and time, that is, distance = speed $\times $ time. Using your notation, we get, $$12S_D -11S_C =12.5$$ $$7S_C -5S_D =3.25$$ Hope you can take it from here by solving these two equations.

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    Sir, I understood this method. Its an easier and more efficient method. However why am I not getting the desired results on solving the equations? – 2017-01-25
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    @user3237657 In fact I rearranged your equations. Your equations are perfectly fine. Use the first two of them to get a relation between $S_C $ and $S_D $. Similarly for the last two. Definitely you will get what I have got. Please check your work again. – 2017-01-25
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    Thank you. This was really helpful – 2017-01-25
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Let $v_c$ and $v_d$ denote the velocity of c and d respectively. We can then write:

$\begin{align} v_c &= \frac{12v_d-12.5}{11} \\ v_d &= \frac{7v_c-3.25}{5} \\ \end{align}$

So this gives us a system of two equations with two variables, which implies we can solve the system. The easiest way to to that here is to isolate one of the two and then substitute it in the other equation. For example, we isolate $v_c$:

$v_d = \frac{7v_c-3.25}{5} \implies v_c = \frac{5v_d+3.25}{7}$.

I believe you can proceed yourself from here! :)

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    Sir Vd/12 has a unit of (time)^-1 and 12.5 has a unit of distance. How can we subtract unlike terms?.....I think Vd/12 should be replaced by Vd*12 and Vc/7 should be replaced by Vc*7. – 2017-01-25
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    Excuse me! You are fully right of course. – 2017-01-25
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    However, on solving these equations I am not getting the desired results :( – 2017-01-25