Question

A takes 3 hour more than B to walk 30 km. But if A doubles his pace, he is ahead of B by $1\dfrac{1}{2}$ hours. Find their speed of walking.

Answer 1

Hint: Here we will proceed by using the substitution method of solving linear equations in two variables. Then by applying the conditions given in the question we will get our answer.

Complete step-by-step solution:
It is given that –
Distance $ = $ 30 km.
Let speed of $A$ be $ = x$ km per hour and speed of B be $ = y$km per hour
We know that, $Time = \dfrac{{Dis\tan ce}}{{Speed}}$ ,
Time taken by $A$ to cover 30 km $ = $$\dfrac{{30}}{x}$
Time taken by $B$ to cover 30 km $ = \dfrac{{30}}{y}$
Now,
$\dfrac{{30}}{x}$ $ = \dfrac{{30}}{4} + 3$
$\dfrac{{30}}{x} - \dfrac{{30}}{y} = 3$ …… (1)
When $A$ doubles his speed, then the time taken by $A$ $ = \dfrac{{30}}{2x}$
$
  \dfrac{{30}}{y} = \dfrac{{30}}{{2x}} + \dfrac{3}{2} \\
  \dfrac{{30}}{y} - \dfrac{{30}}{{2x}} = \dfrac{3}{2} \\
 $
$\dfrac{{30}}{y} - \dfrac{{15}}{x} = \dfrac{3}{2}$ ….. (2)
Let us assume that $\dfrac{1}{x}$ $ = $ p and $\dfrac{1}{y} = $ q
So equation (1) and (2) becomes
$30p - 30q = 3$ …. (iii)
$30q - 15p = \dfrac{3}{2}$ …. (iv)
On adding equation (iii) and (iv) we get,
$30p - 30q = 3$
$ + \left( { - 15p} \right) + 30q = \dfrac{3}{2}$
From $p$ we get,
$15p = \dfrac{9}{2}$
$p = \dfrac{9}{{\left( {2 \times 15} \right)}}$
$p = \dfrac{3}{{10}}$
Substituting the value of $p = \dfrac{3}{{10}}$ in equation (iii) we get
$
  30p - 30q = 3 \\
  30\left( {\dfrac{3}{{10}}} \right) - 3q = 3 \\
  9 - 30q = 3 \\
   - 30q = - 6 \\
  q = \dfrac{1}{5} \\
 $
Now we know that $\dfrac{1}{y} = q$
$\dfrac{1}{y} = \dfrac{1}{5}$
$y = 5$ km per hour.
And, $\dfrac{1}{x} = p$
$\dfrac{1}{x} = \dfrac{3}{{10}}$
$x = \dfrac{{10}}{3}$ km per hour
Hence,
$A$ $ = \dfrac{{10}}{3}$ km per hour.
$B = 5$ km per hour.

Note: Whenever we come up with this type of question, one must know that the method of linear equations by substitution works by solving one of the equations for one of the variable, and then plugging this back into the other equation, “substitution” for the chosen variable and solving for the other. Then you back-solve for the first variable.