Question

Ram starts from home and moves North along a straight road. Shyam starts from the same home and along the same path, after some time and catches up with Ram at $12$ noon. One day, Shyam started one hour later than usual and caught up with Ram at $4:00$ PM the same day. If the ratio speed of Ram to speed of Shyam in lowest form is $\dfrac{r}{s}$ (where $r$ & $s$ are integers). Find the value of ($r + s$). (Ram and Shyam move with their constant speed.)

Answer 1

We have to use the distance formula as the distance that they cover are equal. We have to assume a time when they start and develop an expression with two variables. As their velocities are constant we can compare two equations for two cases and hence find the ratio.

Complete step by step answer:
Case:1 Let the distance travelled by Ram and Shyam be $d$.Let us first consider that case when Ram leaves home. Let Ram start home at time $t$ with a constant velocity $v$.The total time taken by Ram to cover distance $d$ is $12 - t$.Now using distance equation we get,
$d = v \times \left( {12 - t} \right) - - - - - - \left( 1 \right)$
Now, Shyam leaves home after $x$ time from Ram. It means heaves at time $t + x$ with a constant velocity $u$. The total time taken by Shyam to cover distance $d$ is $12 - (t + x)$.Now using distance equation we get,
$d = u \times \left( {12 - t - x} \right) - - - - - - \left( 2 \right)$
Comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,
$v \times \left( {12 - t} \right) = u \times \left( {12 - t - x} \right)$
Simplifying the equation we get,
$\left( {t - 12} \right)\left( {v - u} \right) = ux - - - - - - \left( 3 \right)$

Case:2 Let the total distance covered by Ram and Shyam be $k$and $4:00$ PM is $16$ hours for a $24$ hour clock. Let us first consider the case when Ram leaves home.Let Ram start home at time $t$ with a constant velocity $v$. The total time taken by Ram to cover distance $k$ is $16 - t$. Now using distance equation we get,
$k = v \times \left( {16 - t} \right) - - - - - - \left( 4 \right)$
Now, Shyam leaves home after $1$ hr from usual time $t + x$ time from Ram. It means heaves at time $t + x + 1$ with a constant velocity $u$.The total time taken by Shyam to cover distance $d$ is $16 - (t + x + 1)$. Now using distance equation we get,
$d = u \times \left( {16 - t - x - 1} \right) - - - - - - \left( 5 \right)$
Comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,
$v \times \left( {16 - t} \right) = u \times \left( {16 - t - x - 1} \right)$
Simplifying the equation we get,
$\left( {t - 16} \right)\left( {v - u} \right) = u\left( {x + 1} \right) - - - - - - \left( 6 \right)$

Dividing equation $\left( 6 \right)$ by $\left( 3 \right)$ we get,
$\dfrac{{\left( {t - 16} \right)\left( {v - u} \right)}}{{\left( {t - 12} \right)\left( {v - u} \right)}} = \dfrac{{u\left( {x + 1} \right)}}{{ux}}$
Now by eliminating and simplifying we get,
$\dfrac{{\left( {t - 16} \right)}}{{\left( {t - 12} \right)}} = \dfrac{{\left( {x + 1} \right)}}{x}$
$\Rightarrow t = 12 - 4x - - - - - \left( 7 \right)$
Putting the value of $t$ from equation $\left( 7 \right)$ to equation $\left( 3 \right)$ we get,
$(12 - 4x - 12)\left( {v - u} \right) = ux \\\ \Rightarrow - 4x\left( {v - u} \right) = ux \\\$
Eliminating $x$ and simplifying we get,
$- 4v + 4u = u \\\ \Rightarrow \dfrac{v}{u} = \dfrac{3}{4}$
So, the ratio of velocities,
$\dfrac{r}{s} = \dfrac{v}{u} = \dfrac{3}{4}$
Now we can say that $r = 3$ and $s = 4$

Thus, $\left( {r + s} \right) = 7$.

Note: As the distance is the only variable which remains the same for both so we have to use it here. We must use the variable which remains the same for both purposes. It must be noted here that if a person starts from a particular point and ends up at the same point, there is always a distance that has been covered by him but the displacement remains zero.