Algebra - Word Problems: Derivation of Equations

In accordance with the schedule, a train is to travel the distance between $A$ and $B$, equal to $20$ km, at a constant speed. It traveled half-way with the specified speed and stopped for three minutes. To arrive at point $B$ on time, it had to increase its speed by $10$ kmph for the rest of the way. Next time the train stopped half-way for five minutes. At what speed must it travel the remaining half of the distance to arrive at point $B$ in accordance with the schedule?

Given: $\;$ Distance $d\left(AB\right) = 20$ km

Let the original speed of train $= s$ kmph

Time taken by the train to reach point $B$ $= t_B = \dfrac{20}{s}$ hr $\;\;\; \cdots \; (1)$

Let half-way point be $= A'$

Then, distance $d\left(AA'\right) = 10$ km

Time taken by the train to cover $d\left(AA'\right) = t_1 = \dfrac{10}{s}$ hr $\;\;\; \cdots \; (2)$

Train stops for $t_{stop1} = 3$ min $= \dfrac{3}{60} = \dfrac{1}{20}$ hr $\;\;\; \cdots \; (3)$

New speed of train $= s_1 = s + 10$ kmph

Distance $d\left(A'B\right) = 10$ km is covered with speed $= s_1 = \left(s + 10\right)$ kmph

Time taken to cover distance $d\left(A'B\right) = t_2 = \left(\dfrac{10}{s + 10}\right)$ hr $\;\;\; \cdots \; (4)$

Time taken by the train to reach $B$ $= t_B = t_1 + t_{stop1} + t_2$

$\therefore \;$ We have from equations $(2)$, $(3)$ and $(4)$

$t_B = \dfrac{10}{s} + \dfrac{1}{20} + \dfrac{10}{s + 10}$ hr $\;\;\; \cdots \; (5)$

$\therefore \;$ We have from equations $(1)$ and $(5)$

$\dfrac{20}{s} = \dfrac{10}{s} + \dfrac{1}{20} + \dfrac{10}{s + 10}$

i.e. $\;$ $\dfrac{10}{s} - \dfrac{10}{s + 10} = \dfrac{1}{20}$

i.e. $\;$ $10 \left[\dfrac{1}{s} - \dfrac{1}{s + 10}\right] = \dfrac{1}{20}$

i.e. $\;$ $10 \left[\dfrac{s + 10 - s}{s \left(s + 10\right)}\right] = \dfrac{1}{20}$

i.e. $\;$ $\dfrac{100}{s^2 + 10s} = \dfrac{1}{20}$

i.e. $\;$ $s^2 + 10s - 2000 = 0$

i.e. $\;$ $\left(s + 50\right) \left(s - 40\right) = 0$

i.e. $\;$ $s = -50$ $\;$ or $\;$ $s = 40$

Since the speed of the train cannot be negative

$s = -50$ is not an acceptable solution

$\therefore \;$ Original speed of train $= s = 40$ kmph

$\therefore \;$ Time taken to reach $B$ $= t_B = \dfrac{20}{40} = \dfrac{1}{2}$ hr $\;\;\; \cdots \; (1a)$ [in view of equation $(1)$]

and time taken to cover distance $d\left(AA'\right) = t_1 = \dfrac{10}{40} = \dfrac{1}{4}$ hr $\;\;\; \cdots \; (2a)$ [in view of equation $(2)$]

Now, the train stops for time $= t_{stop2} = 5$ min $= \dfrac{5}{60} = \dfrac{1}{12}$ hr $\;\;\; \cdots \; (6)$

Let the new speed of train be $= s_2$ kmph

Since $A'$ is the halfway point, distance $d\left(A'B\right) = 10$ km

Now, the train covers distance $d\left(A'B\right)$ with speed $s_2$

Time taken by the train to cover $d \left(A'B\right) = t_3 = \dfrac{10}{s_2}$ hr $\;\;\; \cdots \; (7)$

$\therefore \;$ Time taken by the train to reach point $B$ $= t_B = t_1 + t_{stop2} + t_3$

$\therefore \;$ In view of equations $(2a)$, $(6)$ and $(7)$ we have

$t_B = \dfrac{1}{4} + \dfrac{1}{12} + \dfrac{10}{s_2}$ hr $\;\;\; \cdots \; (8)$

Since the train is to arrive at $B$ in accordance with the schedule

$\therefore \;$ we have from equations $(1a)$ and $(8)$,

$\dfrac{1}{4} + \dfrac{1}{12} + \dfrac{10}{s_2} = \dfrac{1}{2}$

i.e. $\;$ $\dfrac{10}{s_2} = \dfrac{1}{2} - \dfrac{1}{4} - \dfrac{1}{12} = \dfrac{1}{6}$

i.e. $\;$ $s_2 = 60$

$\therefore \;$ New speed of train $= s_2 = 60$ kmph

Algebra - Word Problems: Derivation of Equations

A motor boat went down the river for $14$ km and then upstream for $9$ km, having covered the whole way in five hours. Find the speed of the river flow if the speed of the boat in still water is $5$ kmph.

Speed of boat in still water $= u = 5$ kmph

Let speed of river flow $= v$ kmph

Then, speed of boat down-stream $= s_d = \left(u + v\right)$ kmph $= \left(5 + v\right)$ kmph

and speed of boat up-stream $= s_{up} = \left(u - v\right)$ kmph $= 5 - v$ kmph

Distance covered by the boat down-stream $= d_d = 14$ km

Time for which the boat went down-stream $= t_d = \dfrac{d_d}{s_d} = \dfrac{14}{5 + v}$ hr

Distance covered by the boat up-stream $= d_{up} = 9$ km

Time for which the boat went up-stream $= t_{up}= \dfrac{d_{up}}{s_{up}} = \dfrac{9}{5 - v}$ hr

Time taken by the boat to go up-stream and down-stream

$= t_{total} = t_{up} + t_{d} = \dfrac{9}{5-v} + \dfrac{14}{5 + v}$

Given: $\;$ $t_{total} = 5$ hr

$\implies$ $\dfrac{9}{5 - v} + \dfrac{14}{5 + v} = 5$

i.e. $\;$ $\dfrac{45 + 9v + 70 - 14v}{25 - v^2} = 5$

i.e. $\;$ $115 - 5v = 125 - 5v^2$

i.e. $\;$ $5v^2 - 5v - 10 = 0$

i.e. $\;$ $v^2 - v - 2 = 0$

i.e. $\;$ $\left(v - 2\right) \left(v + 1\right) = 0$

i.e. $\;$ $v = 2$ $\;$ or $\;$ $v = -1$

$\because \;$ the speed of river flow cannot be negative

$\implies$ $v = -1$ is not a valid solution

$\therefore \;$ Speed of river flow $= v = 2$ kmph

Algebra - Word Problems: Derivation of Equations

The train left station A for station B. Having traveled $450$ km, which constitutes $75$ percent of the distance between A and B, the train was stopped by a snow drift. Half an hour later the track was cleared and the engine driver, having increased the speed by $15$ kmph, arrived at station $B$ on time. Find the initial speed of the train.

Let distance between stations A and B = $d\left(AB\right) = x$ km

Let the train stop at $A'$

Given: $\;$ $d\left(AA'\right) = 450$ km and $d\left(AA'\right) = 75\% \text{ of } d\left(AB\right)$

i.e. $\;$ $450 = \dfrac{75}{100} \times x$

i.e. $\;$ $x = \dfrac{450 \times 100}{75} = 600$ km

i.e. $\;$ distance between stations A and B $= d\left(AB\right) = 600$ km

$\therefore \;$ Remaining distance $= d\left(A'B\right) = 600 - 450 = 150$ km

Let initial speed of train $= s$ kmph

Then, time taken to cover $d\left(AB\right) = t_{\left(AB\right)} = \dfrac{d\left(AB\right)}{s} = \dfrac{600}{s}$ hr $\;\;\; \cdots \; (1)$

Time taken to cover $d\left(AA'\right) = t_{\left(AA'\right)} = \dfrac{d\left(AA'\right)}{s} = \dfrac{450}{s}$ hr

Train stopped at A' for $\dfrac{1}{2}$ hour

New speed of train $= s_1 = \left(s + 15\right)$ kmph

Time taken to cover distance $d\left(A'B\right) = t_{\left(A'B\right)} = \dfrac{d\left(A'B\right)}{s_1} = \left(\dfrac{150}{s+15} + \dfrac{1}{2}\right)$ hr

Now, time taken to cover $d\left(AB\right) = t_{\left(AB\right)} = t_{\left(AA'\right)} + t_{\left(A'B\right)}$

i.e. $\;$ $t_{\left(AB\right)} = \dfrac{450}{s} + \dfrac{150}{s + 15} + \dfrac{1}{2}$ $\;\;\; \cdots \; (2)$

Since the train arrives at station B on time after stoppage,

we have from equations $\left(1\right)$ and $\left(2\right)$

$\dfrac{450}{s} + \dfrac{150}{s + 15} + \dfrac{1}{2} = \dfrac{600}{s}$

i.e. $\;$ $\dfrac{150}{s + 15} + \dfrac{1}{2} = \dfrac{600}{s} - \dfrac{450}{s} = \dfrac{150}{s}$

i.e. $\;$ $\dfrac{150}{s} - \dfrac{150}{s + 15} = \dfrac{1}{2}$

i.e. $\;$ $150 \times 15 \times 2 = s\left(s + 15\right)$

i.e. $\;$ $s^2 + 15s - 4500 = 0$

i.e. $\;$ $\left(s+75\right)\left(s-60\right) = 0$

i.e. $\;$ $s = -75$ $\;$ or $\;$ $s = 60$

Since the speed of a train cannot be negative $\implies$ $s = -75$ is not a valid solution

$\therefore \;$ The original speed of train $= s = 60$ kmph