Algebra - Word Problems: Derivation of Equations

If a steamer and a motor-launch go down stream, then the steamer covers the distance from A to B $1.5$ times as fast as the motor-launch, the latter lagging behind the steamer $8$ km more each hour. Now if they go up stream, then the steamer covers the distance from B to A twice as fast as the motor-launch. Find the speeds of the steamer and the motor-launch in still water.

Direction A to B is downstream

Direction B to A is upstream

Let speed of steamer in still water $= u$ kmph

Let speed of motor-launch in still water $= v$ kmph

Let speed of stream $= s$ kmph

Speed of steamer downstream $= \left(u + s\right)$ kmph

Speed of motor-launch downstream $= \left(v + s\right)$ kmph

Given: Steamer downstream speed $=$ motor-launch downstream speed $\times 1.5$

i.e. $\;$ $u + s = \left(v + s\right) \times 1.5$

i.e. $\;$ $u + s = 1.5 v + 1.5s$

i.e. $\;$ $u - 1.5 v = 0.5 s$

i.e. $\;$ $s = 2u - 3v$ $\;\;\; \cdots \; (1)$

In view of equation $(1)$,

speed of steamer downstream $= u + 2u - 3v = 3u - 3v$ kmph $\;\;\; \cdots \; (2)$

speed ofmotor-launch downstream $= v + 2u - 3v = 2u - 2v$ kmph $\;\;\; \cdots \; (3)$

speed of steamer upstream $= \left(u - s\right) = u - 2u + 3v = 3v - u$ kmph $\;\;\; \cdots \; (4)$

speed of motor-launch upstream $= \left(v - s\right) = v - 2u + 3v = 4v - 2u$ kmph $\;\;\; \cdots \; (5)$

Given: Steamer upstream speed $=$ motor-launch upstream speed $\times 2$

i.e. $\;$ $3v - u = \left(4v - 2u\right) \times 2$

i.e. $\;$ $3v - u = 8v - 4u$

i.e. $\;$ $3u = 5v$

i.e. $\;$ $v = 0.6 u$ $\;\;\; \cdots \; (6)$

In view of equation $(6)$, we have

from equation $(2)$, speed of steamer downstream $= 3u - \left(3 \times 0.6u\right) = 1.2u$ kmph $\;\;\; \cdots \; (7)$

from equation $(3)$, speed of motor-launch downstream $= 2u - \left(2 \times 0.6u\right) = 0.8u$ kmph $\;\;\; \cdots \; (8)$

Let distance between points A and B be $= d$ km

While going downstream,

time taken by steamer to cover distance AB $= \dfrac{d}{1.2u}$ hr

time taken by motor-launch to cover distance AB $= \dfrac{d}{0.8u}$ hr

$\therefore \;$ Excess time taken by motor-launch to cover AB $= T = \dfrac{d}{0.8u} - \dfrac{d}{1.2u} = \dfrac{d}{2.4u}$ hr $\;\;\; \cdots \; (9)$

For every hour, the motor-launch lags the steamer by $8$ km

$\therefore \;$ For time $T$, the motor-launch lags the steamer by $= \dfrac{d}{2.4u} \times 8 = \dfrac{d}{0.3u}$ km $\;\;\; \cdots \; (10)$

Now, in time $T = \dfrac{d}{2.4u}$ hr,

the steamer covers downstream a distance $= d_1 = 1.2u \times \dfrac{d}{2.4u} = \dfrac{d}{2}$ km

and the motor-launch covers downstream a distance $= d_2 = 0.8u \times \dfrac{d}{2.4u} = \dfrac{d}{3}$ km

$\therefore \;$ In time $T$, the motor-launch lags the steamer by $= d_1 - d_2 = \dfrac{d}{2} - \dfrac{d}{3} = \dfrac{d}{6}$ km $\;\;\; \cdots \; (11)$

We have from equations $(10)$ and $(11)$

$\dfrac{d}{0.3u} = \dfrac{d}{6}$ $\implies$ $u = 20$ kmph

$\therefore \;$We have from equation $(6)$, $v = 0.6 \times 20 = 12$ kmph

Therefore, speed of steamer in still water $= 20$ kmph

and speed of motor-launch in still water $= 12$ kmph

Algebra - Word Problems: Derivation of Equations

A goods train left the town M for the town N at 5 a.m. An hour and a half later a passenger train left M, whose speed was $5$ km/h higher than that of the goods train. At 9.30 p.m. of the same day the distance between the trains was $21$ km. Find the speed of the goods train.

Let the speed of the goods train $= u$ kmph

Speed of passenger train $= \left(u + 5\right)$ kmph

Goods train leaves town M at 5.00 am

Passenger train leaves town M an hour and a half later i.e. at 6.30 am

At 9.30 pm, time for which goods train is traveling $= 16.5$ hrs

At 9.30pm, time for which passenger train is traveling $= 15$ hrs

Distance covered by goods train in $16.5$ hrs $= d_g = u \times 16.5$ km

Distance covered by passenger train in $15$ hrs $= d_p = \left(u + 5\right) \times 15 = \left(15 u + 75\right)$ km

Given: Distance between the trains at 9.30 pm is $21$ km

i.e. $\;$ $d_p - d_g = 21$ $\;$ or $\;$ $d_g - d_p = 21 $

i.e. $\;$ $15u + 75 - 16.5 u = 21$ $\;$ or $\;$ $16.5 u - 15 u - 75 = 21$

i.e. $\;$ $1.5 u = 54$ $\;$ or $\;$ $1.5 u = 96$

i.e. $\;$ $u = \dfrac{54}{1.5} = 36$ kmph $\;$ or $\;$ $u = \dfrac{96}{1.5} = 64$ kmph

Therefore, speed of the goods train $= 36$ kmph $\;$ or $\;$ $64$ kmph

Algebra - Word Problems: Derivation of Equations

The distance between $A$ and $B$ is $30$ km. A bus left $A$ and first travelled at a constant speed. Ten minutes later a helicopter left $A$ and flew along the highroad to $B$. It overtook the bus in five minutes and continued on its way to $B$. Without landing at $B$, the helicopter turned back and again encountered the bus $20$ minutes after it left point $A$. Determine the speeds of the bus and the helicopter.

Given: Distance $AB = d\left(AB\right) = 30$ km

Let speed of bus $= u$ kmph

Helicopter starts after $10$ minutes after the bus

Therefore, time taken by bus $= \text{time taken by helicopter} + 10 \text{ min}$

Helicopter overtook the bus in $5$ minutes

i.e. $\;$ the bus has been traveling for $5 + 10 = 15$ min

Distance covered by the bus in $15$ min i.e. $\left(\dfrac{1}{4} \text{hr}\right)= \dfrac{u}{4}$ km

The helicopter covers a distance $\left(\dfrac{u}{4}\right)$ km in $5$ min i.e. $\dfrac{1}{12}$ hr

Therefore, speed of helicopter $= s_h = \dfrac{u/4}{1/12} = 3u$ kmph

The helicopter meets the bus in $20$ min

i.e. $\;$ the bus has been traveling for $20 + 10 = 30$ min

Let the bus reach point $A_1$ in $30$ min

Distance covered by the bus in $30$ min i.e. $\left(\dfrac{1}{2} \text{hr}\right) = d \left(AA_1\right) = \dfrac{u}{2}$ km

Remaining distance $= d\left(A_1B\right) = 30 - \dfrac{u}{2} km$

As per sum, distance covered by helicopter in $20$ min $= 30 + 30 - \dfrac{u}{2} = 60 - \dfrac{u}{2}$ km $\;\;\; \cdots \; (1)$

Also, distance covered by helicopter in $20$ min i.e. $\left(\dfrac{1}{3} \text{hr}\right) = \dfrac{3u}{3} = u$ km $\;\;\; \cdots \; (2)$

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

$60 - \dfrac{u}{2} = u$

i.e. $\;$ $60 = u + u/2 = 3u /2$ $\implies$ $u = 40$

i.e. $\;$ speed of bus $= 40$ kmph

Therefore, speed of helicopter $= 3u = 3 \times 40 = 120$ kmph

Algebra - Word Problems: Derivation of Equations

A boat goes down the river from point A to point B, which is at the distance of $10$ km from A, and then returns to A. If the actual speed of the boat is $3$ km/h, then it takes $2$ h $30$ min less for the boat to go from A to B than from B to A. What should the actual speed of the boat be for the distance from A to B to be covered in two hours?

Let speed of boat in still water $= u$ kmph

and speed of river $= v$ kmph

Then, speed of boat down river (from A to B) $= \left(u + v\right)$ kmph

and speed of boat up river (from B to A) = $\left(u - v\right)$ kmph

Distance AB $= d \left(AB\right) = 10$ km (given)

Case 1:

Speed of boat in still water $= u = 3$ kmph (given)

Time taken by the boat to go from A to B $= T_1 = \dfrac{10}{u + v} = \dfrac{10}{3 + v}$ hr

Time taken by the boat to go from B to A $= T_2 = \dfrac{10}{u - v} = \dfrac{10}{3 - v}$ hr

As per problem, $T_1 = T_2 - 2.5$ hr

i.e. $\;$ $\dfrac{10}{3 + v} = \dfrac{10}{3 - v} - 2.5$

i.e. $\;$ $\dfrac{10}{3 - v} - \dfrac{10}{3 + v} = 2.5$

i.e. $\;$ $\dfrac{30 + 10v - 30 + 10v}{9 - v^2} = 2.5$

i.e. $\;$ $20v = 22.5 - 2.5 v^2$

i.e. $\;$ $5v^2 + 40v - 45 = 0$

i.e. $\;$ $v^2 + 8v - 9 = 0$

i.e. $\;$ $\left(v + 9\right) \left(v - 1\right) = 0$

i.e. $\;$ $v = -9$ $\;$ or $\;$ $v = 1$

Since speed of river cannot be negative, $\implies$ $v = -9$ is not an acceptable solution

$\therefore \;$ Speed of river $= v = 1$ kmph

Case 2:

Given: Time taken by the boat to cover $d \left(AB\right) = 2$ hr

Speed of boat from A to B $= u + v = u + 1 = \dfrac{\text{distance}}{\text{time}} = \dfrac{10}{2} = 5$ kmph

i.e. $\;$ $u = 5 - 1 = 4$ kmph

Therefore, if the speed of boat in still water $= u = 4$ kmph, then the distance AB will be covered in $2$ hours.

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