Algebra - Word Problems: Derivation of Equations

A pedestrian and a cyclist start simultaneously towards each other from towns $A$ and $B$ which are $40$ km apart and meet two hours after the start. Then they resumed their trips and the cyclist arrives at $A$ $7$ h $30$ min earlier than the pedestrian arrives at $B$. Find the speeds of the pedestrian and the cyclist.

The pedestrain starts from point $A$ and the cyclist from point $B$.

Distance between $A$ and $B$ $= 40$ km (given)

Let the pedestrian and the cyclist meet at point $P$.

Let the speed of the pedestrian be $= S_p$ kmph

and the speed of the cyclist be $= S_c$ kmph

In $2$ hours,

distance covered by the pedestrian $= d_{AP} = S_p \times 2$ km

distance covered by the cyclist $= d_{BP} = S_c \times 2$ km

Since the pedestrian and the cyclist meet after $2$ hours

$\implies$ $2 \; S_p + 2 \; S_c = 40$

i.e. $\;$ $S_p + S_c = 20$

i.e. $\;$ $S_p = 20 - S_c$ $\;\;\; \cdots \; (1)$

Remaining distance covered by the pedestrian $= d_{PB} = 2 \; S_c$ km

Since the speed of the pedestrian is $S_p$

$\therefore \;$ Time taken by the pedestrian from $P$ to $B$ is $= t_{PB} = \dfrac{d_{PB}}{S_p}$

i.e. $\;$ $t_{PB} = \dfrac{2 \; S_c}{S_p}$ $\;\;\; \cdots \; (2)$

Remaining distance covered by the cyclist $= d_{PA} = 2 \; S_p$ km

Since the speed of the cyclist $= S_c$

$\therefore \;$ Time taken by the cyclist from $P$ to $A$ $= t_{PA} = \dfrac{d_{PA}}{S_c}$

i.e. $\;$ $t_{PA} = \dfrac{2 \; S_p}{S_c}$ $\;\;\; \cdots \; (3)$

Given: $\;$ The cyclist arrives at point $A$ $7.5$ hours before the pedestrian arrives at point $B$.

i.e. $\;$ $t_{PA} = t_{PB} - 7.5$

i.e. $\;$ $\dfrac{2 \; S_p}{S_c} = \dfrac{2 \; S_c}{S_p} - 7.5$

i.e. $\;$ $\dfrac{2 \left(20 - S_c\right)}{S_c} = \dfrac{2 \; S_c}{20 - S_c} - 7.5$ $\;$ [in view of equation $(1)$]

i.e. $\;$ $\dfrac{2 \; S_c}{20 - S_c} - \dfrac{2 \left(20 - S_c\right)}{S_c} = 7.5$

i.e. $\;$ $2 \; S_c^2 - 2 \left(400 - 40 \; S_c + S_c^2\right) = 7.5 \left(20 \; S_c - S_c^2\right)$

i.e. $\;$ $-800 + 80 \; S_c = 150 \; S_c - 7.5 \; S_c^2$

i.e. $\;$ $7.5 \; S_c^2 - 70 \; S_c - 800 = 0$ $\;\;\; \cdots \; (4)$

Solving the quadratic equation $(4)$ gives

$S_c = 16$ $\;$ or $\;$ $S_c = \dfrac{-20}{3}$

Since the speed of the cyclist cannot be negative

$\implies$ Speed of cyclist $= S_c = 16$ kmph

Substituting the value of $S_c$ in equation $(1)$ gives

$S_p = 20 - 16 = 4$

i.e. $\;$ Speed of pedestrian $= S_p = 4$ kmph

Algebra - Word Problems: Derivation of Equations

A train left point $A$ at noon sharp. Two hours later another train started from point $A$ in the same direction. It overtook the first train at $8$ p.m. Find the average speeds of the trains if the sum of their average speeds is $70$ kmph.

Let the average speed of the first train be $= s_1$ kmph

and the average speed of the second train be $= s_2$ kmph

Given: $\;$ $s_1 + s_2 = 70$ kmph $\;\;\; \cdots \; (1)$

Time from noon sharp to $8$ p.m. $= 8$ hours

Let the first train cover a distance $d$ km in time $t_1 = 8$ hr

Then $\;$ $d = s_1 \times t_1 = 8 \times s_1$ $\;\;\; \cdots \; (2)$

Since the second train starts $2$ hours later and overtakes the first train at $8$ p.m.,

$\implies$ the second train covers the distance $d$ km in time $t_2 = 6$ hours

$\therefore \;$ Distance covered by the second train $= d = s_2 \times t_2 = 6 \times s_2$ $\;\;\; \cdots \; (3)$

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

$8 \times s_1 = 6 \times s_2$

i.e. $\;$ $s_2 = \dfrac{4 \; s_1}{3}$ $\;\;\; \cdots \; (4)$

Substituting the value of $s_2$ in equation $(1)$ gives

$s_1 + \dfrac{4 \; s_1}{3} = 70$

i.e. $\;$ $\dfrac{7 \; s_1}{3} = 70$ $\implies$ $s_1 = 30$

Substituing the value of $s_1$ in equation $(1)$ gives

$s_2 = 70 - 30 = 40$

$\therefore \;$ Speed of the first train $= s_1 = 30$ kmph

and speed of the second train $= s_2 = 40$ kmph

Algebra - Word Problems: Derivation of Equations

A car travels from point $A$ to point $B$ with a constant speed. If the driver increased the speed of the car by $6$ kmph, it would take him $4$ hours less to cover that distance. And traveling with the speed $6$ kmph lower than the initial speed, it would take him $6$ hours more. Find the distance between $A$ and $B$.

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

Let the time taken to cover the distance $AB = t$ hours

Let the speed of the car be $= s$ kmph

Then, $\;$ $d = s \times t$ $\;\;\; \cdots \; (1)$

Case 1: $\;$ Speed of car $= s_1 = \left(s + 6\right)$ kmph

Time taken to cover the distance $AB$ $= t_1 = \left(t - 4\right)$ hours (given)

Then, distance $AB = d = s_1 \times t_1 = \left(s + 6\right) \left(t - 4\right)$ km $\;\;\; \cdots \; (2)$

Case 2: $\;$ Speed of car $= s_2 = \left(s - 6\right)$ kmph

Time taken to cover distance $AB$ $= t_2 = \left(t + 6\right)$ hours (given)

Then, distance $AB = d = s_2 \times t_2 = \left(s - 6\right) \left(t + 6\right)$ km $\;\;\; \cdots \; (3)$

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

$\left(s + 6\right) \left(t - 4\right) = \left(s - 6\right) \left(t + 6\right)$

i.e. $\;$ $s \cdot t - 4s + 6t - 24 = s \cdot t + 6s - 6t - 36$

i.e. $\;$ $10 s - 12 t - 12 = 0$

i.e. $\;$ $5 s - 6t = 6$ $\;\;\; \cdots \; (4)$

We have from equations $(1)$ and $(2)$

$s \times t = \left(s + 6\right) \left(t - 4\right)$

i.e. $\;$ $s \cdot t = s \cdot t - 4s + 6t - 24$

i.e. $\;$ $-4s + 6t = 24$ $\;\;\; \cdots \; (5)$

Solving equations $(4)$ and $(5)$ simultaneously (adding the two equations) gives

$s = 30$ $\implies$ the speed of the car $= 30$ kmph

Substituing the value of $s$ in equation $(4)$ gives

$5 \times 30 - 6t = 6$

i.e. $\;$ $150 - 6t = 6$

i.e. $\;$ $6t = 144$ $\implies$ $t = 24$ $\;$ i.e. $\;$ time taken to cover distance $AB = 24$ hours

$\therefore \;$ By equation $(1)$, distance $AB = 30 \times 24 = 720$ km

Algebra - Word Problems: Derivation of Equations

Find the speed and the length of a train being given that it traveled with a constant speed past a stationary observer for $7$ seconds and needed $25$ seconds to pass a platform $378$ meter long with the same speed.

Length of the platform $= 378$ m

Let the length of the train $= \ell$ m

$\therefore \;$ Total distance covered by the train in crossing the platform $= \left(378 + \ell\right)$ m

Time taken to cross the platform $= 25$ s

$\therefore \;$ Speed of train $= \left(\dfrac{378 + \ell}{25}\right)$ m/s $\;\;\; \cdots \; (1)$

The train crosses a stationary observer in $7$ s

$\therefore \;$ Speed of train $= \left(\dfrac{\ell}{7}\right)$ m/s $\;\;\; \cdots \; (2)$

Since the train is traveling with the same speed, we have from equations $(1)$ and $(2)$

$\dfrac{378 + \ell}{25} = \dfrac{\ell}{7}$

i.e. $\;$ $2646 + 7 \ell = 25 \ell$

i.e. $\;$ $18 \ell = 2646$ $\implies$ $\ell = 147$ m

$\therefore \;$ Length of the train $= \ell = 147$ m

Substituting the value of $\ell$ in equation $(2)$ gives

Speed of train $= \dfrac{147}{7} = 21$ m/s $= \dfrac{21 \times 3600}{1000}$ kmph $= 75.6$ kmph

Algebra - Word Problems: Derivation of Equations

Two people started simultaneously towards each other from points $A$ and $B$ which are $50$ km apart. They met $5$ hours later. After their meeting, the first person who traveled from $A$ to $B$, decreased his speed by $1$ kmph and the other person who traveled from $B$ to $A$ increased her speed by $1$ kmph. The first person is known to arrive at $B$ two hours earlier than the second person arrived at $A$. Find the initial speed of the first person.

Distance between $A$ and $B$ $= 50$ km

Let the two people meet at point $P$ after $5$ hours.

Let distance $AP = x$ km

Then, distance $PB = \left(50 - x\right)$ km

$\therefore \;$ Speed of person traveling from $A$ $= \left(\dfrac{x}{5}\right)$ kmph

and speed of person traveling from $B$ $= \left(\dfrac{50 - x}{5}\right)$ kmph

New speed of person (who started from $A$) traveling from $P$ towards $B$ [distance $PB$] is

$S_{PB} = \left(\dfrac{x}{5} - 1\right) = \left(\dfrac{x - 5}{5}\right)$ kmph

New speed of person (who started from $B$) traveling from $P$ towards $A$ [distance $PA$] is

$S_{PA} = \left(\dfrac{50 - x}{5} + 1\right) = \left(\dfrac{55 - x}{5}\right)$ kmph

$\therefore \;$ Time taken by person traveling from $A$ to $B$ to arrive at point $B$ is

$= T_{AB} = \dfrac{PB}{S_{PB}} = \dfrac{50 - x}{\left(x - 5\right) / 5} = \left(\dfrac{250 - 5x}{x - 5}\right)$ hr

and time taken by person traveling from $B$ to $A$ to arrive at point $A$ is

$= T_{BA} = \dfrac{PA}{S_{PA}} = \dfrac{x}{\left(55 - x\right) / 5} = \left(\dfrac{5x}{55 - x}\right)$ hr

Given: $\;$ $T_{AB} = \left(T_{BA} - 2\right)$ hr

i.e. $\;$ $\dfrac{250 - 5x}{x - 5} = \dfrac{5x}{55 - x} - 2$

i.e. $\;$ $\dfrac{250 - 5x}{x - 5} = \dfrac{5x - 110 + 2x}{55 - x}$

i.e. $\;$ $\dfrac{250 - 5x}{x - 5} = \dfrac{7x - 110}{55 - x}$

i.e. $\;$ $13750 - 275 x - 250 x + 5x^2 = 7x^2 - 110 x - 35x + 550$

i.e. $\;$ $2x^2 + 380 x - 13200 = 0$

i.e. $\;$ $x^2 + 190 x - 6600 = 0$

i.e. $\;$ $\left(x + 220\right) \left(x - 30\right) = 0$

i.e. $\;$ $x = - 220$ $\;$ or $\;$ $x = 30$

$\because \;$ Distance covered cannot be negative $\implies$ $x = 30$ is the only acceptable solution.

$\therefore \;$ Initial speed of the person who traveled from $A$ to $B$ (i.e. the first person) is

$= \dfrac{x}{5}$ kmph $= \dfrac{30}{5} = 6$ kmph

Algebra - Word Problems: Derivation of Equations

A cyclist covered a distance of $96$ km two hours faster than he assumed. Every hour he traveled $1$ km more than he intended to cover in $1$ hr $15$ min. What was his actual and assumed speed?

Total distance covered by the cyclist $= 96$ km

Let the assumed time taken $= t$ hours

Then, assumed speed of the cyclist $= \left(\dfrac{96}{t}\right)$ kmph

Actual time taken by the cyclist $= \left(t - 2\right)$ hours

$\therefore \;$ Actual speed of the cyclist $= \left(\dfrac{96}{t - 2}\right)$ kmph $\;\;\; \cdots \; (1)$

Let the distance covered by the cyclist in $1$ hr $15$ min $\left(\dfrac{5}{4} \; \text{hr}\right)$ be $= d$ km

Then the distance covered in $1$ hr by the cyclist $= \left(d + 1\right)$ km

Now, distance covered in $\left(\dfrac{5}{4}\right)$ hr is the assumed distance covered with an assumed speed of $\left(\dfrac{96}{t}\right)$ kmph

$\therefore \;$ Assumed distance covered $= d = \dfrac{96}{t} \times \dfrac{5}{4} = \dfrac{120}{t}$ km

Actual distance covered in $1$ hour $= d + 1 = \left(\dfrac{120}{t} + 1\right)$ km

$\therefore \;$ Actual speed in $1$ hr $= \left(\dfrac{120}{t} + 1\right)$ kmph $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$

$\dfrac{96}{t - 2} = \dfrac{120}{t} + 1$

i.e. $\;$ $96 t = \left(120 + t\right) \left(t - 2\right)$

i.e. $\;$ $96 t = 120 t - 240 + t^2 - 2t$

i.e. $\;$ $t^2 + 22t - 240 = 0$

i.e. $\;$ $\left(t + 30\right) \left(t - 8\right) = 0$

i.e. $\;$ $t = -30$ $\;$ or $\;$ $t = 8$

$\because \;$ Time taken cannot be negative

$\implies$ Assumed time of the cyclist $= t = 8$ hr

$\therefore \;$ Assumed speed of the cyclist $= \dfrac{96}{t} = \dfrac{96}{8} = 12$ kmph

and actual speed of the cyclist $= \dfrac{96}{t - 2} = \dfrac{96}{8 - 2} = \dfrac{96}{6} = 16$ kmph

Algebra - Word Problems: Derivation of Equations

A skier had to cover a distance of $30$ km. Having started $3$ minutes later than the fixed time, he glided with the speed which exceeded the assumed speed by $1$ kmph and reached the finish at the time he would reach it if he began to race strictly at the appointed time and ran with the assumed speed. Find the speed of the skier.

Total distance covered by the skier $= d = 30$ km

Let the time taken to cover $d = 30$ km be equal to $t$ hours.

$\therefore \;$ Assumed speed of skier $= s = \dfrac{30}{t}$ kmph

Actual time taken by the skier $= t$ hours $- 3$ min $= \left(t - \dfrac{3}{60}\right)$ hr

$= \left(t - \dfrac{1}{20}\right)$ hr $= \left(\dfrac{20 t - 1}{20}\right)$ hr

$\therefore \;$ Actual speed of skier $= \dfrac{30}{\left(20t - 1\right) / 20} = \dfrac{600}{20t - 1}$ kmph

Given: $\;$ Actual speed $=$ Assumed speed $+ 1$ kmph

i.e. $\;$ $\dfrac{600}{20 t - 1} = \dfrac{30}{t} + 1$

i.e. $\;$ $600 t = \left(30 + t\right) \left(20 t - 1\right)$

i.e. $\;$ $600t = 600t - 30 + 20 t^2 - t$

i.e. $\;$ $20 t^2 - t - 30 = 0$

i.e. $\;$ $\left(4t - 5\right) \left(5t + 6\right) = 0$

i.e. $\;$ $4t - 5 = 0$ $\;$ or $\;$ $5t + 6 = 0$

i.e. $\;$ $t = \dfrac{5}{4}$ $\;$ or $\;$ $t = \dfrac{-6}{5}$

Since time cannot be negative

$\therefore \;$ Time taken to cover $d = 30$ km is $t = \dfrac{5}{4}$ hour

$\therefore \;$ Actual speed of skier $= \dfrac{600}{20 \times \dfrac{5}{4} - 1} = \dfrac{600}{24} = 25$ kmph