Algebra - Word Problems: Derivation of Equations

Two cyclists started simultaneously towards each other from points $A$ and $B$ which are $28$ km apart. An hour later they met and kept pedalling with the same speed without stopping. The first cyclist arrived at $B$ $35$ minutes earlier than the second arrived at $A$. Find the speed of each cyclist.

Given: $\;$ Distance between the points $A$ and $B$ $= d_{AB} = 28$ km

Let cyclist $1$ start from point $A$ towards point $B$

and cyclist $2$ start from $B$ towards $A$.

Let the two cyclists meet at point $P$ after time $= t = 1$ hr

Let cyclist $1$ cover a distance $x$ km in $1$ hour

i.e. $d_{AP} = x$ km

Then cyclist $2$ covers a distance $\left(28 - x\right)$ km in $1$ hour

i.e. $\;$ $d_{BP} = \left(28 - x\right)$ km

$\therefore \;$ Speed of cyclist $1$ $= s_1 = \dfrac{d_{AP}}{t} = \dfrac{x}{1} = x$ kmph

and speed of cyclist $2$ $= s_2 = \dfrac{d_{BP}}{t} = \dfrac{28 - x}{1} = \left(28 - x\right)$ kmph

To reach point $B$,

cyclist $1$ covers a distance $= d_{PB} = \left(28 - x\right)$ km with a speed of $x$ kmph

$\therefore \;$ Time taken by cyclist $1$ to reach point $B$ $= t_1 = \left(\dfrac{28 - x}{x}\right)$ hr

To reach point $A$,

cyclist $2$ covers a distance $= d_{PA} = x$ km with speed $\left(28 - x\right)$ kmph

$\therefore \;$ Time taken by cyclist $2$ to reach point $A$ $= t_2 = \left(\dfrac{x}{28 - x}\right)$ hr

Given: $\;$ Cyclist $1$ arrives at point $B$ $35$ minutes $\left(= \dfrac{35}{60} = \dfrac{7}{12} \text{ hr}\right)$ before cyclist $2$ arrives at point $A$.

i.e. $\;$ $t_1 = t_2 - \dfrac{7}{12}$

i.e. $\;$ $\dfrac{28 - x}{x} = \dfrac{x}{28 - x} - \dfrac{7}{12}$

i.e. $\;$ $\dfrac{x}{28 - x} - \dfrac{28 - x}{x} = \dfrac{7}{12}$

i.e. $\;$ $12 \left(x^2 - 784 + 56x - x^2\right) = 7x \left(28 - x\right)$

i.e. $\;$ $672x - 9408 = 196x - 7x^2$

i.e. $\;$ $7 x^2 + 476 x - 9408 = 0$

i.e. $\;$ $x^2 + 68 x - 1344 = 0$

i.e. $\;$ $\left(x + 84\right) \left(x - 16\right) = 0$

i.e. $\;$ $x = -84$ $\;$ or $\;$ $x = 16$

Since distance covered cannot be negative,

therefore, $x = 16$ is the only acceptable solution.

$\therefore \;$ Speed of cyclist $1$ $= s_1 = x = 16$ kmph

and speed of cyclist $2$ $= s_2 = 28 - x = 28 - 16 = 12$ kmph

Algebra - Word Problems: Derivation of Equations

A fisher had to sail $35$ km to the meeting place and the other fisher had to sail $31 \dfrac{3}{7}$ percent less. To arrive at the place at the same time as the other fisher, the first fisher satrted half an hour earlier and sailed with the speed exceeding by $2$ kmph the speed of the second fisher. Find the speed of each fisher and the time it took each of them to cover the distance.

Let $F_1$ and $F_2$ be the two fishers.

Let the fishers meet at a point $P$.

Let fisher $F_1$ sail $35$ km to the meeting place

Then, distance covered by $F_1$ $= F_1P = 35$ km $\;\;\; \cdots \; (1)$

and distance covered by $F_2$ $= F_2P = 35 - 35 \times \dfrac{31 \dfrac{3}{7}}{100} = 35 - \dfrac{35 \times 220}{700} = 24$ km $\;\;\; \cdots \; (2)$

Let time taken by fisher $F_2$ to reach point $P$ $= t$ hours

and the speed of fisher $F_2$ be $= s$ kmph

$\therefore \;$ Distance covered by $F_2$ to reach point $P$ $= F_2 P$

$F_2 P = s \times t = 24$ $\;\;\; \cdots \; (3)$ $\;$ [in view of $(2)$]

or, $\;$ $s = \dfrac{24}{t}$ $\;\;\; \cdots \; (3a)$

Since fisher $F_1$ started half an hour earlier than $F_2$ to reach point $P$,

time taken by $F_1$ $= t_1 = \left(t + 0.5\right)$ hours $\;\;\; \cdots \; (4a)$

$F_1$ sailed with a speed exceeding by $2$ kmph the speed of $F_2$.

$\therefore \;$ Speed of fisher $F_1$ $= s_1 = \left(s + 2\right)$ kmph $\;\;\; \cdots \; (4b)$

Distance covered by $F_1$ to reach point $P$ $= F_1 P$

$F_1 P = s_1 \times t_1 = \left(s + 2\right) \left(t + 0.5\right) = 35$ $\;$ [in view of $(1)$]

i.e. $\;$ $s \; t + 0.5 s + 2t + 1 = 35$

i.e. $\;$ $s \; t + 0.5 s + 2t = 34$ $\;\;\; \cdots \; (5)$

In view of equations $(3)$ and $(3a)$, equation $(5)$ becomes

$24 + 0.5 \times \dfrac{24}{t} + 2t = 34$

i.e. $\;$ $2 t^2 - 10 t + 12 = 0$

i.e. $\;$ $t^2 - 5t + 6 = 0$

i.e. $\;$ $\left(t - 3\right) \left(t - 2\right) = 0$

i.e. $\;$ $t = 3$ $\;$ or $\;$ $t = 2$

When $t = 3$,

$s = \dfrac{24}{3} = 8$ $\;$ [by equation $(3a)$]

$s_1 = 8 + 2 = 10$ $\;$ [by equation $(4b)$]

and $\;$ $t_1 = 3 + 0.5 = 3.5$ $\;$ [by equation $(4a)$]

When $t = 2$,

$s = \dfrac{24}{2} = 12$ $\;$ [by equation $(3a)$]

$s_1 = 12 + 2 = 14$ $\;$ [by equation $(4b)$]

and $\;$ $t_1 = 2 + 0.5 = 2.5$ $\;$ [by equation $(4a)$]

$\therefore \;$ Speed of $F_1$ $= 10$ kmph; $\;$ time taken by $F_1$ to reach $P$ $= 3.5$ hr

and speed of $F_2$ $= 8$ kmph; $\;$ time taken by $F_2$ to reach $P$ $= 3$ hr

OR

Speed of $F_1$ $= 14$ kmph; $\;$ time taken by $F_1$ to reach $P$ $= 2.5$ hr

and speed of $F_2$ $= 12$ kmph; $\;$ time taken by $F_2$ to reach $P$ $= 2$ hr

Algebra - Word Problems: Derivation of Equations

After their meeting, one ship went south and the other went west. Two hours after their meeting, they were $60$ km apart. Find the speed of each ship if the speed of one of them is known to be $6$ kmph higher than that of the other.

Let the speed of ship going south $= S_s$ kmph

Let the speed of ship going west $= S_w$ kmph

Let $\;$ $S_s = \left(S_w + 6\right)$ kmph $\;\;\; \cdots \; (1)$

Initially, both the ships meet at point $O$

After $2$ hours, ship going west covers a distance $= OW$

and the ship going south covers a distance $= OS$

$\therefore \;$ Speed of ship going west $= S_w = \dfrac{OW}{2}$

$\implies$ $OW = 2 \; S_w$ $\;\;\; \cdots \; (2a)$

and speed of ship going south $= S_s = \dfrac{OS}{2}$

$\implies$ $OS = 2 \; S_s$ $\;\;\; \cdots \; (2b)$

Since both the ships are traveling perpendicular to one-another from the point $O$, $\triangle OWS$ is a right angled triangle.

Given: $\;$ Distance between the ships $= WS = 60$ km

We have from right triangle $OWS$,

$WS^2 = OW^2 + OS^2$

i.e. $\;$ $60^2 = \left(2 \; S_w\right)^2 + \left(2 \; S_s\right)^2$ $\;$ [in view of equations $(2a)$ and $(2b)$]

i.e. $\;$ $3600 = 4 \; S_w^2 + 4 \; S_s^2$

i.e. $\;$ $900 = S_w^2 + S_s^2$ $\;\;\; \cdots \; (3)$

$\therefore \;$ In view of equation $(1)$ equation $(3)$ becomes

$900 = S_w^2 + \left(S_w + 6\right)^2$

i.e. $\;$ $900 = S_w^2 + S_w^2 + 12 \; S_w + 36$

i.e. $\;$ $2 \; S_w^2 + 12 \; S_w - 864 = 0$

i.e. $\;$ $S_w^2 + 6 \; S_w - 432 = 0$

i.e. $\;$ $\left(S_w + 24\right) \left(S_w - 18\right) = 0$

i.e. $\;$ $S_w = -24$ $\;$ or $\;$ $S_w = 18$

Since the spped of a ship cannot be negative

$\implies$ Speed of the ship sailing west from $O$ is $= S_w = 18$ kmph

Substituting the value of $S_w$ in equation $(1)$ gives $\;$ $S_s = 18 + 6 = 24$

$\implies$ Speed of the ship sailing south from the point $O$ is $= S_s = 24$ kmph

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