Algebra - Arithmetic Progression

In an arithmetic sequence, the sum of the first $m$ terms is equal to the sum of the first $n$ terms.
Prove that the sum of the first $m + n$ terms is equal to zero.

Let first term of the arithmetic sequence $ = a$

and common difference $= d$

Sum of first $m$ terms of the arithmetic sequence $= S_m = \dfrac{m}{2} \left[2a + \left(m - 1\right) d \right]$

Sum of first $n$ terms of the arithmetic sequence $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right) d \right]$

Given: $\;$ $S_m = S_n$

i.e. $\;\;\;$ $\dfrac{m}{2} \left[2a + \left(m - 1\right) d \right] = \dfrac{n}{2} \left[2a + \left(n - 1\right) d \right]$

i.e. $\;\;\;$ $2am + m \left(m - 1\right) d = 2an + n \left(n - 1\right) d$

i.e. $\;\;\;$ $2a \left(m - n\right) = \left(n^2 - n - m^2 + m\right) d$

i.e. $\;\;\;$ $2a \left(m - n\right) = \left[\left(n^2 - m^2\right) - \left(n - m\right) \right] d$

i.e. $\;\;\;$ $2a \left(m - n\right) = \left[\left(n + m\right) \left(n - m\right) - \left(n - m\right) \right] d$

i.e. $\;\;\;$ $\left(n - m\right) \left(n + m - 1\right) d - 2a \left(m - n\right) = 0$

i.e. $\;\;\;$ $\left(n - m\right) \left[2a + \left(m + n - 1\right)d\right] = 0$

$\because$ $\;\;\;$ $n \neq m \implies n - m \neq 0$

$\therefore$ $\;\;\;$ $2a + \left(m + n - 1\right) d = 0$ $\;\;\; \cdots \; (1)$

Now, $\;$ $S_{m + n} = \left(\dfrac{m + n}{2}\right) \left[2a + \left(m + n - 1\right) d \right]$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;\;\;$ In view of equation $(1)$, equation $(2)$ becomes

$S_{m + n} = \left(\dfrac{m + n}{2}\right) \times 0$

i.e. $\;\;\;$ $S_{m + n} = 0$

Hence proved.

Algebra - Arithmetic Progressions

Sum of the first six terms of an arithmetic sequence is $9$. Sum of the first twelve terms is $90$. Find the sum of the thirteenth and the seventeenth terms of this sequence.

Let first term of the arithmetic sequence $ = a$

and common difference $= d$

Sum of $n$ terms of an arithmetic sequence $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right) d \right]$

Sum of first six terms $= S_6 = \dfrac{6}{2} \left[2a + 5d\right] = 6a + 15d$

Given: $\;$ $S_6 = 9$

i.e. $\;\;\;$ $6a + 15d = 9$ $\;\;\; \cdots \; (1)$

Sum of first twelve terms $= S_{12} = \dfrac{12}{2} \left[2a + 11d\right] = 12a + 66d$

Given: $\;$ $S_{12} = 90$

i.e. $\;\;\;$ $12a + 66d = 90$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously gives

$a = \dfrac{-7}{2}$ $\;$ and $\;$ $d = 2$

Now, $\;$ $n^{th}$ $\;$ term of an arithmetic sequence $= t_n = a + \left(n - 1\right) d$

$\therefore \;$ $13^{th}$ $\;$ term of the arithmetic sequence $= t_{13} = a + 12d$

and $\;$ $17^{th}$ $\;$ term of the arithmetic sequence $= t_{17} = a + 16d$

$\therefore \;$ Sum of the thirteenth and the seventeenth terms of the sequence

$= t_{13} + t_{17} = a + 12 d + a + 16 d = 2a + 28d$

Substituting the values of $a$ and $d$ gives

$t_{13} + t_{17} = 2 \times \left(\dfrac{-7}{2}\right) + 28 \times 2 = 49$

Matrices

Solve for matrices $A$ and $B$:
$2A + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix}$ and $A - 2B = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}$

Given matrix equations:

$2A + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix}$ $\;\;\; \cdots \; (1)$; $\;\;$ $A - 2B = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$

$A = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} + 2 B$ $\;\;\; \cdots \; (3)$

$\implies$ $2A = \begin{bmatrix} 8 & 6 \\ 2 & 2 \end{bmatrix} + 4 B$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(1)$ becomes

$\begin{bmatrix} 8 & 6 \\ 2 & 2 \end{bmatrix} + 4 B + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix}$

i.e. $\;$ $5 B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 6 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 3 - 8 & -4 - 6 \\ 2 - 2 & 7 - 2 \end{bmatrix} = \begin{bmatrix} -5 & -10 \\ 0 & 5 \end{bmatrix}$

$\implies$ $B = \begin{bmatrix} -1 & -2 \\ 0 & 1 \end{bmatrix}$ $\;\;\; \cdots \; (5)$

Substituting the value of $B$ from equation $(5)$ in equation $(3)$ gives

$A = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} + 2 \begin{bmatrix} -1 & -2 \\ 0 & 1 \end{bmatrix}$

i.e. $\;$ $A = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} -2 & -4 \\ 0 & 2 \end{bmatrix}$

i.e. $\;$ $A = \begin{bmatrix} 4 - 2 & 3 - 4 \\ 1 + 0 & 1 + 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

$\therefore \;$ $A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$; $\;$ $B = \begin{bmatrix} -1 & -2 \\ 0 & 1 \end{bmatrix}$

Algebra - Airthmetic Progressions

The eighth term of an A.P is half its second term and the eleventh term exceeds one third of its fourth term by $1$. Find the $15^{th}$ term of the A.P.

Let the first term of the A.P be $\;$ $t_1 = a$ $\;$ and the common difference $= d$

$n^{th}$ $\;$ term of A.P $\;$ $= t_n = a + \left(n - 1\right) d$

$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d$;

$4^{th}$ term $= t_4 = a + 3d$;

$8^{th}$ term $= t_8 = a + 7d$; $\;\;$ and

$11^{th}$ term $= t_{11} = a + 10d$

As per the question,

$t_8 = \dfrac{1}{2} \times t_2$

i.e. $\;$ $a + 7d = \dfrac{1}{2} \times \left(a + d\right)$

i.e. $\;$ $2a + 14 d = a + d$ $\;\;\;$

i.e. $\;$ $a + 13d = 0$ $\implies$ $a = -13d$ $\;\;\; \cdots \;\;\; (1)$

Also, as per the problem

$t_{11} = \left(\dfrac{1}{3} \times t_4\right) + 1$

i.e. $\;$ $a + 10 d = \dfrac{1}{3} \times \left(a + 3d\right) + 1$

i.e. $\;$ $3a + 30d = a + 3d + 3$

i.e. $\;$ $2a + 27d = 3$ $\;\;\; \cdots \;\;\; (2)$

In view of equation $(1)$, equation $(2)$ becomes

$2 \times \left(-13 d\right) + 27 d = 3$ $\;\;\;$ $\implies$ $d = 3$

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

$a = -13 \times 3 = -39$

Now, $15^{th}$ term of A.P $= t_{15} = a + 14d$

i.e. $\;$ $t_{15} = -39 + \left(14 \times 3\right) = 3$

Trigonometry - Trigonometric Functions

If $\;$ $\text{cosec} \theta - \sin \theta = a^3$, $\;$ $\sec \theta - \cos \theta = b^3$,
then prove that $\;$ $a^2 \; b^2 \left(a^2 + b^2\right) = 1$

Given: $\;$ $\text{cosec } \theta - \sin \theta = a^3$

i.e. $\;$ $\dfrac{1}{\sin \theta} - \sin \theta = a^3$

i.e. $\;$ $a^3 = \dfrac{1 - \sin^2 \theta}{\sin \theta}$

i.e. $\;$ $a^3 = \dfrac{\cos^2 \theta}{\sin \theta}$ $\;\;\; \cdots \; (1)$

and $\;$ $\sec \theta - \cos \theta = b^3$

i.e. $\;$ $\dfrac{1}{\cos \theta} - \cos \theta = b^3$

i.e. $\;$ $b^3 = \dfrac{1 - \cos^2 \theta}{\cos \theta}$

i.e. $\;$ $b^3 = \dfrac{\sin^2 \theta}{\cos \theta}$ $\;\;\; \cdots \; (2)$

Dividing equations $(1)$ and $(2)$ gives

$\dfrac{b^3}{a^3} = \dfrac{\sin^2 \theta}{\cos \theta} \div \dfrac{\cos^2 \theta}{\sin \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \dfrac{\sin^2 \theta}{\cos \theta} \times \dfrac{\sin \theta}{\cos^2 \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \dfrac{\sin^3 \theta}{\cos^3 \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \tan^3 \theta$

i.e. $\;$ $\dfrac{b}{a} = \tan \theta$

$\implies$ $\sin \theta = \dfrac{b}{\sqrt{a^2 + b^2}}$ $\;\;\; \cdots \; (3)$

and $\;$ $\cos \theta = \dfrac{a}{\sqrt{a^2 + b^2}}$ $\;\;\; \cdots \; (4)$

Substituting the values of $\sin \theta$ and $\cos \theta$ from equations $(3)$ and $(4)$ in equation $(1)$ gives

$a^3 = \dfrac{a^2}{a^2 + b^2} \div \dfrac{b}{\sqrt{a^2 + b^2}}$

i.e. $\;$ $a^3 = \dfrac{a^2}{a^2 + b^2} \times \dfrac{\sqrt{a^2 + b^2}}{b}$

i.e. $\;$ $a \; b = \dfrac{1}{\sqrt{a^2 + b^2}}$

i.e. $\;$ $a^2 \; b^2 \left(a^2 + b^2\right) = 1$

Hence proved

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity:
$\dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right) = \dfrac{\sin \beta}{\sin \alpha}$

LHS $= \dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right)$

$= \dfrac{\sin \left[\left(\alpha + \alpha\right) + \beta\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha }$

$= \dfrac{\sin \left[\alpha + \left(\alpha + \beta\right)\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$

$= \dfrac{\sin \alpha \cos \left(\alpha + \beta\right) + \cos \alpha \sin \left(\alpha + \beta\right) - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$

$= \dfrac{\sin \left(\alpha + \beta\right) \cos \alpha - \cos \left(\alpha + \beta\right) \sin \alpha}{\sin \alpha}$

$= \dfrac{\sin \left[\left(\alpha + \beta\right) - \alpha\right]}{\sin \alpha}$

$= \dfrac{\sin \beta}{\sin \alpha} = $ RHS

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity:
$\sec \left(\dfrac{\pi}{4} + \alpha\right) \sec \left(\dfrac{\pi}{4} - \alpha\right) = 2 \sec 2 \alpha$

LHS $=\sec \left(\dfrac{\pi}{4} + \alpha\right) \sec \left(\dfrac{\pi}{4} - \alpha\right)$

$= \dfrac{1}{\cos \left(\dfrac{\pi}{4} + \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\cos \left[\left(\dfrac{\pi}{2} - \dfrac{\pi}{4}\right) + \alpha \right] \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\cos \left[\dfrac{\pi}{2} - \left(\dfrac{\pi}{4} - \alpha\right)\right] \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\sin \left(\dfrac{\pi}{4} - \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{2}{2 \sin \left(\dfrac{\pi}{4} - \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{2}{\sin \left[2 \left(\dfrac{\pi}{4} - \alpha\right)\right]}$

$= \dfrac{2}{\sin \left(\dfrac{\pi}{2} - 2 \alpha\right)}$

$= \dfrac{2}{\cos 2 \alpha}$

$= 2 \sec 2 \alpha = $ RHS

Algebra - Word Problems: Derivation of Equations

Two cyclists left the same point simultaneously and traveled in the same direction. The speed of the first was $15$ km/h and that of the second was $12$ km/h. Half an hour later, another cyclist left the same point and traveled in the same direction. Some time later, he overtook the second cyclist and another hour and a half later he overtook the first cyclist. Find the speed of the third cyclist.

Let cyclists $C_1$ and $C_2$ leave point $A$ simultaneously and travel in the same direction.

After $\dfrac{1}{2}$ hr, let cyclist $C_3$ start from point $A$.

Speed of $C_1 = u_1 = 15$ kmph

and speed of $C_2 = u_2 = 12$ kmph

Let $C_1$ and $C_2$ start from point $A$ at time $t = 0$

In time $= \dfrac{1}{2}$ hr, distance covered by $C_1 = 15 \times \dfrac{1}{2} = 7.5$ km

In time $= \dfrac{1}{2}$ hr, distance covered by $C_2 = 12 \times \dfrac{1}{2} = 6$ km

Let speed of third cyclist $C_3 = u_3$ kmph

Let $C_3$ overtake $C_2$ after a time $T$ hr

Distance covered by $C_2$ in time $T$ $= \left(12 \times T\right)$ km

$\because \;$ $C_3$ starts $\dfrac{1}{2}$ hr later,

distance covered by $C_2$ in time $T$ $= \left(12T + 6\right)$ km $\;\;\; \cdots \; (1)$

Distance covered by $C_3$ in time $T$ $= u_3 \times T$ km $\;\;\; \cdots \; (2)$

$\because \;$ $C_3$ overtakes $C_2$, we have from equations $(1)$ and $(2)$

$12 T + 6 = u_3 T$ $\implies$ $T = \dfrac{6}{u_3 - 12}$ $\;\;\; \cdots \; (3)$

Now, $C_3$ overtakes $C_1$ after a time $\left(T + 1.5\right)$ hr

Distance covered by $C_1$ in time $\left(T + 1.5\right)$ hr is $= 15 \left(T + 1.5\right) = \left(15 T + 22.5\right)$ km

$\because \;$ $C_3$ starts $\dfrac{1}{2}$ hr later,

distance covered by $C_1$ in time $\left(T + 1.5\right)$ hr

$= \left(15T + 22.5 + 7.5\right) = 15 T + 30$ km $\;\;\; \cdots \; (4)$

Distance covered by $C_3$ in time $\left(T + 1.5\right)$ hr

$= u_3 \times \left(T + 1.5\right) = u_3 T + 1.5 u_3$ km $\;\;\; \cdots \; (5)$

$\because \;$ $C_3$ overtakes $C_1$, we have from equations $(4)$ and $(5)$

$15 T + 30 = u_3 T + 1.5 u_3$

i.e. $\;$ $T = \dfrac{1.5 u_3 - 30}{15 - u_3}$ $\;\;\; \cdots \; (6)$

$\therefore \;$ We have from equations $(3)$ and $(6)$,

$\dfrac{6}{u_3 - 12} = \dfrac{1.5 u_3 - 30}{15 - u_3}$

i.e. $\;$ $1.5 u_3^2 - 42 u_3 + 270 = 0$ $\;\;\; \cdots \; (7)$

Solving quadratic equation $(7)$ gives $\;$ $u_3 = 10$ $\;$ or $\;$ $u_3 = 18$

Speed of $C_3$ has to be greater than the speeds of $C_1$ and $C_2$ for $C_3$ to overtake them.

$\implies$ $u_3 = 10$ is not an acceptable solution.

$\therefore \;$ Speed of cyclist $C_3$ $= u_3 = 18$ kmph

Algebra - Word Problems: Derivation of Equations

Two people left simultaneously two points: one left point $A$ for point $B$ and the other left $B$ for $A$. Each of them walked at a constant speed and, having arrived at the point of destination, went back at once. First time they met $12$ km from $B$, and the second time, six hours after the first meeting, $6$ km from $A$ . Find the distance between A and B and the speeds of the two people.

Let person $P_1$ start from point $A$ and person $P_2$ start from point $B$ simultaneously

Let speed of $P_1$ $= u$ kmph

and speed of $P2$ $= v$ kmph

Let distance $AB = x$ km

For first meet at point $M_1$, $12$ km from point $B$:

Let $P_1$ and $P_2$ meet for the first time at point $M_1$

Then, distance $\left(BM_1\right) = 12$ km and distance $\left(M_1 A \right) = x - 12$ km

Time taken by $P_1$ to cover distance $M_1 A$ $= t_1 = \dfrac{M_1 A}{u} = \dfrac{x - 12}{u}$ hr $\;\;\; \cdots \; (1)$

Time taken by $P_2$ to cover distance $BM_1$ $= t_2 = \dfrac{B M_1}{v} = \dfrac{12}{v}$ hr $\;\;\; \cdots \; (2)$

Since $P_1$ and $P_2$ meet at point $M_1$, $t_1 = t_2$

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

$\dfrac{x - 12}{u} = \dfrac{12}{v}$

i.e. $\;$ $\dfrac{u}{v} = \dfrac{x - 12}{12}$ $\;\;\; \cdots \; (3)$

For second meet at point $M_2$, $6$ km from point $A$:

Let $P_1$ and $P_2$ meet for the second time at point $M_2$

Then, distance $\left(AM_2\right) = 6$ km and distance $\left(M_2 B \right) = x - 6$ km

Time taken by $P_1$ to meet at $M_2$ $= t_3 = \dfrac{AB + M_2 B}{u} = \dfrac{x + x - 6}{u} = \dfrac{2x - 6}{u}$ hr $\;\;\; \cdots \; (4)$

Time taken by $P_2$ to meet at $M_2$ $= t_4 = \dfrac{AB + A M_2}{v} = \dfrac{x + 6}{v}$ hr $\;\;\; \cdots \; (5)$

Since $P_1$ and $P_2$ meet at point $M_2$, $t_3 = t_4$

$\therefore \;$ We have from equations $(4)$ and $(5)$

$\dfrac{2x - 6}{u} = \dfrac{x + 6}{v}$

i.e. $\;$ $\dfrac{u}{v} = \dfrac{2x - 6}{x + 6}$ $\;\;\; \cdots \; (6)$

$\therefore \;$ We have from equations $(3)$ and $(6)$

$\dfrac{x - 12}{12} = \dfrac{2x - 6}{x + 6}$

i.e. $\;$ $x^2 - 12x + 6x - 72 = 24x - 72$

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

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

i.e. $\;$ $x = 0$ $\;$ or $\;$ $x = 30$

Since distance between points $A$ and $B$ cannot be $0$ km

$\therefore \;$ Distance between points $A$ and $B$ is $= x = 30$ km $\;\;\; \cdots \; (7)$

Also, second meet time $=$ first meet time $+ 6$ hr

$\therefore \;$ We have from equations $(1)$ and $(4)$, for $P_1$

$\dfrac{2x - 6}{u} = \dfrac{x - 12}{u} + 6$

i.e. $\;$ $\dfrac{\left(2 \times 30 \right) - 6}{u} = \dfrac{30 - 12}{u} + 6$ $\;\;\;$ [in view of equation $(7)$]

i.e. $\;$ $\dfrac{54}{u} = \dfrac{18}{u} + 6$

i.e. $\;$ $\dfrac{36}{u} = 6$ $\implies$ $u = 6$ $\;\;\; \cdots \; (8)$

and we have from equations $(2)$ and $(5)$, for $P_2$

$\dfrac{x + 6}{v} = \dfrac{12}{v} + 6$

i.e. $\;$ $\dfrac{30 + 6}{v} = \dfrac{12}{v} + 6$ $\;\;\;$ [in view of equation $(7)$]

i.e. $\;$ $\dfrac{36}{v} - \dfrac{12}{v} = 6$

i.e. $\;$ $\dfrac{24}{v} = 6$ $\implies$ $v = 4$ $\;\;\; \cdots \; (9)$

$\therefore \;$ Speed of $P_1$ $= u = 6$ kmph

and speed of $P_2$ $= v = 4$ kmph

Algebra - Word Problems: Derivation of Equations

Point C is at a distance of $12$ km from point B down the river. A fisher left point A, which is somewhat farther up the river than point B, for point C in a boat. He arrived at C four hours later and covered the return trip in six hours. Fixing a motor to the boat and thus trebling its speed relative to the water, the fisher covered the distance from A to B in $45$ minutes. Determine the speed of the river flow, considering it to be constant.

Direction A to B to C is downstream

Direction C to B to A is upstream

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

Let speed of river flow $= v$ kmph

Downstream speed of boat $= u + v$ kmph $\;\;\; \cdots \; (1)$

Upstream speed of boat $= u - v$ kmph $\;\;\; \cdots \; (2)$

Let distance AB $= x$ km

Given: Distance BC $= 12$ km

Then, distance AC $= x + 12$ km

Time taken by the boat to arrive from point A to point C (downstream) $= 4$ hours

Therefore, downstream speed of boat $= \dfrac{x + 12}{4}$ kmph $\;\;\; \cdots \; (3)$

Therefore, we have from equations $(1)$ and $(3)$

$u + v = \dfrac{x + 12}{4}$

i.e. $\;$ $4u + 4v = x + 12$ $\implies$ $x = 4u + 4v - 12$ $\;\;\; \cdots \; (4)$

Time taken by the boat to arrive from point C to point A (upstream) $= 6$ hours

Therefore, upstream speed of boat $= \dfrac{x + 12}{6}$ kmph $\;\;\; \cdots \; (5)$

Therefore, we have from equations $(2)$ and $(5)$

$u - v = \dfrac{x + 12}{6}$

i.e. $\;$ $6u - 6v = x + 12$ $\implies$ $x = 6u - 6v - 12$ $\;\;\; \cdots \; (6)$

Therefore, we have from equations $(4)$ and $(6)$

$4u + 4v - 12 = 6u - 6v - 12$

i.e. $\;$ $2u = 10 v$ $\implies$ $u = 5v$ $\;\;\; \cdots \; (7)$

After fixing motor, new speed of boat $= 3u = 3 \times 5v = 15v$ kmph [in view of equation $(7)$]

Therefore, speed of boat downstream from A to B $= 15v + v = 16v$ kmph

Time taken by the boat to cover distance AB $= \dfrac{x}{16v}$ hr $\;\;\; \cdots \; (8)$

Given: Time now taken to cover distance AB $= 45$ minutes $= \dfrac{45}{60} = \dfrac{3}{4}$ hr $\;\;\; \cdots \; (9)$

We have from equations $(8)$ and $(9)$

$\dfrac{x}{16v} = \dfrac{3}{4}$ $\implies$ $x = \dfrac{3}{4} \times 16v = 12v$ km $\;\;\; \cdots \; (10)$

In view of equations $(7)$ and $(10)$, equation $(4)$ becomes

$12v = 20v + 4v - 12$

i.e. $\;$ $12v = 12$ $\implies$ $v = 1$ kmph

i.e. Speed of river flow $= v = 1$ kmph

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

Algebra - Word Problems: Derivation of Equations

Two ships left a sea port simultaneously in two mutually perpendicular directions. Half an hour later, the shortest distance between them was $15$ km and another $15$ minutes later, one ship was $4.5$ km farther from the port than the other. Find the speed of each ship.

In the figure

$P$: Port

$S_1, \; S_2$: Two ships traveling in mutually perpendicular directions from port $P$

$PS_1, \; PS_2$: Position of the two ships after time $t = 0.5$ hours ($30$ minutes)

$PS_1 = x$ km, $\;$ $PS_2 = y$ km

$\therefore \;$ Speed of ship $S_1$ $= s_1 = \dfrac{x}{0.5} = 2x$ kmph

and speed of ship $S_2$ $= s_2 = \dfrac{y}{0.5} = 2y$ kmph

$S_1 S_2$: Shortest distance between ships $S_1$ and $S_2$ $= 15$ km

i.e. $\;$ $x^2 + y^2 = 15^2$ $\;\;\; \cdots \; (1)$ $\;\;$ $\left(\because \; S_1 \perp S_2\right)$

When time $= t_1 = 30 + 15 = 45$ min $= \dfrac{3}{4}$ hour

distance of ship $S_1$ from port $P$ $= d_1 = 2x \times \dfrac{3}{4} = \dfrac{3x}{2}$ km

and distance of ship $S_2$ from port $P$ $= d_2 = 2y \times \dfrac{3}{4} = \dfrac{3y}{2}$ km

Given: $\;$ $d_1 \; \text{km} = \left(d_2 + 4.5\right) \; \text{km}$

i.e. $\;$ $\dfrac{3x}{2} = \dfrac{3y}{2} + 4.5$

i.e. $\;$ $3x = 3y + 9$

i.e. $\;$ $x = y + 3$ $\;\;\; \cdots \; (2)$

$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes

$\left(y + 3\right)^2 + y^2 = 15^2$

i.e. $\;$ $y^2 + 6y + 9 + y^2 = 225$

i.e. $\;$ $2 y^2 + 6y - 216 = 0$

i.e. $\;$ $y^2 + 3y - 108 = 0$

i.e. $\;$ $\left(y - 9\right) \left(y + 12\right) = 0$

i.e. $\;$ $y = 9$ $\;$ or $\;$ $y = -12$

$\therefore \;$ We have from equation $(2)$,

when $\;$ $y = 9$, $\;$ $x = 9 + 3 = 12$

when $\;$ $y = -12$, $\;$ $x = -12 + 3 = -9$

$\because \;$ Distance covered by the ship cannot be negative,

$\therefore \;$ $x = -9$ and $y = -12$ are not valid solutions.

$\therefore \;$ Speed of ship $S_1 = s_1 = 2x = 2 \times 12 = 24$ kmph

and speed of ship $S_2 = s_2 = 2y = 2 \times 9 = 18$ kmph

Algebra - Word Problems: Derivation of Equations

Three cars leave $A$ for $B$ in equal time intervals. They reach $B$ simultaneously and then leave for point $C$ which is $120$ km from $B$. The first car arrives there an hour after the second car, and the third car, having reached $C$, immediately reverses the direction and $40$ km from $C$ meets the first car. Find the speed of the first car.

Let the first car be $C_1$, second car be $C_2$ and the third car be $C_3$.

Let $C_1$ cover distance $AB$ in time $t_1$ hours.

Let $C_2$ start after $C_1$ at point $A$ after $t$ hours

and $C_3$ start after $C_2$ at point $A$ after another $t$ hours.

Since cars $C_1, \; C_2, \; C_3$ reach point $B$ simultaneously

$\implies$ car $C_2$ covers distance $AB$ in $\left(t_1 - t\right)$ hours

and car $C_3$ covers distance $AB$ in $\left(t_1 - 2t\right)$ hours

$\therefore \;$ Speed of first car $C_1$ is $= s_1 = \dfrac{AB}{t_1}$ kmph

Speed of second car $C_2$ is $= s_2 = \dfrac{AB}{t_1 - t}$ kmph

Speed of third car $C_3$ is $s_3 = \dfrac{AB}{t_1 - 2t}$ kmph

Given: $\;$ Distance $BC = 120$ km

Time taken by $C_1$ (with speed $s_1$) to cover distance $BC$

$= T_1 = \dfrac{BC}{s_1} = \dfrac{120}{AB / t_1} = \dfrac{120 \times t_1}{AB}$ hours

Time taken by $C_2$ (with speed $s_2$) to cover distance $BC$

$= T_2 = \dfrac{BC}{s_2} = \dfrac{120}{AB / \left(t_1 - t\right)} = \dfrac{120 \times \left(t_1 - t\right)}{AB}$ hours

Given: $\;$ $T_1 = \left(T_2 + 1\right)$ hours

i.e. $\;$ $\dfrac{120 \times t_1}{AB} = \dfrac{120 \times \left(t_1 - t\right)}{AB} + 1$

i.e. $\;$ $120 t_1 = 120 t_1 - 120 t + AB$

i.e. $\;$ $AB = 120 t$ $\implies$ $t = \dfrac{AB}{120}$ $\;\;\; \cdots \; (1)$

Let cars $C_1$ and $C_3$ meet at a point $P$.

Then, distance $CP = 40$ km and distance $BP = 80$ km

Since $C_1$ and $C_3$ meet at point $P$

$\implies$ Time taken by $C_1$ to cover distance $BP$

$\hspace{2cm}$ $=$ Time taken by $C_3$ to cover distance $BC + CP$

i.e. $\;$ $\dfrac{BP}{s_1} = \dfrac{BC + CP}{s_3}$

Substituting the values of $BP$, $BC$, $CP$, $s_1$ and $s_3$ gives

$\dfrac{80}{AB / t_1} = \dfrac{120 + 40}{AB / \left(t_1 - 2t\right)}$

i.e. $\;$ $\dfrac{80 t_1}{AB} = \dfrac{160 \left(t_1 - 2t\right)}{AB}$

i.e. $\;$ $t_1 = 2\left(t_1 - 2t\right)$

i.e. $\;$ $t_1 = 2 t_1 - 4t$

i.e. $\;$ $t_1 = 4t = 4 \times \dfrac{AB}{120}$ $\;\;\;$ [by equation $(1)$]

i.e. $\;$ $t_1 = \dfrac{AB}{30}$ hours

$\therefore \;$ Speed of first car $C_1$ $= s_1 = \dfrac{AB}{t_1} = \dfrac{AB}{AB / 30} = 30$ kmph

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