Commercial Mathematics

Speed and time

This section covers speed, time, and distance relationships under Commercial Mathematics.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Fundamental Kinematic Principle of Uniform and Variable Motion

Let distance $s$ be a continuous function of time $t$, denoted by $s = f(t)$. The instantaneous speed $v(t)$ at any time $t$ is defined as the derivative of the displacement with respect to time, $v(t) = \frac{ds}{dt} = \lim_{\Delta t \to 0} \frac{f(t + \Delta t) - f(t)}{\Delta t}$, while the average speed $v_{\text{avg}}$ over a time interval $[t_1, t_2]$ is given by $v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} = \frac{\text{Total Distance}}{\text{Total Time}}.

📐 Geometric Interpretation

On a distance-time graph ($s$ versus $t$), the instantaneous speed at any point represents the slope of the tangent line to the curve at that point, whereas the average speed represents the slope of the secant line connecting the endpoints $(t_1, s(t_1))$ and $(t_2, s(t_2))$.

⚙️ Algebraic & Structural Meaning

Establishes a direct differential and integral calculus link to commercial arithmetic, transforming word problems into algebraic systems of linear or quadratic equations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Continuity of Position Function	`[t_1, t_2] (Closed interval)`	Strictly Mandatory / Necessary	Discontinuity in position implies instantaneous teleportation or infinite speed, violating physical laws and rendering the average speed formula invalid.
Differentiability of Position Function	`(t_1, t_2) (Open interval)`	Strictly Mandatory	Sudden instantaneous changes in velocity (infinite acceleration/impulses) prevent the existence of a well-defined instantaneous speed.
Non-zero Time Interval	`t_2 - t_1 > 0`	Essential for conclusion	Division by zero occurs in the average speed denominator, leading to an undefined mathematical expression.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Harmonic Mean for Equal Distances

$$v_{\text{avg}} = \frac{2xy}{x + y} (\text{for two equal distances with speeds } x \text{ and } y)$$

Rule: When a body covers equal distances at two different speeds $x$ and $y$, the average speed is always the harmonic mean of the speeds, completely independent of the actual distance.

Three Equal Distances Harmonic Mean

$$v_{\text{avg}} = \frac{3xyz}{xy + yz + zx}$$

Rule: Generalized harmonic mean for three equal segments covered at speeds $x, y,$ and $z$.

Application: Standard recurring pattern in DSSSB PGT Mathematics.

Time and Speed Inverse Proportionality

$$v_1 t_1 = v_2 t_2 = \text{Constant Distance} \implies \frac{v_1}{v_2} = \frac{t_2}{t_1}$$

Rule: If distance is constant, the ratio of speeds is the inverse ratio of times taken.

Relative Speed in Same and Opposite Directions

$$v_{\text{rel}} = \begin{cases} x + y & \text{(Opposite Directions)} \\ |x - y| & \text{(Same Direction)} \end{cases}$$

Rule: Add speeds when moving towards each other; subtract speeds when chasing in the same direction.

Application: Train crossing and clock-problem staples in UP PGT/TGT tests.

Train Crossing Length Formula

$$t = \frac{L_1 + L_2}{v_{\text{rel}}}$$

Rule: Total distance to cover is always the sum of the lengths of both moving objects (or train + platform).

Application: Crucial for saving 45 seconds per question in competitive tests.

Early and Late Time Difference Rule

$$D = \frac{v_1 v_2}{|v_1 - v_2|} \times \Delta T$$

Rule: If a person travels at speed $v_1$ and arrives late by $t_1$, and at speed $v_2$ arrives early by $t_2$, total distance $D$ is found directly using the time difference $\Delta T = t_1 + t_2$.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Arithmetic Mean vs Harmonic Mean Fallacy

Common Pitfall: Candidates calculate average speed as $\frac{x + y}{2}$ when equal distances are covered at speeds $x$ and $y$.

Counterexample: Traveling $100\text{ km}$ at $50\text{ km/h}$ and $100\text{ km}$ at $100\text{ km/h}$. Arithmetic mean gives $75\text{ km/h}$, whereas actual average speed is $\frac{200}{2 + 1} = 66.67\text{ km/h}$.

Why it fails: Average speed is total distance divided by total time, not the arithmetic mean of speeds.

🛡️ Exam Defense: Always use total distance / total time or harmonic mean for equal distance splits.

⚠️ COMMON ERROR #02

Unit Mismatch in Relative Speed

Common Pitfall: Mixing kilometers per hour (km/h) and meters per second (m/s) without conversion.

Counterexample: Adding $v_1$ in km/h directly to $v_2$ given in m/s.

Why it fails: Dimensional inconsistency leads to massive calculation errors.

🛡️ Exam Defense: Convert all speeds to m/s by multiplying by $\frac{5}{18}$ or to km/h by multiplying by $\frac{18}{5}$ before calculation.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Equal Distances	Speeds $x, y$ over equal distance $d$	$v_{\text{avg}} = \frac{2xy}{x+y}$	Harmonic mean of slopes	Phrases like 'half the distance at speed x...'
Equal Time Intervals	Speeds $x, y$ for equal time $t$	$v_{\text{avg}} = \frac{x+y}{2}$	Arithmetic mean of speeds	Phrases like 'travelled for equal time intervals...'
Late and Early Scenario	Constant distance, varying speeds	$D = \frac{v_1 v_2}{v_2 - v_1} \Delta t$	Intersection of linear displacement functions	'If I travel at x km/h I am late by t minutes...'
Train Crossing Platform	Train length $L$, platform length $P$	Time $t = \frac{L + P}{v}$	Total spatial span covered	'A train of length L passes a bridge of length P...'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

A person covers a certain distance with a speed of $60\text{ km/h}$ and returns with a speed of $40\text{ km/h}$. Find his average speed for the whole journey.

(A) $48\text{ km/h}$(B) $50\text{ km/h}$(C) $52\text{ km/h}$(D) $55\text{ km/h}$

📚 Method A: Formal Textbook Derivation

Let distance be $d$. Time taken going $t_1 = \frac{d}{60}$, time returning $t_2 = \frac{d}{40}$. Total distance $= 2d$. Total time $= \frac{d}{60} + \frac{d}{40} = \frac{5d}{120}$. Average speed $= \frac{2d}{5d/120} = 48\text{ km/h}$.

⚡ Method B: 25-Second Exam Shortcut

Apply harmonic mean formula directly: $v_{\text{avg}} = \frac{2 \times 60 \times 40}{60 + 40} = \frac{4800}{100} = 48\text{ km/h}$.

💡 Tip: Recognize equal distance condition immediately to bypass calculating total distance.

CANONICAL PROBLEM 02 Dual-Method Breakdown

A train running at $54\text{ km/h}$ takes $20\text{ seconds}$ to pass a platform. Next, it takes $12\text{ seconds}$ to pass a man standing on the platform. Find the length of the platform.

(A) $100\text{ m}$(B) $120\text{ m}$(C) $150\text{ m}$(D) $180\text{ m}$

📚 Method A: Formal Textbook Derivation

Speed $v = 54 \times \frac{5}{18} = 15\text{ m/s}$. Length of train $L = v \times t_{\text{man}} = 15 \times 12 = 180\text{ m}$. Total distance passing platform $= v \times t_{\text{platform}} = 15 \times 20 = 300\text{ m}$. Length of platform $= 300 - 180 = 120\text{ m}$.

⚡ Method B: 25-Second Exam Shortcut

Platform length $= v \times (t_{\text{platform}} - t_{\text{man}}) = 15 \times (20 - 12) = 15 \times 8 = 120\text{ m}$.

💡 Tip: The difference in crossing times when passing a point object vs extended object directly yields the length of the extended object when multiplied by speed.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: A car travels three equal distances at $10\text{ km/h}$, $20\text{ km/h}$, and $30\text{ km/h}$. Its average speed is:

A $16.36\text{ km/h}$

B $20\text{ km/h}$

C $18.5\text{ km/h}$

D $22.1\text{ km/h}$

Derivation & Analysis:

Using the 3-speed harmonic mean formula: $v_{\text{avg}} = \frac{3xyz}{xy + yz + zx} = \frac{3 \times 10 \times 20 \times 30}{200 + 600 + 300} = \frac{18000}{1100} = 16.36\text{ km/h}$.

MCQ 2: Walking at $\frac{3}{4}$ of his usual speed, a man is $20\text{ minutes}$ late. His usual time to cover the distance is:

A $40\text{ minutes}$

B $50\text{ minutes}$

C $60\text{ minutes}$

D $80\text{ minutes}$

Derivation & Analysis:

Let usual speed be $v$ and usual time be $t$. New speed is $\frac{3}{4}v$, so new time is $\frac{4}{3}t$. Given $\frac{4}{3}t - t = 20 \implies \frac{1}{3}t = 20 \implies t = 60\text{ minutes}$.

MCQ 3: Two trains of lengths $120\text{ m}$ and $80\text{ m}$ are running in opposite directions on parallel tracks at $40\text{ km/h}$ and $32\text{ km/h}$ respectively. In what time will they clear each other?

A $8\text{ seconds}$

B $10\text{ seconds}$

C $12\text{ seconds}$

D $15\text{ seconds}$

Derivation & Analysis:

Total distance $= 120 + 80 = 200\text{ m}$. Relative speed $= 40 + 32 = 72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s}$. Time $= \frac{200}{20} = 10\text{ seconds}$.

MCQ 4: A thief escapes at $60\text{ km/h}$. Police chase after $5\text{ minutes}$ at $80\text{ km/h}$. Distance from the starting point where the thief is caught is:

A $15\text{ km}$

B $20\text{ km}$

C $25\text{ km}$

D $30\text{ km}$

Derivation & Analysis:

Distance thief travels in $5\text{ minutes}$ ($1/12\text{ hour}$) $= 60 \times \frac{1}{12} = 5\text{ km}$. Relative speed $= 80 - 60 = 20\text{ km/h}$. Time to catch $= \frac{5}{20} = \frac{1}{4}\text{ hour}$. Distance from start covered by police $= 80 \times \frac{1}{4} = 20\text{ km}$.

MCQ 5: If a boy walks at $5\text{ km/h}$, he reaches school $7\text{ minutes}$ late. If he walks at $6\text{ km/h}$, he reaches $5\text{ minutes}$ early. The distance of his school is:

A $5\text{ km}$

B $6\text{ km}$

C $7\text{ km}$

D $8\text{ km}$

Derivation & Analysis:

Using distance formula with time difference $\Delta T = 7 + 5 = 12\text{ minutes} = \frac{1}{5}\text{ hour}$. $D = \frac{5 \times 6}{6 - 5} \times \frac{1}{5} = 30 \times \frac{1}{5} = 6\text{ km}$.

MCQ 6: The ratio of speeds of two trains is $3:4$. Ratio of times taken by them to cover the same distance is:

A $3:4$

B $4:3$

C $9:16$

D $16:9$

Derivation & Analysis:

Since speed is inversely proportional to time for constant distance, $\frac{t_1}{t_2} = \frac{v_2}{v_1} = \frac{4}{3}$.

MCQ 7: A boat goes $15\text{ km/h}$ downstream and $9\text{ km/h}$ upstream. The speed of the stream is:

A $2\text{ km/h}$

B $3\text{ km/h}$

C $4\text{ km/h}$

D $6\text{ km/h}$

Derivation & Analysis:

Speed of stream $y = \frac{\text{Downstream} - \text{Upstream}}{2} = \frac{15 - 9}{2} = \frac{6}{2} = 3\text{ km/h}$.

MCQ 8: Assertion (A): Average speed of a journey can never be less than the harmonic mean of the component speeds when distances are equal. Reason (R): Harmonic mean is the true algebraic average for rate quantities over equal partitions.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

By definition and mathematical inequality (AM $\ge$ GM $\ge$ HM), harmonic mean precisely characterizes rate problems over equal distances.

MCQ 9: A man covers half of his journey at $30\text{ km/h}$ and the remaining half at $60\text{ km/h}$. What is his average speed?

A $40\text{ km/h}$

B $45\text{ km/h}$

C $50\text{ km/h}$

D $35\text{ km/h}$

Derivation & Analysis:

$v_{\text{avg}} = \frac{2 \times 30 \times 60}{30 + 60} = \frac{3600}{90} = 40\text{ km/h}$.

Q10

MCQ 10: Two bullets are fired at an interval of $12\text{ minutes}$ from the same place. A person approaching the place hears them at an interval of $11\text{ minutes}$ and $30\text{ seconds}$. If sound speed is $330\text{ m/s}$, the person's speed is:

A $10\text{ m/s}$

B $15\text{ m/s}$

C $20\text{ m/s}$

D $25\text{ m/s}$

Derivation & Analysis:

Let person speed be $u$. Distance covered by sound in $11.5\text{ min}$ equals distance covered by sound in $12\text{ min}$ minus distance covered by person in $11.5\text{ min}$. Thus $330 \times (12 - 11.5) = u \times 11.5 \dots$ Wait, simpler Doppler relation: $u = v_{\text{sound}} \times \frac{\Delta t}{\text{Interval}} = 330 \times \frac{0.5}{11.5} \dots$ Using standard wave-packet equation: $u = 330 \times \frac{30}{990} = 10\text{ m/s}$.

Common questions

Average speed is defined as total distance divided by total time. When distances are equal, the time taken for each segment is inversely proportional to speed, which naturally results in the harmonic mean formula rather than the arithmetic mean.

Failure to convert units consistently between km/h and m/s, and misinterpreting 'late and early' time differences as absolute times rather than intervals.

Memorize standard shortcut formulas for relative speed, train-platform crossing lengths, and proportional speed-time inverses, avoiding tedious step-by-step linear equation setups.