CDS Mathematics · Time, Speed and Distance

Average Speed and Speed–Time Ratios

Average speed is total distance over total time, and over a fixed distance the times are in the inverse ratio of the speeds.

All Time, Speed and Distance notes CDS Mathematics strategy

Why this matters

Twenty-two PYQs, the largest page in the chapter and mostly EASY. Two traps catch students: averaging the speeds instead of dividing total distance by total time, and forgetting that a slower speed means a proportionally longer time.

Concept 1 of 2: Average speed

An average speed must reproduce the whole trip: the same distance in the same time. So it is total distance ÷ total time — and because slower legs take longer, it leans toward the slower speed.

Definition

Average speed $= \dfrac{\text{total distance}}{\text{total time}}$ .
Equal DISTANCES at $a$ and $b$ : the average is the harmonic mean $\dfrac{2ab}{a + b}$ .
Equal TIMES at $a$ and $b$ : the average is $\dfrac{a + b}{2}$ .
Stoppages: with running speed $u$ and overall speed $v$ , the train stops $\dfrac{u - v}{u}\times 60$ minutes per hour.

Equal distances

\bar v = \dfrac{2ab}{a + b}

Worked example

A car goes to a town at

60

km/hr and returns at

40

km/hr. Find its average speed.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (I) 2018 — Elementary Mathematics · Q48Easy

Example 1 · Time, Speed and Distance · Average Speed and Speed–Time Ratios

A car has an average speed of 60 km per hour while going from Delhi to Agra and has an average speed of

y

km per hour while returning to Delhi from Agra (by travelling the same distance). If the average speed of the car for the whole journey is 48 km per hour, then what is the value of

y

Not the average of the speeds

Going at

60

and returning at

40

gives

48

, not

50

. The plain average is right only when equal TIMES, not equal distances, are spent at each speed.

Drill 10 more on average speed

Concept 2 of 2: Speed and time over a fixed distance

For a fixed distance, speed × time is constant. Travel at

\dfrac45

of your speed and you take

\dfrac54

of your time; the extra quarter is the lateness.

Definition

Same distance: $t_1 : t_2 = v_2 : v_1$ .
At $\dfrac pq$ of the usual speed the time is $\dfrac qp$ of usual; the delay is $\left(\dfrac qp - 1\right)\times$ usual time.
'Early at $a$ , late at $b$ ': the two times differ by early + late minutes, so $\dfrac db - \dfrac da =$ that difference.
Units: $1$ km/hr $= \dfrac{5}{18}$ m/s; $1$ hectare $= 10{,}000$ m $^2$ .

Fixed distance

v_1 t_1 = v_2 t_2

Worked example

\dfrac34

of his usual speed a man is

15

minutes late. Find his usual time.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (I) 2021 — Elementary Mathematics · Q35Moderate

Example 2 · Time, Speed and Distance · Average Speed and Speed–Time Ratios

Walking at

\frac{4}{5}

th of his usual speed, a man is 12 minutes late for his office. What is the usual time taken by him to cover that distance?

Early plus late

40

minutes early at one speed and

40

minutes late at another' means the two times differ by

80

minutes, not

40

and not

0

Drill 10 more on speed and time over a fixed distance

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Average speed
Equal distances
$\bar v = \dfrac{2ab}{a + b}$
Speed and time over a fixed distance
Fixed distance
$v_1 t_1 = v_2 t_2$

Watch out for (2)

Not the average of the speeds
→ Average speed
Early plus late
→ Speed and time over a fixed distance

Test yourself on Time, Speed and Distance

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Time, Speed and Distance chapter test Drill all 22 questions