CDS Mathematics · Time, Speed and Distance

Speed Changes and Equations

'Had the speed been x faster, it would have taken y less' gives d/v − d/(v + x) = y, a quadratic in the speed.

All Time, Speed and Distance notes CDS Mathematics strategy

Why this matters

Nine PYQs, a quarter of them HARD. They all share one equation shape; the only choices are which unknown to call v and remembering to convert minutes to hours. Checking the answer by plugging it back takes seconds.

Concept 1 of 2: Faster speed, less time

The distance is fixed, so two speeds give two times whose difference is known. That difference, written out, is an equation in the speed.

Definition

$\dfrac dv - \dfrac{d}{v + x} = y$ simplifies to $dx = y\,v(v + x)$ .
Convert the time difference to hours before substituting ( $30$ minutes $= \dfrac12$ ).
Keep the positive root; check it in the original statement.
The general answer: distance $= \dfrac{x\,t(t - y)}{y}$ when the original time is $t$ .

Speed change

\dfrac dv - \dfrac{d}{v + x} = y \;\Rightarrow\; v(v + x) = \dfrac{dx}{y}

Worked example

A bus takes

1

hour less for a

240

km trip if its speed is raised by

20

km/hr. Find its usual speed.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (II) 2022 — Elementary Mathematics · Q34Moderate

Example 1 · Time, Speed and Distance · Speed Changes and Equations

In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 100 km/hr and time of flight increased by 30 minutes. What was the original average speed of the aircraft ?

Slower means more time

If the speed is REDUCED, the new time is longer: write

\dfrac{d}{v - x} - \dfrac dv = y

. Reversing the order gives a negative time difference and nonsense roots.

Drill 6 more on faster speed, less time

Concept 2 of 2: Two conditions, two unknowns

When a journey is split between two modes, or two people's times are compared twice, each condition gives an equation in the reciprocals of the speeds. Treat

\dfrac1a

and

\dfrac1b

as the unknowns and the system is linear.

Definition

Write each condition as time = distance ÷ speed, summed over the legs.
Substitute $p = \dfrac1a$ , $q = \dfrac1b$ : the equations become linear in $p$ and $q$ .
Subtract one equation from the other to eliminate one unknown.

Linear in reciprocals

\dfrac{d_1}{a} + \dfrac{d_2}{b} = T_1, \quad \dfrac{e_1}{a} + \dfrac{e_2}{b} = T_2

Worked example

A trip of

300

km takes

5

hours with

100

km by train and the rest by car, and

4

hours with

200

km by train and the rest by car. Find both speeds.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (I) 2019 — Elementary Mathematics · Q84Hard

Example 2 · Time, Speed and Distance · Speed Changes and Equations

It takes 11 hours for a 600 km journey if 120 km is done by train and the rest by car. It takes 40 minutes more if 200 km are covered by train and the rest by car. What is the ratio of speed of the car to that of the train ?

Solve for the reciprocals

The equations are linear in

\dfrac1a

and

\dfrac1b

, not in

a

and

b

. Clearing denominators first produces a messy quadratic system that is easy to get wrong.

Drill 1 more on two conditions, two unknowns

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)

Faster speed, less time
Speed change
$\dfrac dv - \dfrac{d}{v + x} = y \;\Rightarrow\; v(v + x) = \dfrac{dx}{y}$
Two conditions, two unknowns
Linear in reciprocals
$\dfrac{d_1}{a} + \dfrac{d_2}{b} = T_1, \quad \dfrac{e_1}{a} + \dfrac{e_2}{b} = T_2$

Watch out for (2)

Slower means more time
→ Faster speed, less time
Solve for the reciprocals
→ Two conditions, two unknowns

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.

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