SSC CGL Maths · Time, Speed and Distance

Speed Changes and Equations

On a fixed route, speed and time change in opposite directions by the same ratio, and two speeds over the same route differ in time by a gap that fixes the distance.

All Time, Speed and Distance notes

Why this matters

Fourteen questions in the bank, nine EASY and five MODERATE, half of them from the 2025 shifts: a speed raised by a percentage, a student late at one speed and early at another, a rider who stops every 10 km. Each has a one-line route once you see which quantity stays fixed.

Concept 1 of 3: Speed and time on the same route

On the same route the distance is fixed, so speed times time stays the same. Go faster and the time shrinks in exactly the same ratio. A speed one quarter higher does not save a quarter of the time: it saves one fifth.

Definition

Same distance: $s_1 t_1 = s_2 t_2$ , so $\frac{t_2}{t_1} = \frac{s_1}{s_2}$ .
Fraction rule: if the speed becomes $\frac{a}{b}$ of itself, the time becomes $\frac{b}{a}$ of itself. A 25% rise in speed ( $\times \frac{5}{4}$ ) makes the time $\frac{4}{5}$ , a saving of 20%.
Time up by $\frac{1}{n}$ means speed down by $\frac{1}{n + 1}$ . Time up by $\frac{1}{4}$ means speed down by $\frac{1}{5}$ , which is 20%.
Same time: distance and speed rise by the same ratio. A route 20% longer in the same time needs 20% more speed.
New speed for a new time: find the distance once, old speed × old time, then divide by the new time.
Limit: the fraction rule holds only when the distance is the same both times.

Fixed distance

s_1 t_1 = s_2 t_2 \qquad \frac{t_2}{t_1} = \frac{s_1}{s_2}

$s_1, t_1$ the usual speed and time
$s_2, t_2$ the new speed and time

Worked example

A train takes 5 hours for a journey at 64 km/h. If its speed is raised by 25%, how much time does it save?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 15 Sep 2025, Shift 3 · Q68Easy

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

A delivery van travels at a constant speed of 60 km/h to cover a certain route in 1 hour and 45 minutes. If traffic increases the time taken by 15 minutes, by what percentage must the van's average speed decrease to complete the route?

Taking a 25% rise in speed as a 25% cut in time

On the same route the time falls in the inverse ratio: speed

\times \frac{5}{4}

makes time

\times \frac{4}{5}

, so the cut is 20%, not 25%.

Giving the new speed when the increase is asked

When a question asks by how much the speed must rise, find the new speed and then subtract the old one. The new speed itself is usually a printed wrong option.

Drill 4 more on speed and time on the same route

Concept 2 of 3: Two speeds over the same distance and the time between them

When the same distance is covered at two speeds, the slower trip takes longer. The difference between the two times is the only extra fact you need: it fixes the distance.

Definition

Equation: $\frac{d}{s_1} - \frac{d}{s_2} = \Delta t$ , with $s_1$ the slower speed and $\Delta t$ the time gap in hours.
Fast route: $d = \frac{s_1 s_2 \, \Delta t}{s_2 - s_1}$ : the product of the speeds times the time gap, over the difference of the speeds.
Late and early: late by $x$ minutes at one speed and early by $y$ minutes at the other gives a time gap of $x + y$ . Late by $x$ at one speed and late by $y$ at the other gives $x - y$ .
The usual, on time, journey time is then either trip's time corrected by its lateness or earliness.
Limit: the fast route needs the same distance both times and the gap in the same unit of time as the speeds.

Distance from a time gap

d = \frac{s_1 s_2 \, \Delta t}{s_2 - s_1}

$s_1$ the slower speed
$s_2$ the faster speed
$\Delta t$ how much longer the slower trip takes

Worked example

Walking at 5 km/h a boy reaches the station 6 minutes late; at 6 km/h he is 4 minutes early. How far is the station?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 13 Sep 2025, Shift 2 · Q70Easy

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

A person travels from city A to city B. If he travels at a speed of 50 km/h, he arrives 1 hour early. If he travels at a speed of 30 km/h, he arrives 1 hour late. What is the distance (in km) between city A and city B?

Leaving the time gap in minutes

With speeds in km/h the time gap must be in hours. 20 minutes is

\frac{1}{3}

hour; putting 20 into the formula makes the distance 60 times too big.

Subtracting a late time from an early time

Late by 15 minutes at one speed and early by 5 minutes at the other is a gap of 20 minutes, not 10. Add the two when one trip is late and the other is early.

Drill 4 more on two speeds over the same distance and the time between them

Concept 3 of 3: Halts, rests and the speed needed to arrive on time

A stop costs time but covers no ground. Split the trip into moving time, worked out from distance and speed, and stopped time, counted stop by stop.

Definition

Total time = moving time + stopped time. Moving time is distance ÷ speed; stopped time is the number of stops times the length of each stop.
Count the stops with care. A rest after every $k$ km on a trip of $d$ km, where $k$ divides $d$ , happens $\frac{d}{k} - 1$ times: there is no rest at the finish. If $k$ does not divide $d$ , count the whole number of $k$ km stretches done before the end.
Same journey, different halts: subtract the halts from the known total to get the driving time, find the distance, then build the other traveller's time.
Arriving on time: the speed needed for the rest of a trip is the distance left divided by the time left.

Rests after every k km

\text{number of rests} = \frac{d}{k} - 1 \quad (k \text{ divides } d)

Worked example

A cyclist rides at 15 km/h and rests for 8 minutes after every 6 km. How long does he take for a 30 km ride?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 20 Jul 2023, Shift 4 · Q58Moderate

Example 3 · Time, Speed and Distance · Speed Changes and Equations

Ram rides at the rate of 36 km/h but stops for five minutes to take a drink at the end of every 10 km. Find the time that Ram will take to cover a distance of 84 km.

Counting a rest at the finish

A rest after every 8 km on a 40 km trip happens 4 times, not 5: the trip ends at the 40th kilometre and no rest is taken there.

Using the whole trip to find the catching up speed

To arrive on time after a slow start, divide the distance still left by the time still left. Dividing the whole distance by the whole time gives the planned average, not the speed needed now.

Drill 3 more on halts, rests and the speed needed to arrive on time

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 (3)

Speed and time on the same route
Fixed distance
$s_1 t_1 = s_2 t_2 \qquad \frac{t_2}{t_1} = \frac{s_1}{s_2}$
Two speeds over the same distance and the time between them
Distance from a time gap
$d = \frac{s_1 s_2 \, \Delta t}{s_2 - s_1}$
Halts, rests and the speed needed to arrive on time
Rests after every k km
$\text{number of rests} = \frac{d}{k} - 1 \quad (k \text{ divides } d)$

Watch out for (6)

Taking a 25% rise in speed as a 25% cut in time
→ Speed and time on the same route
Giving the new speed when the increase is asked
→ Speed and time on the same route
Leaving the time gap in minutes
→ Two speeds over the same distance and the time between them
Subtracting a late time from an early time
→ Two speeds over the same distance and the time between them
Counting a rest at the finish
→ Halts, rests and the speed needed to arrive on time
Using the whole trip to find the catching up speed
→ Halts, rests and the speed needed to arrive on time

Test yourself on Time, Speed and Distance

20 past SSC CGL questions from this chapter, timed at 12 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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