AFCAT Numerical Ability · Speed, Time and Distance

Speed and Time in Inverse Proportion

Over a fixed distance, time varies inversely with speed: ratios of speed turn upside down into ratios of time, a percent change in speed gives a different percent change in time, and a known trip gives the distance for a new speed.

All Speed, Time and Distance notes

Why this matters

Most of the page is EASY, and the one AFCAT question here (2018) is a driver behind schedule. The mark is lost by treating a percent rise in speed as the same percent cut in time: 25% faster saves 20% of the time, not 25%.

Concept 1 of 3: Speed ratios turn into time ratios

Over the same distance, twice the speed means half the time. So speeds in the ratio

a : b

give times in the ratio

b : a

, and no distance is needed.

Definition

Same distance: $\dfrac{t_1}{t_2} = \dfrac{v_2}{v_1}$ .
At $\tfrac pq$ of the usual speed, the time is $\tfrac qp$ of the usual time.
Three speeds $a : b : c$ : the times are $\tfrac1a : \tfrac1b : \tfrac1c$ , cleared of fractions by the LCM.
When distance and time both change, speed is distance over time: half the distance in double the time is a quarter of the speed.

Same distance

\dfrac{t_1}{t_2} = \dfrac{v_2}{v_1}

Worked example

Three cars have speeds in the ratio

4 : 5 : 10

. Find the ratio of the times they take over the same road.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Speed, Time and Distance · Speed and Time in Inverse Proportion

The ratio of speeds of three cars is 2:3:4. What is the ratio of the time taken by these cars to travel the same distance?

Turn the ratio upside down

Speeds

4 : 5

do not give times

4 : 5

. The faster car takes less time, so the times are

5 : 4

Drill 14 more on speed ratios turn into time ratios

Concept 2 of 3: Through the distance: a new speed or a new time

When actual speeds or times are asked, find the fixed distance from the trip you know, then use it at the new speed or in the new time.

Definition

$\text{distance} = \text{old speed} \times \text{old time}$ . Then new time $= \dfrac{\text{distance}}{\text{new speed}}$ and new speed $= \dfrac{\text{distance}}{\text{new time}}$ .
Keeping to a schedule: required speed $= \dfrac{\text{distance still to go}}{\text{time still left}}$ .
'Increases its speed by 20%' means the new speed is $1.2$ times the old.

Keeping to a schedule

\text{required speed} = \dfrac{\text{distance left}}{\text{time left}}

Worked example

A bus must cover 280 km in 4 h. After 2 h it has covered only 110 km. At what speed must it run for the rest to arrive on time?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

AFCAT · 2018 · AFCAT (I) 2018 — Numerical Ability · 04-208Moderate

Example 2 · Speed, Time and Distance · Speed and Time in Inverse Proportion

A car driver leaves Bengaluru at 8:30 am and expects to reach a place 300 km from Bengaluru at 12:30 pm. At 10:30, he finds that he has covered only 40% of the distance. By how much does he have to increase the speed of the car in order to keep to his schedule?

Use the distance left, not the whole route

After 110 of 280 km in 2 h, the bus needs 170 km in the 2 h left, which is 85 km/h. Dividing the whole 280 km by 2 h gives 140 km/h.

Drill 12 more on through the distance: a new speed or a new time

Concept 3 of 3: A percent change in speed and the change in time

Speed times time is the fixed distance, so when speed is multiplied by a fraction, time is multiplied by the same fraction upside down. The two percentages are not equal.

Definition

Speed up by $r\%$ : time falls by $\dfrac{r}{100 + r} \times 100\%$ .
Slow down by $r\%$ : time rises by $\dfrac{r}{100 - r} \times 100\%$ .
Fractions are quickest: 25% faster is $\tfrac54$ of the speed, so $\tfrac45$ of the time, which is 20% less.
Time available falls from $T$ to $T'$ : speed must rise by $\dfrac{T - T'}{T'} \times 100\%$ .
A change in speed itself, such as 80 to 64 km/h, is measured on the old speed: 16 of 80, so 20%.

Speed and time move inversely

\text{speed} \times \tfrac{a}{b} \;\Rightarrow\; \text{time} \times \tfrac{b}{a}

Worked example

A car's speed rises by 25%. By what percent does its time for a fixed trip fall?

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Speed, Time and Distance · Speed and Time in Inverse Proportion

If speed is decreased by

30\%

, by what percentage should time be increased to cover the same distance?

The two percentages differ

25% faster saves 20% of the time, not 25%. And 20% slower costs 25% more time, not 20%.

Drill 7 more on a percent change in speed and the change in 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 ratios turn into time ratios
Same distance
$\dfrac{t_1}{t_2} = \dfrac{v_2}{v_1}$
Through the distance: a new speed or a new time
Keeping to a schedule
$\text{required speed} = \dfrac{\text{distance left}}{\text{time left}}$
A percent change in speed and the change in time
Speed and time move inversely
$\text{speed} \times \tfrac{a}{b} \;\Rightarrow\; \text{time} \times \tfrac{b}{a}$

Watch out for (3)

Turn the ratio upside down
→ Speed ratios turn into time ratios
Use the distance left, not the whole route
→ Through the distance: a new speed or a new time
The two percentages differ
→ A percent change in speed and the change in time

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