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Speed, Distance & Time · Topic 3 of 5

⏱️ Speed Change & Rest Stops

3 exam question types, fully solved

These problems keep the distance fixed and change the speed — so a faster speed always cuts the time, and the early/late gap or the extra speed pins down the answer. Rest stops and there-and-back trips are the same idea with one extra step.

Fixed distance ⇒ speed and time flip

When the distance does not change, speed and time are inversely proportional. So a speed ratio becomes a time ratio — just flip it.

• Late + early: “10 min late” then “5 min early” means the two arrival times differ by min — you add the two gaps, never subtract. Direct formula: .

• Rest periods: , and the number of rests is — you never rest at the finish line.

• There-and-back: the two legs share the same distance, so add their times for a total or take their difference when one leg is slower.

🔁Same Road, Two SpeedsSpeed up ⇒ time flips down
🛣 Same road: 120 km
🚗 Red car speed: 40 km/h
🚙 Blue car speed: 60 km/h
⏱ 0 min🏁40 km/h60 km/hSame 120 km — arrival gap: 1 h
⏱ drag time
Red car time
3 h
120 ÷ 40
Blue car time
2 h
120 ÷ 60
Gap (late/early)
1 h
the Δt in the formula
Speed ratio 2 : 3 → time ratio flips to 3 : 2. Distance fixed means speed and time are inverse — check: 3 h : 2 h really is 3 : 2.

Cutting time in the ratio needs the speed ratio .

📝Practice Questions

Q1A car travels from town A to town B at 60 km/h and returns at 40 km/h. The return trip takes 1 hour longer than the onward trip. What is the one-way distance?

Q2A man walking at 3 km/h reaches his office 10 minutes late, but walking at 4 km/h he reaches 5 minutes early. How far is his office?

Q3A cyclist covers 5 km at 15 km/h and rests for 5 minutes after every 1 km. What is the total time taken?

📚

Real exam questions — Speed Change & Rest Stops

3 question types · 9 solved examples from real SSC papers

How a change of speed shifts your arrival time — and how rest stops and return legs add up.

How to solve this type
When the distance does not change, speed and time are inversely proportional: go faster and the time falls in the same ratio. So if the speed ratio is a:ba:b, the time ratio is b:ab:a (flip it).
The trick is reading the time gap correctly. "Late" means extra time taken, "early" means time saved, so a man who is 1010 min late at one speed and 55 min early at another has a total time difference of 10+5=1510+5=15 min — you ADD them, you never subtract.
Method: write the speed ratio, flip it for the time ratio, set the difference of the time units equal to the given gap, then use distance=speed×time\text{distance}=\text{speed}\times\text{time}.
Direct formula: Distance=S1×S2S2−S1×Δt\text{Distance}=\dfrac{S_1\times S_2}{S_2-S_1}\times \Delta t, with the gap Δt\Delta t in hours.

A man walking at 3 km/h reaches his office 10 minutes late. Walking at 4 km/h, he reaches 5 minutes early. Find the distance to his office.

A4 kmB2 kmC1.5 kmD3 km

A train travels 480 km at a uniform speed. If its speed were increased by 6 km/h, the journey would take 4 hours less. Find the original speed.

A30 km/hB24 km/hC20 km/hD36 km/h

A man covers a certain distance in 50 minutes. If he increases his speed by 3 km/h, he covers it in 40 minutes. Find the distance.

A10 kmB8 kmC12 kmD6 km