6.4 Time, Distance, Work & Rate Problems

6.4 Time, Distance, Work & Rate Problems

Quick Answer: Remember three linked formulas: D = S × T, S = D ÷ T, T = D ÷ S. For work: Work = Rate × Time, and rates of people working together add. Always match units (km with hours, or metres with seconds) before substituting.

Rate problems look intimidating because they mix quantities, but they rest on the same arithmetic as earlier sections. This section uses everyday travel between Trinidad and Tobago locations, ordinary household or office tasks, and simple production rates—never fire-pump discharge or fireground hydraulics.

Speed, distance, and time

Core relationships

Example — Port of Spain to Arima style trip

A driver travels 26 km in 40 minutes. What is the average speed in km/h?

Convert 40 minutes to hours: 40/60 = 2/3 hour.
Speed = 26 ÷ (2/3) = 26 × 3/2 = 39 km/h.

If you leave time as 40 minutes, you must use km per minute and convert: 26/40 = 0.65 km/min; × 60 = 39 km/h. Same result.

Example — finding time

Distance 90 km, constant speed 60 km/h.
Time = 90 ÷ 60 = 1.5 hours = 1 hour 30 minutes.

Example — finding distance

Speed 45 km/h for 2 hours 20 minutes.
2 h 20 min = 2 + 20/60 = 2 + 1/3 = 7/3 hours.
Distance = 45 × 7/3 = 15 × 7 = 105 km.

Multi-leg journeys and total time

Example — two towns with a break

A traveller drives 48 km at 64 km/h, rests 15 minutes, then drives 36 km at 48 km/h. Total time door-to-door?

Leg 1 time = 48 ÷ 64 = 0.75 h = 45 minutes.
Rest = 15 minutes.
Leg 2 time = 36 ÷ 48 = 0.75 h = 45 minutes.
Total = 45 + 15 + 45 = 105 minutes = 1 hour 45 minutes.

Total distance = 48 + 36 = 84 km.
Average speed for the whole outing including rest = total distance ÷ total time = 84 ÷ (105/60) = 84 ÷ 1.75.
1.75 = 7/4; 84 ÷ 7/4 = 84 × 4/7 = 12 × 4 = 48 km/h.

Average speed excluding rest would use only 90 minutes = 1.5 h: 84 ÷ 1.5 = 56 km/h. Read whether rest counts.

The average-speed trap

Wrong: average of 40 km/h and 60 km/h is always 50 km/h.
Right: average speed = total distance ÷ total time.

Equal distance, different speeds

Go 60 km at 40 km/h and return 60 km at 60 km/h.

Time out = 60/40 = 1.5 h.
Time back = 60/60 = 1 h.
Total distance = 120 km. Total time = 2.5 h.
Average speed = 120 ÷ 2.5 = 48 km/h (not 50).

Equal time, different speeds

If you truly spend the same time at each speed, the average of the two speeds works. Entrance questions more often use equal distances—so default to total distance / total time.

Unit-rate applications in travel and cost

Fuel and distance

Car uses 1 L per 12 km. Fuel for 150 km = 150 ÷ 12 = 12.5 L.
If fuel costs $7 per L, cost = 12.5 × 7 = $87.50.

Cost per kilometre of a trip

Trip costs $45 in total for 90 km.
Cost per km = 45 ÷ 90 = $0.50/km.
At that rate, 40 km would be 40 × 0.50 = $20 (if costs scale linearly—only when the question implies they do).

Work and rate (people finishing tasks)

Think of a completed job as 1 whole.

If a person finishes a job in n hours, their rate is 1/n job per hour.

One worker

Amina paints a room in 5 hours. Rate = 1/5 room per hour.
In 2 hours she completes 2 × 1/5 = 2/5 of the room.
Time to finish remaining 3/5: (3/5) ÷ (1/5) = 3 hours.

Two workers together

Ravi mows a lawn in 4 hours; Sasha in 6 hours. Together?

Rate Ravi = 1/4 per hour.
Rate Sasha = 1/6 per hour.
Combined = 1/4 + 1/6. Common denominator 12: 3/12 + 2/12 = 5/12 lawn per hour.
Time together = 1 ÷ (5/12) = 12/5 hours = 2.4 hours = 2 hours 24 minutes.

Partial help

If they work together for 1 hour, work done = 5/12. Remaining = 7/12.
If only Ravi continues: time = (7/12) ÷ (1/4) = (7/12) × 4 = 7/3 hours ≈ 2 hours 20 minutes.

Work problems with “people × days”

If 6 people take 10 days to complete a project (same steady rate, no complications):

Total person-days = 6 × 10 = 60.
With 8 people: days = 60 ÷ 8 = 7.5 days.

Assumptions (usually implied on entrance papers): people work at equal rates; work adds; no diminishing returns. If the question states otherwise, follow the question.

Example — inverse proportion

4 workers need 9 days → person-days = 36.
How many workers to finish in 6 days? 36 ÷ 6 = 6 workers.

Pipes and tanks (household sense only)

Same math as work rates: filling adds, emptying subtracts.

A household tank is filled by a tap in 6 hours and emptied by an open outlet in 9 hours (if both open, net fill).

Fill rate = 1/6 per hour. Empty rate = 1/9 per hour.
Net = 1/6 − 1/9 = 3/18 − 2/18 = 1/18 per hour.
Time to fill with both open = 18 hours.

This is ordinary fraction arithmetic—not fire-service pump theory.

Scheduling with unit rates

Example — photocopying forms

A machine copies 24 pages per minute. How long for 360 pages?

Time = 360 ÷ 24 = 15 minutes.

Two machines same rate together: 48 pages/min → 360 ÷ 48 = 7.5 minutes.

Example — packing boxes

Worker packs 15 boxes/hour. For 90 boxes: 90 ÷ 15 = 6 hours.
If a second worker packs 10 boxes/hour alongside: combined 25/hour → 90 ÷ 25 = 3.6 hours = 3 hours 36 minutes.

Putting multi-step and rates together

Story

A delivery van leaves Chaguanas at 9:00 a.m. at 50 km/h toward a depot 80 km away. After 1 hour it stops for 12 minutes, then continues at 40 km/h. When does it arrive?

Hour 1 distance = 50 km; remaining = 30 km.
Stop = 0.2 h.
Time for remaining = 30 ÷ 40 = 0.75 h = 45 minutes.
Timeline: 9:00 → 10:00 (drive), 10:00 → 10:12 (stop), 10:12 → 10:57 a.m. (arrive).

Total elapsed = 1 h + 12 min + 45 min = 1 h 57 min.

Checklist for rate items

1. Convert all times to hours (or all to minutes)—stay consistent.
2. Write knowns: D, S, T or Work, Rate, Time.
3. Solve for the unknown with the matching formula.
4. For average speed, recompute total D and total T from scratch.
5. For combined work, convert each person to a unit rate, then add.
6. End by converting decimal hours back to hours and minutes if options use clock form.

Synthesis across Chapter 6

Mastery means you can read a civilian story, choose the structure, keep units honest, and show every arithmetic step—exactly what multiple-choice Mathematics items on a public-service entrance paper reward.