7.1 Word Problems I: Rate, Distance, Work, and Mixture
Key Takeaways
- Distance = Rate x Time; keep units consistent (km/h needs time in hours, so 30 minutes becomes 0.5 hour).
- For equal distances at two different speeds, average speed uses total distance over total time, never the plain average of the two speeds.
- A job finished in T hours means a rate of 1/T per hour; combine two workers with T = (a x b) / (a + b).
- Fixed work equals workers x time, so 5 workers x 12 days = 60 worker-days; doubling the crew to 10 halves the time to 6 days.
- In a mixture, pure component = concentration x volume, and the pure amounts add before you divide by the total volume.
Turning Words into Equations
Numerical word problems on the CSE Professional level reward one repeatable habit: read the sentence, name the unknown, translate each fact into an equation, then solve using the mental-math tools from Chapter 6. There is no calculator, so every technique below leans on clean numbers, cancellation, and quick estimation to check the answer. This section covers the three problem families the Civil Service Commission (CSC) uses most: rate-distance-time, work-rate, and mixture. Each one collapses to a single relationship you can memorize.
Rate, Distance, and Time
The master equation is Distance = Rate x Time (D = R x T). Rearranged, R = D / T and T = D / R. The single biggest trap is units: if speed is in kilometers per hour, time must be in hours, not minutes.
Worked Example 1. A car travels 240 km in 4 hours. Average speed = D / T = 240 / 4 = 60 km/h. Shortcut without a calculator: 24 / 4 = 6, then reattach the zero to get 60.
Worked Example 2 (unit care). A jeepney covers 15 km in 30 minutes. Convert 30 minutes to 0.5 hour first. Speed = 15 / 0.5 = 30 km/h. If you carelessly divide by 30, you get 0.5, which is wrong by a factor of 60.
Worked Example 3 (average speed). A commuter drives 60 km at 60 km/h, then another 60 km at 40 km/h. The average speed is NOT (60 + 40) / 2 = 50. Use total distance over total time. Leg 1 time = 60 / 60 = 1 hour; leg 2 time = 60 / 40 = 1.5 hours. Total distance 120 km over total time 2.5 hours = 48 km/h. For equal distances, the average speed is always below the plain average.
Worked Example 4 (gap between two movers). Two buses leave the same terminal in opposite directions at 50 km/h and 70 km/h. After 3 hours they are (50 + 70) x 3 = 360 km apart. If instead they travel the same direction, the gap grows at only 70 - 50 = 20 km/h, so after 3 hours they are 60 km apart. Rule: opposite directions ADD the rates; same direction SUBTRACTS them.
| Find | Formula | Example |
|---|---|---|
| Distance | D = R x T | 60 km/h x 2 h = 120 km |
| Rate | R = D / T | 240 km / 4 h = 60 km/h |
| Time | T = D / R | 300 km / 60 km/h = 5 h |
| Combined (toward each other) | R1 + R2 | 50 + 70 = 120 km/h |
| Combined (one chasing) | R1 - R2 | 70 - 50 = 20 km/h |
Work-Rate Problems
The key idea: if a worker finishes a job in T hours, the rate is 1/T of the job per hour. Add the rates when workers cooperate; the combined time is the reciprocal of the summed rate.
Worked Example 5 (the tested pipe item). Pipe A fills a tank in 6 hours (rate 1/6) and Pipe B in 12 hours (rate 1/12). Together: 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4 of the tank per hour. Combined time = reciprocal = 4 hours. Two-worker shortcut: T = (a x b) / (a + b) = (6 x 12) / (6 + 12) = 72 / 18 = 4 hours.
Worked Example 6 (more workers, inverse proportion). If 5 workers finish a job in 12 days, how long will 10 workers take? Total work is fixed at 5 x 12 = 60 worker-days. With 10 workers: 60 / 10 = 6 days. Workers and time are inversely proportional, so their product stays constant. Doubling the crew halves the time.
Worked Example 7 (one works against you). A tap fills a tank in 4 hours (+1/4 per hour) while a leak empties it in 6 hours (-1/6 per hour). Net rate = 1/4 - 1/6 = 3/12 - 2/12 = 1/12 per hour, so the tank fills in 12 hours. When something drains or subtracts, subtract its rate.
Mixture and Solution Problems
Mixtures track the amount of one component across a blend. Use pure amount = concentration x volume, and remember the pure amounts add before dividing by the total.
Worked Example 8. Mix 4 liters of 25% alcohol with 6 liters of 50% alcohol. Pure alcohol = 0.25 x 4 + 0.50 x 6 = 1 + 3 = 4 liters. Total volume = 10 liters. Final concentration = 4 / 10 = 40%.
Worked Example 9 (price blend, a weighted average). A store blends 3 kg of rice at PHP 40/kg with 2 kg at PHP 55/kg. Total cost = 3 x 40 + 2 x 55 = 120 + 110 = 230 pesos for 5 kg, so the blend costs 230 / 5 = PHP 46/kg. This is the same weighted-average machinery used to compute exam scores and grades.
Common Traps
Watch for three recurring mistakes: (1) averaging speeds or concentrations without weighting by distance or volume; (2) mixing units, such as minutes with hours or milliliters with liters; and (3) adding times instead of rates in work problems, which is why two workers who each take 6 hours finish in 3 hours together, not 12. Always sanity-check: a combined crew must finish FASTER than the fastest single worker, and an average speed must fall BETWEEN the two given speeds.
One pipe fills a tank in 3 hours and another fills it in 6 hours. Working together, how long will they take to fill the tank?
A train travels 350 kilometers in 5 hours. What is its average speed?
Two liters of a 30% salt solution are mixed with three liters of an 80% salt solution. What is the salt concentration of the mixture?