Problem Solving: Rate, Work, Average & Multi-Step Word Problems

Problem Solving: Rate, Work, Average & Multi-Step Word Problems

Forty of the 64 Mathematics questions are Problem Solving items, and almost all of them are word problems. The arithmetic inside them is ordinary; the difficulty is modeling - turning a two-sentence situation into an equation before the clock runs out.

Every word problem yields to the same four-step routine.

[1. Name the unknown]  ──>  Write "let n = ..." in the margin
        │
[2. Choose the model]  ──>  d = rt, rate x time = job, total = mean x count, part = percent x whole
        │
[3. Check the units]   ──>  Minutes with minutes, dollars with dollars
        │
[4. Answer the question asked] ──> Largest? Remaining? Per item? Reread the final clause

Step 4 is where most avoidable points are lost. Solving for n correctly and then reporting n when the question asked for n + 4 is the single most common word-problem error.

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Distance, Rate and Time

The relationship is distance = rate x time, rearranged as needed:

• rate = distance / time
• time = distance / rate

Worked Example 1: Single Traveler

A cyclist rides 36 miles at a steady 12 miles per hour. How long does the ride take?

time = 36 / 12 = 3 hours.

Worked Example 2: Two Legs at Different Speeds

A driver travels 120 miles at 60 miles per hour, then 60 miles at 30 miles per hour. What is the average speed for the whole trip?

Do not average 60 and 30. Compute totals:

• First leg time: 120 / 60 = 2 hours. Second leg time: 60 / 30 = 2 hours.
• Total distance 180 miles; total time 4 hours.
• Average speed = 180 / 4 = 45 miles per hour.

The average-speed trap: average speed is total distance over total time, never the mean of the two speeds. The naive average, 45 here by coincidence of equal times, is usually wrong - if the legs had taken unequal times the two methods would diverge.

Worked Example 3: Unit Mismatch

A runner covers 1,800 meters in 6 minutes. What is the speed in meters per second?

Convert first: 6 minutes = 360 seconds. Speed = 1,800 / 360 = 5 meters per second.

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Combined Work

When two workers or machines share a job, add their rates, not their times.

If one worker finishes a job in a hours and another in b hours, their combined rate is 1/a + 1/b jobs per hour, and the combined time is the reciprocal of that sum.

Worked Example 4

Machine A fills an order in 6 hours; Machine B fills the same order in 12 hours. Working together, how long do they take?

• Rates: 1/6 and 1/12 of the job per hour.
• Combined rate: 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4 of the job per hour.
• Time = 1 / (1/4) = 4 hours.

Sanity check: the answer must be less than the faster worker's time alone (6 hours) and more than half of it. Four hours passes both tests; nine hours - the trap answer from averaging - fails immediately.

The Worker-Hours Shortcut

For identical workers, use total worker-hours instead.

If 5 painters finish a mural in 12 hours, how long will 4 painters take?

Total work = 5 x 12 = 60 worker-hours. With 4 painters: 60 / 4 = 15 hours. Fewer workers means more time - an inverse relationship, so an answer smaller than 12 would be nonsense.

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Averages

The key identity is total = mean x count. Work through the total in both directions.

Worked Example 5: Missing Value

Four quiz scores average 84. What must the fifth score be for the five-quiz average to reach 86?

• Current total: 84 x 4 = 336.
• Required total: 86 x 5 = 430.
• Needed score: 430 - 336 = 94.

Worked Example 6: Weighted Average

A class of 20 students averages 78 on a test; a second class of 30 students averages 88. What is the combined average?

• Totals: 20 x 78 = 1,560 and 30 x 88 = 2,640, summing to 4,200.
• Students: 50. Combined average = 4,200 / 50 = 84.

Averaging 78 and 88 to get 83 ignores the larger second class. Weighted averages always lean toward the bigger group.

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Consecutive Integers

The sum of three consecutive odd integers is 87. Find the largest. n + (n + 2) + (n + 4) = 87, so 3n + 6 = 87, giving 3n = 81 and n = 27. The integers are 27, 29, 31, and the largest is 31.

Notice that both even and odd runs use steps of 2. The difference is only where you start, which is why the setup is identical.

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Mixture and Multi-Step Percent Problems

Worked Example 7

A jacket is marked down 30 percent to a sale price of $84. What was the original price?

The sale price is 70 percent of the original, so original = 84 / 0.70 = $120. Dividing rather than multiplying is the move students miss: 30 percent of 84 is 25.20, and adding it back gives $109.20, which is wrong.

Worked Example 8

A 40-ounce solution is 25 percent salt. How many ounces of pure water must be added to dilute it to 20 percent salt?

• Salt present: 0.25 x 40 = 10 ounces, and adding water does not change it.
• Final solution must satisfy 10 = 0.20 x total, so total = 50 ounces.
• Water added = 50 - 40 = 10 ounces.

The technique generalizes: track the quantity that does not change - here the salt - and let the total absorb the difference.

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Back-Solving

Every item is multiple choice, so the answer is printed. When the algebra is uncertain, test the choices.

The sum of a number and twice its reciprocal is 3. Which of 1, 2, 3, or 4 could the number be? Test 2: 2 + 2(1/2) = 2 + 1 = 3. Match.

Start with a middle value when the choices are ordered numerically: if it is too large you can often discard everything above it in one move.

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The Final Reread

Before bubbling, reread the last clause of the problem and confirm you answered that question - the largest of three integers rather than the smallest, the amount remaining rather than the amount spent, the cost per item rather than the total. Ten seconds spent here recovers more points on the Mathematics subtest than any additional formula.