Classic word problems (number, age, distance, value & relationship)

Worked example

6.2.ex1 Example 1: number (consecutive integers). "Three consecutive integers sum to 33. Find them." First name the unknown: let n = the smallest. Since each is one more than the last, the next two are n + 1 and n + 2. So all three are built from n. Now the equation: the three add up to 33. $$\begin{aligned} &n + (n + 1) + (n + 2) = 33 \\ &\Rightarrow\; 3n + 3 = 33 \\ &\Rightarrow\; 3n = 30 \\ &\Rightarrow\; n = 10 \end{aligned}$$ So the integers are 10, 11, 12. Check in words: 10 + 11 + 12 = 33. The story holds.

6.2.ex2 Example 2: number (consecutive even). "Three consecutive even integers sum to 78." Let n = the smallest even integer. Consecutive even integers step by 2, so the next two are n + 2 and n + 4. $$\begin{aligned} &n + (n + 2) + (n + 4) = 78 \\ &\Rightarrow\; 3n + 6 = 78 \\ &\Rightarrow\; n = 24 \end{aligned}$$ Integers: 24, 26, 28. Check: 24 + 26 + 28 = 78, and all three are even.

6.2.ex3 Example 3: age. "Sam is 3 years older than Pat. Together their ages total 27. How old is each?" Two unknowns, so define one and describe the other with it: let p = Pat's age, then Sam's age is p + 3. Now everything is in one letter, p. $$\begin{aligned} &p + (p + 3) = 27 \\ &\Rightarrow\; 2p + 3 = 27 \\ &\Rightarrow\; 2p = 24 \\ &\Rightarrow\; p = 12 \end{aligned}$$ Pat is 12, and Sam is 12 + 3 = 15. Check in words: Sam (15) is 3 more than Pat (12), and 12 + 15 = 27. Both parts of the story check out.

6.2.ex4 Example 4: distance (d = rt). "A car travels at 60 mph. How long to go 180 miles?" Let t = the time in hours. Use d = rt with the distance d = 180 and the rate r = 60: $$60t = 180 \;\Rightarrow\; t = 3$$ 3 hours. Check: 60 × 3 = 180 miles.

6.2.ex5 Example 5: distance (two objects). "Two cars leave the same point in opposite directions, one at 50 mph and one at 70 mph. After how many hours are they 360 miles apart?" Both cars travel for the same time, so one letter covers both: let t = the time in hours. Going opposite ways, the distance between them is the sum of the two distances, 50t + 70t. $$\begin{aligned} &50t + 70t = 360 \\ &\Rightarrow\; 120t = 360 \\ &\Rightarrow\; t = 3 \end{aligned}$$ 3 hours. Check: in 3 hours one car goes 150 miles, the other 210; 150 + 210 = 360.

6.2.ex6 Example 6: value (coins). "A jar has 15 coins, all dimes and quarters, worth $2.55 total. How many of each?" Let d = the number of dimes. The rest of the 15 coins are quarters, so that's 15 − d. Each dime is worth $0.10 and each quarter $0.25, so the total value in dollars is: $$0.10d + 0.25(15 - d) = 2.55$$ $$\begin{aligned} &0.10d + 3.75 - 0.25d = 2.55 \\ &\Rightarrow\; -0.15d = -1.20 \\ &\Rightarrow\; d = 8 \end{aligned}$$ So 8 dimes and 15 − 8 = 7 quarters. Notice everything stayed in dollars from the start; mixing in cents partway is where value problems usually go wrong. Check in words: that's 15 coins, worth 8(0.10) + 7(0.25) = 0.80 + 1.75 = $2.55.

6.2.ex7 Example 7: value (tickets). "100 tickets sold: adult $8, child $5, for $680 total. How many adult tickets?" Let a = the number of adult tickets; the rest are child tickets, 100 − a. Value, all in dollars: $$\begin{aligned} &8a + 5(100 - a) = 680 \\ &\Rightarrow\; 8a + 500 - 5a = 680 \\ &\Rightarrow\; 3a = 180 \\ &\Rightarrow\; a = 60 \end{aligned}$$ So 60 adult and 40 child. Check: 60(8) + 40(5) = 480 + 200 = $680.

The hard part is the second unknown. It is tempting to give it a new letter, like q for quarters or s for Sam. But then you have two letters and only one equation, which you cannot solve. Instead, use the clue in the problem. "3 older than" means + 3, and "the rest of the 15" means 15 − d.

In distance problems with two things moving, watch the direction. If they go opposite ways, add their distances. If they go the same way, subtract them. Either way, both travel for the same amount of time, so use one t for both. A quick sketch of two arrows from a point makes which-is-which obvious.

Here is one short example to try before the practice problems. "Two consecutive integers sum to 11." Let n = the smaller, so the next is n + 1; then n + (n + 1) = 11, which gives 2n + 1 = 11, so n = 5. The integers are 5 and 6, and 5 + 6 = 11. The numbers are small, but the steps are the same as every example above.