AMC 10 · 2008 · #15
Grade 7 rate-ratioYesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
Pick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Three people drove yesterday. Compared to Ian, Han drove 1 hour longer at a speed 5 mph faster, and Jan drove 2 hours longer at a speed 10 mph faster. Han's trip was 70 miles longer than Ian's. Find how many more miles Jan drove than Ian.
Givens: Distance = speed × time for each driver.; Han's speed = Ian's speed + 5 mph; Han's time = Ian's time + 1 hour.; Jan's speed = Ian's speed + 10 mph; Jan's time = Ian's time + 2 hours.; Han drove 70 miles more than Ian.
Unknowns: How many more miles Jan drove than Ian.
Understand
Restated: Three people drove yesterday. Compared to Ian, Han drove 1 hour longer at a speed 5 mph faster, and Jan drove 2 hours longer at a speed 10 mph faster. Han's trip was 70 miles longer than Ian's. Find how many more miles Jan drove than Ian.
Givens: Distance = speed × time for each driver.; Han's speed = Ian's speed + 5 mph; Han's time = Ian's time + 1 hour.; Jan's speed = Ian's speed + 10 mph; Jan's time = Ian's time + 2 hours.; Han drove 70 miles more than Ian.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #13 Convert to Algebra, #16 Change Focus / Count the Complement
Nothing gives Ian's actual speed or time, so name them with variables and write every distance in terms of them. When the two extra-distance expressions are expanded, the unknown product cancels and the same combination shows up in both. So the plan is to build the expressions, then shift focus to that shared combination instead of chasing the individual variables.
Execute — Answer: D
6.RP.A.3 Step 1 Distance equals speed times time
- Each driver's distance is their speed multiplied by their time.
- This single relationship links every number in the problem, so write it down before naming anything.
💡 Miles come from how fast you go times how long you go.
6.EE.B.6 Step 2 Name Ian's speed and time
- Let s be Ian's speed in mph and t be Ian's time in hours.
- Then Ian drove s·t miles.
- Han drove (s+5)(t+1) miles and Jan drove (s+10)(t+2) miles, using the two comparisons.
💡 Give the unknowns letters so every trip becomes one clean expression.
7.EE.A.1 Step 3 Use Han's 70-mile clue
- Han's extra distance over Ian is (s+5)(t+1) minus s·t.
- Expanding, the s·t terms cancel and you are left with s + 5t + 5.
- Setting this equal to 70 gives s + 5t = 65.
💡 The messy s·t term disappears, leaving a tidy fact about s + 5t.
6.EE.A.3 Step 4 Write Jan's gap the same way
- Jan's extra distance over Ian is (s+10)(t+2) minus s·t.
- Expanding and cancelling s·t gives 2s + 10t + 20, which is exactly 2(s + 5t) + 20.
- The same block s + 5t appears again.
💡 Treat s + 5t as one chunk instead of solving for s and t separately.
7.EE.B.4 Step 5 Substitute and finish
- From Han's clue, s + 5t = 65.
- Substitute it into Jan's expression: 2(65) + 20 = 150.
- So Jan drove 150 more miles than Ian.
- The answer is (D).
💡 One known combination is all you need to nail the final number.
6.RP.A.3 Each driver's distance is their speed multiplied by their time. This single rela 6.EE.B.6 Let s be Ian's speed in mph and t be Ian's time in hours. Then Ian drove s·t mil 7.EE.A.1 Han's extra distance over Ian is (s+5)(t+1) minus s·t. Expanding, the s·t terms 6.EE.A.3 Jan's extra distance over Ian is (s+10)(t+2) minus s·t. Expanding and cancelling 7.EE.B.4 From Han's clue, s + 5t = 65. Substitute it into Jan's expression: 2(65) + 20 = Review
Reasonableness: Jan drove more hours and more speed above Ian than Han did, so Jan's gap should exceed Han's 70 miles, and 150 does. A quick concrete check confirms it: if Ian went 45 mph for 4 hours, then s + 5t = 45 + 20 = 65 holds; Ian drove 180, Han (50)(5)=250 which is 70 more, and Jan (55)(6)=330 which is 150 more. Answer (D) fits.
Alternative: Pick convenient numbers for Ian that satisfy s + 5t = 65 (say s = 45, t = 4), compute all three distances directly, and read off Jan minus Ian. Because the relationship is linear, any valid choice gives the same 150, which is why the algebra never needed s and t on their own.
CCSS standards used (min grade 7)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Turning the speed-and-time description into distance = speed × time for each driver.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming Ian's speed s and time t and writing each person's distance as an expression.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Expanding (s+5)(t+1) - s·t and simplifying it to s + 5t = 65.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Rewriting Jan's gap 2s + 10t + 20 as 2(s + 5t) + 20 to expose the shared block.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Substituting s + 5t = 65 into Jan's expression to get the final 150.)
⭐ You don't always need each unknown by itself; sometimes finding one combination, like s + 5t, is enough to answer the question.
⭐ You don't always need each unknown by itself; sometimes finding one combination, like s + 5t, is enough to answer the question.
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