Andrea and Lauren are 20 kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of 1 kilometer per minute. After 5 minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?
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Toolkit + CCSS Solution
Understand
Restated: Andrea and Lauren start 20 kilometers apart and bike toward each other. Andrea's speed is three times Lauren's, and while both ride the gap between them shrinks by 1 kilometer every minute. After 5 minutes Andrea gets a flat tire and stops, waiting for Lauren, who keeps riding. Find the total time, counted from the start, until Lauren reaches Andrea.
Givens: They begin 20 kilometers apart and ride toward each other.; Andrea rides 3 times as fast as Lauren.; While both ride, the distance between them decreases at 1 kilometer per minute.; After 5 minutes Andrea stops and waits; Lauren keeps riding.
Unknowns: The total number of minutes, measured from the start, until Lauren reaches Andrea.
Understand
Restated: Andrea and Lauren start 20 kilometers apart and bike toward each other. Andrea's speed is three times Lauren's, and while both ride the gap between them shrinks by 1 kilometer every minute. After 5 minutes Andrea gets a flat tire and stops, waiting for Lauren, who keeps riding. Find the total time, counted from the start, until Lauren reaches Andrea.
Givens: They begin 20 kilometers apart and ride toward each other.; Andrea rides 3 times as fast as Lauren.; While both ride, the distance between them decreases at 1 kilometer per minute.; After 5 minutes Andrea stops and waits; Lauren keeps riding.
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems
This is a rate problem, so I track speeds and how the closing rate changes. The given 1 km per minute is the two riders' speeds added together. I split the trip into two phases, both biking and then Lauren alone, because the closing rate is different in each, and units of km and minutes keep every step honest.
Execute — Answer: D
#8 Analyze the Units 6.RP.A.3Step 1
Split the closing rate
While both bike, the gap closes at 1 km each minute, and that 1 km/min is the two speeds added together.
Andrea is 3 times as fast as Lauren, so the rate splits into 4 equal parts: 3 parts for Andrea and 1 part for Lauren.
Each part is 1/4 km/min.
So Lauren rides at 1/4 km/min and Andrea at 3/4 km/min.
💡 A combined closing speed divides between the two riders in the same ratio as their speeds.
#7 Identify Subproblems 6.RP.A.3Step 2
Phase 1: both bike 5 minutes
For the first 5 minutes both ride, so the gap shrinks at 1 km/min.
In 5 minutes it drops by 5 km.
Starting from 20 km, the distance left when Andrea stops is 20 minus 5, which is 15 km.
$$20-(1)(5)=15\text{ km}$$
💡 Distance closed is just the closing rate times the time.
#8 Analyze the Units 6.NS.A.1Step 3
Phase 2: Lauren alone
Now Andrea waits, so only Lauren closes the gap, at 1/4 km/min, with 15 km still to cover.
Time equals distance divided by speed: 15 divided by 1/4.
Dividing by 1/4 is the same as multiplying by 4, so it takes 15 times 4, which is 60 minutes.
$$15\div\tfrac{1}{4}=15\times 4=60\text{ min}$$
💡 Once Andrea stops, the gap only closes as fast as Lauren rides.
#7 Identify Subproblems 4.NBT.B.4Step 4
Add the two phases
The whole trip is the 5 minutes both biked plus the 60 minutes Lauren rode alone.
That gives 5 plus 60, which is 65 minutes from the start until Lauren reaches Andrea, so the answer is (D).
$$5+60=65\text{ min}$$
💡 Total time is the time spent in each phase added together.
[1]
#8 6.RP.A.3While both bike, the gap closes at 1 km each minute, and that 1 km/min is the tw
[2]
#7 6.RP.A.3For the first 5 minutes both ride, so the gap shrinks at 1 km/min. In 5 minutes
[3]
#8 6.NS.A.1Now Andrea waits, so only Lauren closes the gap, at 1/4 km/min, with 15 km still
[4]
#7 4.NBT.B.4The whole trip is the 5 minutes both biked plus the 60 minutes Lauren rode alone
Review
Reasonableness: Check with distances instead. Andrea rides only 5 minutes at 3/4 km/min, covering 3.75 km, so Lauren must cover the remaining 20 minus 3.75, which is 16.25 km. Lauren rides at 1/4 km/min, needing 16.25 times 4, which is 65 minutes. Same answer, so 65 is consistent. It is also sensibly larger than the 20 minutes it would take if both kept riding at 1 km/min, since Andrea quitting slows the closing.
Alternative: Work in hours and km/h. The combined speed is 1 km/min, which is 60 km/h; split 3 to 1 gives Andrea 45 km/h and Lauren 15 km/h. In the first 5 minutes (1/12 hour) the gap drops 5 km to 15 km. Lauren then needs 15 km divided by 15 km/h, which is 1 hour, or 60 minutes. Total 5 plus 60 is 65 minutes.
CCSS standards used (min grade 6)
6.RP.A.3 Use ratio and rate reasoning to solve real-world problems (Splitting the 1 km/min closing rate into Andrea's and Lauren's speeds in a 3-to-1 ratio, and finding distance closed from rate and time.)
6.NS.A.1 Divide a quantity by a fraction (Dividing the remaining 15 km by Lauren's speed of 1/4 km/min to get 60 minutes.)
4.NBT.B.4 Fluently add multi-digit whole numbers (Adding the 5-minute and 60-minute phases to get the total 65 minutes.)
⭐ When one mover stops, the gap only closes as fast as whoever is still moving, so find the rate again for each phase.
⭐ When one mover stops, the gap only closes as fast as whoever is still moving, so find the rate again for each phase.