NYT Math Problem: I Overtake My Brother in 15 Minutes?

  • Context: High School 
  • Thread starter Thread starter blumfeld0
  • Start date Start date
Click For Summary
SUMMARY

The New York Times math problem involves two brothers walking home, where one brother takes 30 minutes and the other takes 40 minutes. The solution reveals that the first brother overtakes the second brother in 15 minutes, despite the latter having a 5-minute head start. This is due to their constant walking speeds and the relative distances covered, which can be calculated algebraically using their speeds and the time difference. The equations derived from their speeds confirm that they meet at the halfway point after 15 minutes of the first brother's walking time.

PREREQUISITES
  • Understanding of basic algebraic equations
  • Knowledge of speed, distance, and time relationships
  • Familiarity with constant speed motion concepts
  • Ability to interpret word problems in mathematics
NEXT STEPS
  • Study algebraic methods for solving motion problems
  • Learn about relative speed concepts in physics
  • Explore practical applications of speed, distance, and time in real-life scenarios
  • Practice additional word problems involving multiple moving objects
USEFUL FOR

Students, educators, and anyone interested in enhancing their problem-solving skills in mathematics, particularly in motion and speed-related scenarios.

blumfeld0
Messages
146
Reaction score
0
Hi. Has anyone seen this?
http://www.nytimes.com/ref/science/20070313_PROF_GRAPHIC.html

I got all the problems except the one that says

"I walk from home to school in 30 minutes and my brother takes 40 minutes. My brother left 5 minutes before I did. In how many minutes will I overtake him?"

I said we need to know how fast each brother is walking.
apparently the answer is 15 minutes??
Is it because their difference in 10 minutes + 5 minutes head start?
if so, that makes no sense to me.

why is the answer 15 minutes?thanks
 
Mathematics news on Phys.org
They are walking at a constant speed apparently. If it takes you 30 minutes to walk home, then you will be halfway there in 15 minutes. If it takes your brother 40 minutes to walk home then he will be halfway there in 20 minutes. So if your brother starts walking home, and then you start walking home five minutes later, you and your brother will meet at the halfway mark. If that does not make sense, consider a concrete example. Suppose your brother starts walking home at 1. Then he will be halfway home at 1:20 (1:00 + 20 minutes). If you start walking at 1:05 (5 minutes after your brother), then you will be halfway home also at 1:20 (1:05 + 15 minutes).

You could also solve it algebraically.
 
You go from point A to point B in 30 minutes, and your brother does it in 40 minutes. You want to find out at what time are your positions equal.
If you say that the distance from home to school is d meters, then your speed is d/30 m/min. Your brother's speed is d/40 m/min. You might know that your position is your speed times your time, or x = vt. If we say you leave at time t=0 min, then your brother left at time t=-5 min (five minutes earlier). So,

xyou = vyout = d/30 * t
and
xbrother = vbrothert = d/40 * (t+5) (since your brother has been traveling for five more minutes than you have)

At the point where you overtake your brother, your x positions are equal. So, set the two equations equal to one another to obtain

d/30 * t = d/40 * (t+5)

Now solve for t. (d's on both sides cancel)

t/30 = (t+5)/40
t/(t+5) = 30/40
(t+5)/t = 40/30
1 + 5/t = 40/30
5/t = 4/3 - 1
t = 5/(4/3 - 1) = 15 min.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 30 ·
2
Replies
30
Views
10K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 37 ·
2
Replies
37
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K