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

  • Thread starter blumfeld0
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In summary, the conversation discusses a math problem where one person walks from home to school in 30 minutes while their brother takes 40 minutes. The question is how long it will take for the first person to catch up to their brother if they start walking 5 minutes later. The answer is 15 minutes, which can be solved either by understanding their positions will be equal at the halfway mark or by using an algebraic equation.
  • #1
blumfeld0
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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
 
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  • #2
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.
 
  • #3
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.
 

1. How does the NYT math problem work?

The NYT math problem involves two siblings, one of whom is traveling at a constant speed while the other is stationary. The problem asks at what point the moving sibling will overtake the stationary sibling, given the distance between them and the speed of the moving sibling.

2. What is the formula for solving the NYT math problem?

The formula for solving the NYT math problem is t = d/(v1-v2), where t is the time it takes for the moving sibling to overtake the stationary sibling, d is the distance between the siblings, and v1 and v2 are the speeds of the moving and stationary siblings, respectively.

3. How do you determine the distance between the siblings in the NYT math problem?

The distance between the siblings in the NYT math problem can be determined by measuring the distance between the starting points of the siblings or by using a known distance and the ratio of the speeds to calculate the distance between them.

4. Can the NYT math problem be solved using other units of measurement?

Yes, the NYT math problem can be solved using any units of measurement as long as the units are consistent. For example, if the distance between the siblings is measured in miles, then the speeds of the siblings should also be measured in miles per hour.

5. Are there any real-life applications of the NYT math problem?

Yes, the NYT math problem has real-life applications in situations where two objects are traveling at different speeds and one overtakes the other. For example, it could be used to calculate when a faster runner will catch up to a slower runner in a race.

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