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What Time Did Ajay Start?

Date: 08/06/2003 at 23:10:17
From: H Kalloli
Subject: Word Problem: Very difficult

Dear Dr. Math:

I have a problem which has really bothered me.

Three persons, Ajay, Bhanu, and Chandu, plan to travel from P and reach 
Q at 4.00 PM. 

Bhanu starts exactly 24 minutes after Ajay starts from P. 
Had Chandu started 12 minutes after Bhanu, all of them would have 
reached Q at 4.00 PM. But before Chandu leaves P, he receives a call 
from Ajay and is told to start immediately toward Q. At exactly the 
same time Ajay reverses his direction and travels toward P. 

All three of them meet at 3.24 PM at R (somewhere between P and Q). 
Each of them has travelled at his own uniform speed.

1) At what time did Ajay start from P?
2) In the entire journey, if Ajay travelled 9km more than Chandu, 
what is the speed of Bhanu?

I know that all word problems related to constant speed, distance, 
and time can be solved by using one simple formula: distance = 
speed * time. Yet I seem to lose track. For example, if a person 
travels two equal distances at speeds V1 and V2, then the average 
velocity is the harmonic mean of V1 and V2. Why such a strange 
relationship? Kindly help me by suggesting a general strategy to 
solve such problems.

Date: 08/07/2003 at 23:01:22
From: Doctor Jeremiah
Subject: Re: Word Problem: Very difficult

Hello. Thanks for writing. This really does seem like a difficult one, 
doesn't it?

We can use the original case where they meet at Q at 4:00 to get the 
relative velocities:

Name PQ as the distance from P to Q and PR as the distance from P to 
R. Now say Ajay starts at time T. Then Bhanu would start at time T+24 
(24 minutes later) and Chandu would start at T+36 (12 minutes after 

Ajay's speed is PQ/(4:00-T)
Bhanu's speed is PQ/(4:00-(T+24))
Chandu's speed is PQ/(4:00-(T+36))

In the second case where Ajay turns around, let's say that he travels X 
minutes before doing so. Since he started out at time T, he turned 
around at time T+X, and arrived at R at 3:24

In this second case Chandu starts at T+X and arrives at R at 3:24, so 
Chandu travels a distance of PR in 3:24-(T+X) minutes, which would 
imply a speed of PR/(3:24-(T+X)).  But we already know Chandu's speed 
is PQ/(4:00-(T+36)) and since they must be equal:

PQ/(4:00-(T+36)) = PR/(3:24-(T+X))

Bhanu travels a distance of PR in 3:24-(T+24) minutes, which implies 
a speed of PR/(3:24-(T+24)). But we already know Bhanu's speed is 
PQ/(4:00-(T+24)) and therefore they must be equal:

PQ/(4:00-(T+24)) = PR/(3:24-(T+24))

Ajay travels for X minutes before he turns around, and since he 
travels at a speed of PQ/(4:00-T) he must travel a distance of X PQ/
(4:00-T) before turning around. He turns around at T+X and arrives at 
R at 3:24, which is a difference of 3:24-(T+X) minutes. After he turns 
around he travels a distance of X PQ/(4:00-T) - PR and that would
imply a speed of (X PQ/(4:00-T) - PR)/(3:24-(T+X)).  But we already 
know Ajay's speed is PQ/(4:00-T) and therefore they must be equal:

PQ/(4:00-T) = (X PQ/(4:00-T) - PR)/(3:24-(T+X))

So now we have three equations:

PQ/(4:00-(T+36)) = PR/(3:24-(T+X))
PQ/(4:00-(T+24)) = PR/(3:24-(T+24))
PQ/(4:00-T) = (X PQ/(4:00-T) - PR)/(3:24-(T+X))

It seems as if there is not enough information. There are three 
equations and four unknowns. But as we will see later on there is 
enough information to solve for T, which is what the first question 
asks for.

At this point it's not very solvable unless we convert 4:00 to 
something else. If we make it minutes since midnight we get:

PQ/(924-T) = PR/(924-T-X)
PQ/(936-T) = PR/(900-T)
PQ/(960-T) = (X PQ/(960-T) - PR)/(924-T-X)

Then, if we use the first two:

PR = (924-T-X) PQ/(924-T)
PR = (900-T) PQ/(936-T)

These can be solved to get:

X = (924-T) - (924-T)(900-T)/(936-T)
X = 36(924-T)/(936-T)

Now if we use these three:

PQ/(960-T) = (X PQ/(960-T) - PR)/(924-T-X)
PQ/(936-T) = PR/(900-T)
X = 36(924-T)/(936-T)

We can solve them to get:

T = 2:48, 3:36
X = 27 minutes

Now, T cannot be 3:36 because they meet at 3:24 so T must be 2:48.

Let's check our work; here are our three equations:

PQ/(4:00-(T+36)) = PR/(3:24-(T+X))
PQ/(4:00-(T+24)) = PR/(3:24-(T+24))
PQ/(4:00-T) = (X PQ/(4:00-T) - PR)/(3:24-(T+X))

If we stick in T=2:48 and X=27 we get PQ = PQ, so we know
that these answers are right.

See if you can get the answer to the second question.
Let me know if you have trouble.

- Doctor Jeremiah, The Math Forum 
Associated Topics:
High School Basic Algebra
Middle School Algebra
Middle School Calendars/Dates/Time
Middle School Word Problems

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