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Re: Lionel left his house and walked towards Walt's house, which is 48 [#permalink]
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Answer is 24.

My way of solving such questions is altogether different. I used diagrams and divided the distance in fractions of 4 and 6. Anyway.

Answering your question, they have taken t-2 because W left his lous 2 hours after L started. Means L walked for 2 more hours. Dont confuse it with actual time of the day.

Simplyfying it,

- Keep the first 2 hours aside for sometime. L walked 8 miles in those 2 hours. So distance left is 48 - 8 = 40 miles
- L walks 4 miles per hour and W walks 6 miles per hour. The meet when they have walked for similar time. Let the time be T

putting this in equation, 4T + 6T = 40 so T = 4.
Tells us, both walked for 4 hours when they met.

So L walked for 4*4 = 16 miles . Adding his initial 8 miles, he waked for 24 miles in total.
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Re: Lionel left his house and walked towards Walt's house, which is 48 [#permalink]
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shrive555 wrote:
Lionel left his house and walked towards Walt's house, 48 miles away. Two hours later, Walt left his house and ran towards Lionel's house. If Lionel's speed was 4 miles per hour and Walt's 6 miles per hour, how many miles had Lionel walked when he met Walt?


For Li : R= 4, Time = t => Distance = 4t

For Wal: R= 6, Time = t-2 => Distance = 6(t-2)

one can easily solve this question with this information.

My question is : how Time for Wal is " t-2" ?? According to question, lets say, Li left home at 4pm, two hours later Wal left his house, it means 6pm. wouldn't it be ( t+2) in algebraic way.

Thanks


t is not the "time" in literal sense. It is the time taken Walt. As Walt started 2 hours late, the time he takes for the travel is 2 hours less than Lionel's. Thus, (t-2)

e.g.
A starts at 1:00PM and walks
B starts at 4:00PM, runs swiftly, and catches A at 5:00PM

By 5:00 PM:
A would have walked for t hours i.e. 4 hours
B would have run for (t-3) hours i.e. (4-3)=1 hour
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Re: Lionel left his house and walked towards Walt's house, which is 48 [#permalink]
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Lionel left his house and walked towards Walt's house, 48 miles away. Two hours later, Walt left his house and ran towards Lionel's house. If Lionel's speed was 4 miles per hour and Walt's 6 miles per hour, how many miles had Lionel walked when he met Walt?

A. 12
B. 16
C. 20
D. 24
E. 28

Hi - can someone please explain the step by step approach to solve this question? Its in the GMATCub Test m03.

Unless, my approach is wrong this is how I solved this:

Lionel Speed = 4mph

Walt speed = 6 mph

Let T be the time taken by Lionel, then T-2 will be the time taken by Walt

4T + 6(T-2) = 48 [ Rate = Distance/Time and therefore Distance = Time * Rate]. I solved this to get

T=6

Lionel's speed = 4

Distance Lionel covered = 4*6 = 24
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Re: Lionel left his house and walked towards Walt's house, which is 48 [#permalink]
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First step - same as Bunuel suggested
In the first 2 hours Lionel at the rate of 4 miles per hour covered distance=rate*time=4*2=8 miles. So, the distance between him and Walt was 48-8=40 miles when Walt left his house.

Then apply the formula
Time to meet = Distance between L and W/(Rate of L + Rate of B)
Time to meet = 40/(4+6)
Time to meet = 40/10
Time to meet = 4 hrs

Lionel will travel 16 miles in 4 hrs and he has initially traveled is 8miles. So 16+8=24 miles

D
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Re: Lionel left his house and walked towards Walt's house, which is 48 [#permalink]
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