Blackbox wrote:

JeffTargetTestPrep - Could you please explain this question in line with your explanation

here. Thanks in advance.

Here is what I did as per your approach:

S(Teena) = 55

D = x+7.5

T = t

S(Joe) = 40

D = x

T = t

x+7.5 =55t --- (1)

x = 40t --- (2)

Therefore, t = 1/2 (OR) 30 mins, which is when they will both be caught up. Total distance scaled by them is

x = 20 milesNow, to the remaining part of the question, where it says Teena is 15 miles ahead of Joe - which translates to:

D = 20 + 15 +7.5 = 42.5 miles

T = t

S = 55 mph

T = 42.5/55*60 (because they wanted the time in minutes) ~ 46 minutes. Where am I going wrong?

Hi

Blackbox,

Please find highlighted portion.

In the time that Teena catches up with Joe,

she travels about 20 + 7.5 miles that she was away.

Therefore, Teena travels a distance of 27.5 and not 20.

Now that both of them are at the same point, we need to use the relative speed

Relative speed at which Teena is moving ahead is 55-40 or 15mph(as they are travelling in the same direction)

and it will take her an hour(60 minutes) to overtake Joe by 15 miles.

Since, we have been asked to find the total time, it would be 30+60(90) minutes

Hope that helps you!

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