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This type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30-40 seconds for a difficult one.

Teena covers 55 miles in 60 mins.
Joe covers 40 miles in 60 mins

So teena gains 15 miles every 60 mins

Teena need to cover 7.5 +15 miles.
Teena can cover 7.5 miles in 30 mins
Teena will cover 15 miles in 60 mins

So answer 30+60= 90 mins.
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Super shortcut
Time =initial separation/Relative speed
=7.5/(55-40)
=7.5/15
=1/2 HRS OR 30 MIN
NOW 7.5 miles....30 min
15 miles.....60 min
Total time (30+60)=90 min

Posted from my mobile device
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Bunuel
If Teena is driving at 55 miles per hour and is currently 7.5 miles behind Joe, who is driving at 40 miles per hour in the same direction then in how many minutes will Teena be 15 miles ahead of Joe?

A. 15
B. 60
C. 75
D. 90
E. 105

let h=hours for T to get 15 miles ahead of J
h(55-40 mph)=7.5+15 miles
h=1.5 hours=90 minutes
D
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It is a good idea to first establish catch up or shrink rate after one unit of time



Tina is catching up to Joe at 15 mph.
In 1 hour she would have passed Joe and gone ahead by 7.5 miles. Hence to reach 15 miles ahead of Joe, she needs half an hour more.

Answer is 1.5 hrs or 90 minutes, D
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If Teena is driving at 55 miles per hour and is currently 7.5 miles behind Joe, who is driving at 40 miles per hour in the same direction then in how many minutes will Teena be 15 miles ahead of Joe?

Speed (Teena) = \(55 \frac{Miles}{Hour}\)

Speed (Joe) = \(40 \frac{Miles}{Hour}\)

Distance Between Teena and JOe = 7.5 Miles, Total Distance Teena has to cover = 7.15 + 15 = 22.5 Miles

As both Teena and Joe are moving in the same direction, we have to substract the higher speed from the lower speed to get the relative speed

= \(55 \frac{Miles}{Hour}\) - \(40 \frac{Miles}{Hour}\) = \(15 \frac{Miles}{Hour}\)

So, lets calculate the time required for Teena to cover 22.5 miles

\(Time = \frac{Distance}{Speed}\)

\(= \frac{22.5}{15}\)

= 1.5 Hrs

As the time asked in the question is in minutes we will multiply it by 60

\(= 1.5 * 60\)

\(= 90 Min\)


Hence, the Answer is D
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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 miles

Now, 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?
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Joe is driving at 40 mph, Teena at 55 mph. So in one hour teena will gain 15 miles over joe (55-40). Teena starts at -7.5 miles wrt joe. and we need to find when teena is +15 miles wrt joe. So basically she has traveled an equivalent of 90 min (60 min for 15 miles and 30 min for 7.5 miles) to reach 15 miles ahead of joe.
Hence answer is D!
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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 miles

Now, 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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Teena is approaching Joe with a relative velocity of 15mph. Distance she has to cover is 7.5m+15m = 22.5miles.
Time = Distance/Speed = 22.5 m/15 mph = 1.5 Hours = 90 minutes.
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Bunuel
If Teena is driving at 55 miles per hour and is currently 7.5 miles behind Joe, who is driving at 40 miles per hour in the same direction then in how many minutes will Teena be 15 miles ahead of Joe?

A. 15
B. 60
C. 75
D. 90
E. 105


solution:
the relative speed=55-40=15 miles per hour

total relative distance to travel=7.5+15=22.5

so it takes 30 min or .5 hour to travel 7.5 miles and 60 min or 1 hour to travel 15 miles.

so 30+60=90 minutes

Option D
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Bunuel
If Teena is driving at 55 miles per hour and is currently 7.5 miles behind Joe, who is driving at 40 miles per hour in the same direction then in how many minutes will Teena be 15 miles ahead of Joe?

A. 15
B. 60
C. 75
D. 90
E. 105

Hi BrentGMATPrepNow, Getting a bit confused that how's this possible that Teena is driving at a faster speed at 55 miles per hour and yet still behind Joe (driving at 40 mph)? Is there an issue with wordings here or have I missed something? Thanks Brent
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Bunuel
If Teena is driving at 55 miles per hour and is currently 7.5 miles behind Joe, who is driving at 40 miles per hour in the same direction then in how many minutes will Teena be 15 miles ahead of Joe?

A. 15
B. 60
C. 75
D. 90
E. 105

We can use the formula of:

time = change in distance/change in rate

time = (15 + 7.5)/(55 - 40)

time = 22.5/15 = 225/150 = 15/10 = 1.5 hours = 90 minutes

Answer: D
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Kimberly77
Bunuel
If Teena is driving at 55 miles per hour and is currently 7.5 miles behind Joe, who is driving at 40 miles per hour in the same direction then in how many minutes will Teena be 15 miles ahead of Joe?

A. 15
B. 60
C. 75
D. 90
E. 105

Hi BrentGMATPrepNow, Getting a bit confused that how's this possible that Teena is driving at a faster speed at 55 miles per hour and yet still behind Joe (driving at 40 mph)? Is there an issue with wordings here or have I missed something? Thanks Brent

If you've ever been passed on the highway, you'll see that it's possible for faster car to be behind you a slower car (but then pass that slower car later on).
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BrentGMATPrepNow
Kimberly77
Bunuel
If Teena is driving at 55 miles per hour and is currently 7.5 miles behind Joe, who is driving at 40 miles per hour in the same direction then in how many minutes will Teena be 15 miles ahead of Joe?

A. 15
B. 60
C. 75
D. 90
E. 105

Hi BrentGMATPrepNow, Getting a bit confused that how's this possible that Teena is driving at a faster speed at 55 miles per hour and yet still behind Joe (driving at 40 mph)? Is there an issue with wordings here or have I missed something? Thanks Brent

If you've ever been passed on the highway, you'll see that it's possible for faster car to be behind you a slower car (but then pass that slower car later on).

I see thanks BrentGMATPrepNow. So it's just a matter of 2 cars driving at different speeds at diffferent times... :lol: GMAT wordings is draining my brain really
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Kimberly77

I see thanks BrentGMATPrepNow. So it's just a matter of 2 cars driving at different speeds at diffferent times... :lol: GMAT wordings is draining my brain really

The two cars are driving at different speeds, BUT they aren't driving at different times. They are driving at the same time.

Imagine this: It's noon, and you are driving north on a highway at a speed of 50 mph.
At the same time (noon), a motorist is 20 miles behind you and is driving north at a speed of 70 mph.
In one hour (at 1pm), the motorists will catch up to you.
At 3pm, the motorist will be 40 miles ahead of you.
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