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Let v be the speed of the taxi

3v = (v-30)8

=> v = 48 mph

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subhashghosh
I'm also getting 48 by both methods :

Speed of taxi = x

Speed of bus = x - 30

5(x-30) + 3 (x-30) = 3x

8x - 240 = 3x

5x = 240
x = 48

Relative Speed method :

5(x-30)/30 = 3

x = 18 + 30 = 48


Yes it's 48.

A taxi leaves Point A 5 hours after a bus left the same spot. The bus is traveling 30 mph
slower than the taxi. Find the speed of the taxi, if it overtakes the bus in three hours.

(A) 44

(B) 46

(C) 48

(D) 50

(E) 52


The idea in this question is to set the distance for each of them to be equivalent. Each of them has a distinct "r*t". Set these equal to each other because the distance covered for each will be the same at the moment that the taxi overcomes the bus.
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meprarth
A taxi leaves Point A 5 hours after a bus left the same spot. The bus is traveling 30 mph slower than the taxi. Find the speed of the taxi, if it overtakes the bus in three hours.

(A) 44
(B) 46
(C) 48
(D) 50
(E) 52

(T-B)(3) = 5B

and B= T-30

Hence, T =48

Answer is C

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Why is the bus only doing 18 mph? Is it some kind of special GMAT bus?
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gracie
Why is the bus only doing 18 mph? Is it some kind of special GMAT bus?
That is an interesting question. May be it is a sign that you should not board this bus :)

As for solving the question.

Let us assume that the speed of the Taxi = x
Hence the speed of the bus = x - 30

At the point of overtaking, both the bus and the taxi would have covered equal distance, so we can safely equate them
Also, note that the Taxi has been running for 3 hours, whereas the bus has been running for 8 hours.

\(x*3 = (x-30)*8\)
\(3x = 8x - 240\)
\(5x = 240\)
Or \(x = 48\). Option C
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Let Speed of Bus =Sb
and Speed of Taxi =St

St = Sb + 30

Distance travelled by Bus in 8 hours is same as the distance travelled by taxi in 3 hours

8 Sb = 3 St
=> 8 Sb = 3Sb + 90
=> Sb= 18

St=48 :-D
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let Vt be speed of taxi.
therefore bus speed is : Vb=Vt-30

Since bus left point A, 5 hours before the taxi left. so the distance travelled by bus will be
d=5(Vt-30)=5Vt-150 ........(a)
GIven that, taxi overtakes the bus after 3 hours. So,
t=d/ (relative speed) where relative speed is (Vt-Vb)= Vt-(Vt-30)=30

3= (5Vt-150) / 30
90+150= 5Vt
Vt=48

Answer is C
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meprarth
A taxi leaves Point A 5 hours after a bus left the same spot. The bus is traveling 30 mph slower than the taxi. Find the speed of the taxi, if it overtakes the bus in three hours.

(A) 44
(B) 46
(C) 48
(D) 50
(E) 52

Experts please help TeamGMATIFY GMATPill Bunuel

I solved it in an interesting way, I am not sure if it is right or wrong.

Relative speed = 30km/hr
Relative distance covered in 3 hours by the car = 30*3 = 90 kms
this 90 kms were what the bus drove in that 5 hours headstart.

Speed of the bus = 90/5 = 18km/hr
Speed of the car = 18+30 = 48km/hr
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(5+3)(x-30)= 3x
8(x-30)= 3x
8x-240=3x
8x-240-3x=0
5x-240=0
x=240/5
x= 48

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Sorry to revive an old thread but for the second method why do we divide by 30? Thank you.
subhashghosh
I'm also getting 48 by both methods :

Speed of taxi = x

Speed of bus = x - 30

5(x-30) + 3 (x-30) = 3x

8x - 240 = 3x

5x = 240
x = 48

Relative Speed method :

5(x-30)/30 = 3

x = 18 + 30 = 48
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BelisariusTirto
Sorry to revive an old thread but for the second method why do we divide by 30? Thank you.
subhashghosh
A taxi leaves Point A 5 hours after a bus left the same spot. The bus is traveling 30 mph slower than the taxi. Find the speed of the taxi, if it overtakes the bus in three hours.

(A) 44
(B) 46
(C) 48
(D) 50
(E) 52

I'm also getting 48 by both methods :

Speed of taxi = x

Speed of bus = x - 30

5(x-30) + 3 (x-30) = 3x

8x - 240 = 3x

5x = 240
x = 48

Relative Speed method :

5(x-30)/30 = 3

x = 18 + 30 = 48

In that method, we divide the distance between the bus and the taxi, 5(x - 30) miles, by the relative speed of the bus and the taxi, which is x - (x - 30) = 30 mph. This gives the time it takes for the taxi to catch up to the bus, which is 3 hours.

Hope it's clear.
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