A man covers a certain distance in a train. If the train mov

Question

A man covers a certain distance in a train. If the train moved 4km/hr faster, it would take 30 min less. If it moved 2km/hr slower, it would take 20 mins more. Find the distance?

A. 200 km
B. 50 km
C. 20 km
D. 60 km
E. 80 km

Bunuel · Answer

(R + 4)(T - 1/2) = (R - 2)(T+1/3).
(R + 4)(T - 1/2) = RT.

Solving gives R = 20 kmh and T = 3 hours. The distance = RT = 60 km.

Answer: D.

inter · Answer

Can anybody explain the second equation (R + 4)(T - 1/2) = RT (I understand the first - it is straight of steam) or show a different way to solve?

ENGRTOMBA2018 · Answer

Both the equations are from the give information and applying the formula Distance = Speed * Time

(R + 4)(T - 1/2) = (R - 2)(T+1/3) ....(1)

(R + 4)(T - 1/2) = RT.....(2)

R and T are the rate and time taken respectively.

You need to remember that the total distance traveled remains the same. Only the rates (R) and time taken (T) changes.

RT= Distance covered with original rate and time taken

(R + 4)(T - 1/2) = Distance covered with original rate+4 kmph and time taken is original time - 30 minutes.

Equation 2, equates the total distance traveled with 'original conditions' to the total distance covered with modified rates. The total distance remains the same.

Hope this helps.

inter · Answer

Thanks a lot. Still, those two systems of equations look rather complex to solve

ENGRTOMBA2018 · Answer

Not really. When you solve the 2 equation above, you get,

6T-4/3 = 5R/6 from simplifying equation 1

4T-2=R/2 from simplifying equation 2

You can now multiply equation 2 by 5 to get

5(4T-2=R/2 ) = 20T-10=5R/2 and then subtract this new equation from equation 1 to get T=3, followed by R=20 to give you distance = R*T=20*3=60 km.

This much of calculation will be required sometimes in GMAT. If you are not comfortable with this, then practice solving algebraic equations.

Hope this helps.

inter · Answer

6T-4/3 = 5R/6 Yep - this is what I call "complex"

. I though there might be another shortcut.

mvictor · Answer

oh man..today i am so slow....

wrote tens of time the same equation and couldn't solve it...what a shame...though it is very simple...

suppose we have R - rate, T-time. RT=distance
(R+4)(T-1/2) = RT
RT - R/2 +4T -2 = RT
-R/2 + 4T -2 = 0 | multiply by 2 everything
-R + 8T - 4 =0 -> R=8T-4

second equation:
(R-2)(T+1/3) = RT
RT +R/3 -2T -2/3 = RT
R/3 -2T -2/3 = 0 | multiply everything by 3
R - 6T - 2 = 0
R=6T+2
now..we have 2 equations for R
8T-4 = 6T+2
2T=6
T=3.
R=6*3+2 = 20

RT = 20 * 3 = 60

D

gracie · Answer

(r+4)(t-1/2)=r*t=distance
(r-2)(t+1/3)=r*t=distance
  8t-r-4=0
  -6t+r-2=0
adding   and  ,
2t-6=0
t=3
r=20
r*t=20*3=60 km distance
D.

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