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Difficulty:
95%
(hard)
Question Stats:
31% (03:22) correct
69%
(02:59)
wrong
based on 167
sessions
History
Date
Time
Result
Not Attempted Yet
Two trains A and B start from two points P1 and P2 respectively at the same time and travel towards each other. The difference between their speed is 10 kmph and train A takes one hour more to cover the distance between P1 and P2 as compared to train B. Also by the time they meet, train B has covered 200/9 km more as compared to train A. What is the distance between P1 and P2?
A) 150
B) 200
C) 250
D) 300
E) Data insufficient
A) 150
B) 200
C) 250
D) 300
E) Data insufficient
01 Jun 2020, 19:53
Kudos
Bookmarks
Speed of train A = v
Time taken by train A = t
\(v*t = (v+10)(t-1)\)
\(10t - v= 10\)
\(10t-10 = v\).....(1)
Both train meet after = \(\frac{20}{9}\) hrs
\(v*\frac{20}{9} + (v+10)*\frac{20}{9} = vt\)
\(40v+200 = 9vt\).....(2)
From (1) and (2)
\(40t-40+20= 9t^2-9t\)
\(9t^2-49t+20\)
\((t-5)(9t-4)=0\)
t= 5
or t = 4/9(<20/9) Not possible
\(v= 10*5-10 = 40\)
Distance \(= 40*5=200\)Km
Time taken by train A = t
\(v*t = (v+10)(t-1)\)
\(10t - v= 10\)
\(10t-10 = v\).....(1)
Both train meet after = \(\frac{20}{9}\) hrs
\(v*\frac{20}{9} + (v+10)*\frac{20}{9} = vt\)
\(40v+200 = 9vt\).....(2)
From (1) and (2)
\(40t-40+20= 9t^2-9t\)
\(9t^2-49t+20\)
\((t-5)(9t-4)=0\)
t= 5
or t = 4/9(<20/9) Not possible
\(v= 10*5-10 = 40\)
Distance \(= 40*5=200\)Km
carpedeim
01 Jun 2020, 20:13
Kudos
Bookmarks
Let the distance between P1 and P2 = D km
Let A speed = x km/hr
B speed =x +10 km /hr
When they both move towards each other B covers 200/9 km extra and B's extra speed over A is 10km/hr
Therefore they have both travel 20/ 9 hrs before they meet.
Therefore
20/9×x +20/9×(x+10)=D ----(1)
Also
D/x- D/(x+10)=1
D=x(x+10)÷10
Putting this value in eq 1
9x^2-310x -2000=0
Solving x=40 km/hr
Therefore
D= 40×50÷10
=200 km
Posted from my mobile device
Let A speed = x km/hr
B speed =x +10 km /hr
When they both move towards each other B covers 200/9 km extra and B's extra speed over A is 10km/hr
Therefore they have both travel 20/ 9 hrs before they meet.
Therefore
20/9×x +20/9×(x+10)=D ----(1)
Also
D/x- D/(x+10)=1
D=x(x+10)÷10
Putting this value in eq 1
9x^2-310x -2000=0
Solving x=40 km/hr
Therefore
D= 40×50÷10
=200 km
Posted from my mobile device
