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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Two trains A and B start from two points P1 and P2 respectively at the

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Intern  B
Joined: 20 May 2020
Posts: 2
Two trains A and B start from two points P1 and P2 respectively at the  [#permalink]

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6 00:00

Difficulty:   95% (hard)

Question Stats: 24% (03:43) correct 76% (03:01) wrong based on 33 sessions

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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

Originally posted by carpedeim on 01 Jun 2020, 09:24.
Last edited by Bunuel on 01 Jun 2020, 09:32, edited 1 time in total.
Renamed the topic.
DS Forum Moderator V
Joined: 19 Oct 2018
Posts: 1980
Location: India
Re: Two trains A and B start from two points P1 and P2 respectively at the  [#permalink]

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1
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

carpedeim wrote:
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
Senior Manager  S
Joined: 18 Dec 2017
Posts: 292
Re: Two trains A and B start from two points P1 and P2 respectively at the  [#permalink]

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1
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
Intern  S
Joined: 24 Nov 2019
Posts: 30
Re: Two trains A and B start from two points P1 and P2 respectively at the  [#permalink]

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carpedeim wrote:
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

Let the distance is D km
P1 speed = x kmph
P2 speed =x +10 kph
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
Senior Manager  S
Joined: 26 May 2020
Posts: 266
Concentration: General Management, Technology
WE: Analyst (Computer Software)
Re: Two trains A and B start from two points P1 and P2 respectively at the  [#permalink]

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carpedeim wrote:
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

Difficult one to complete in 2 minutes during exam condition .

Let the distance between P1 and P2 = D km
Let A speed = V km/hr
B speed = V +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
If A 's speed is zero B will move in 2V+10 to reach D distance before they meet .
so time to meet = D / (2V+10)

Hence ((V+10)D) / (2V+10) - VD/ (2V+10) = 200/9 => 90D = 400V+2000 -----------------------eq 1

Again
D/V- D/(V+10)=1 => 90D = 9V^2+ 90 V --------------------------------------------------------------eq 2

Solving eq 1 and 2

9V^2-310V -2000=0 => V = 40 KM / Hr

Therefore
D = 9V^2+ 90 V / 90
=200 km ..

The calculation part in this ques is a pain . Can any one help me with a better method to solve this calculation in a quick time

_________________
Thank you.
Regards,
Ashish A Das.

The more realistic you are during your practice, the more confident you will be during the CAT.
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 11036
Location: United States (CA)
Re: Two trains A and B start from two points P1 and P2 respectively at the  [#permalink]

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1
carpedeim wrote:
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

Solution:

We can let r = the speed of train A; thus, r + 10 = the speed of train B (notice that train A is the slower train). We can let t = the time it takes train A to cover the distance between P1 and P2, so t - 1 = the time it takes train B to cover the same distance. Finally, we can let m = the time it takes for the two trains to meet. We can create the equations:

rt = (r + 10)(t - 1) = (r + r + 10)m

and

(r + 10)m - rm = 200/9

Notice that the first equation is just different ways to express the distance between P1 and P2 and the second second equation expresses the difference in distance traveled between the two trains when they meet. Anyway, solving the second equation, we have:

rm + 10m - rm = 200/9

10m = 200/9

m = 20/9

Solving the left part and the middle part of the first equation, we have:

rt = rt - r + 10t - 10

r + 10 = 10t

(r + 10)/10 = t

r/10 + 1 = t

Solving the left part and the right part of the first equation, we have:

rt = (r + r + 10)m

Substituting m = 20/9 and t = r/10 + 1 into the above equation, we have:

r(r/10 + 1) = (2r + 10)(20/9)

r^2/10 + r = 40r/9 + 200/9

Multiplying the equation by 90, we have:

9r^2 + 90r = 400r + 2000

9r^2 - 310r - 2000 = 0

(r - 40)(9r + 50)

r = 40 or r = -50/9

Since r can’t be negative, r = 40. Since t = r/10 + 1, t = 40/10 + 1 = 5. Since the distance between P1 and P2 is rt, the distance is 40(5) = 200.

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# Two trains A and B start from two points P1 and P2 respectively at the  