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Two friends phil and andrew started running simultaneously from a poin 04 Nov 2019, 03:44
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Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?

(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr

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Two friends phil and andrew started running simultaneously from a poin 18 Jul 2020, 03:26
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Bunuel
Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?

(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr

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The above solutions consider relative speeds of the Andrew and Phil. However, if you are someone like me who gets confused by relative speeds, check out this simple solution which employs the basic distance formula we all are well versed with:

distance = speed * time
OR
time = \(\frac{distance }{ speed}\\
\)
Let's talk about Andrew first. Let his speed be 5x (where x is a real constant).
When Andrew travels for the first 2 hours at 5x speed, his distance covered will be 2*5x = 10x
Now, he stops after covering 10x distance and starts running backwards till he meets Phil at 10 km away from P.
This means that the distance covered by Andrew during the backward run is 10x-10. Also, as mentioned in the question, speed is 1/5th the original speed - > which is \(\frac{5x}{5}\) = x.

Now, let's move to Phil.
He only covers a distance of 10km with speed 2x (where x is a real constant). Thus, the time he takes is \(\frac{10}{2x}\) = \(\frac{5}{x}\)

Remember that the total time taken by Phil to cover 10km is equal to the total time taken by Andrew to cover 10x (when moving forward)+ 10x - 10 (when moving backward)

Now, this helps me set up the following equation:
\(\frac{5}{x} = \frac{10x }{ 5x} + \frac{10x - 10 }{ x}\\
\)

Solving this equation, we get:

5 = 2x + 10x - 10
12x = 15
x = 5/4

Phil's speed = 2x = 2*(5/4) = 5/2 = 2.5 km/hr
Answer (C)
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Two friends phil and andrew started running simultaneously from a poin Updated on: 04 Nov 2019, 05:58
Originally posted by nick1816 on 04 Nov 2019, 04:41.
Last edited by nick1816 on 04 Nov 2019, 05:58, edited 1 time in total.
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speed of Phil=2x
speed of Andrew=5x

Relative distance between Phil and Andrew after 2 hours= 2*(5x-2x)=6x

speed of A after he turned back= 5x/5=x

Time taken by them to meet= \(\frac{6x}{x+2x}\)= 2

Distance of meeting point from P= Total distance traveled by Phil = (2+2)*2x=8x

8x=10
2x= 2.5


Bunuel
Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?

(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr

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Two friends phil and andrew started running simultaneously from a poin 04 Nov 2019, 05:53
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nick1816
see highlighted part shouldnt it be (2+2)*2x = 8x?

nick1816
speed of Phil=2x
speed of Andrew=5x

Relative distance between Phil and Andrew after 2 hours= 2*(5x-2x)=6x

speed of A after he turned back= 5x/5=x

Time taken by them to meet= \(\frac{6x}{x+2x}\)= 2

Distance of meeting point from P= Total distance traveled by Phil = (2*2)*2x=8x

8x=10
2x= 2.5


Bunuel
Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?

(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr

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Two friends phil and andrew started running simultaneously from a poin 04 Nov 2019, 05:57
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That was typo. You have a good eye ;)

Archit3110
nick1816
see highlighted part shouldnt it be (2+2)*2x = 8x?

nick1816
speed of Phil=2x
speed of Andrew=5x

Relative distance between Phil and Andrew after 2 hours= 2*(5x-2x)=6x

speed of A after he turned back= 5x/5=x

Time taken by them to meet= \(\frac{6x}{x+2x}\)= 2

Distance of meeting point from P= Total distance traveled by Phil = (2*2)*2x=8x

8x=10
2x= 2.5


Bunuel
Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?

(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr

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Two friends phil and andrew started running simultaneously from a poin 04 Nov 2019, 06:57
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speed of P= 2x
speed of A = 5x
relative speed = 3x
2 hrs later the relative distance between P & A = 6x
after 2 hrs A speed ; 5x/5 = x
so relative speed of P & A = 3x
time at which they meet = 6x/3x ; 2 hrs
total distance covered by P = (2+2)*2 x; 8x
this 8x = 10km point where A met P
so 2x= 2.5
IMO C


Bunuel
Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?

(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr

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Two friends phil and andrew started running simultaneously from a poin 21 Jul 2020, 04:31
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Phil speed = 2x
Andrew speed = 5x
Time going in same direction = 2hrs
Time going in opposite direction = y hrs

Phil distance = 2 hrs * 2x + y hrs * 2x = 10 or 4x + 2xy = 10
Andrew distnace = 2hrs * 5x - y hrs * 5/5 x = 10 or 10x - xy = 10

We now have a simple 2 variable system:
4x + 2xy = 10
10x - xy = 10

Solving for x = 5/4, Phil's speed is 2x, which is 10/4 or 2.5

Answer C
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Two friends phil and andrew started running simultaneously from a poin 13 Oct 2020, 23:21
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The one thing that wasn't clear at the outset (but now I see how I misinterpreted the question) is that Phil never makes it to the end of the running track.

Andrew makes it to the END of this running track of Unknown Distance in 2 hours.

Andrew then turns around and Meets Phil 10 km from where they started.

This means, in total, Phil must have ONLY traveled 10 km from the beginning.

(I was working the Problem thinking that somehow Phil made it to the End of the running track, then turned around trying to catch up to Andrew ---- WRONG)



Let Speed of Phil = 2x

Let Speed of Andrew = 5x


in 2 hours, we can use the Ratio Speeds to determine the Distance each traveled because over a CONSTANT 2 hours: Speed is Directly Proportional to Distance Traveled.

Andrew covers the ENTIRE Distance of the running track. We'll call this Distance of the running track = (Speed of Andrew) * (2 hours) = 5x * 2 = 10x DISTANCE


in the Same 2 hours,

Distance PHIL Covers = 2x * (2 hours) = 4x DISTANCE




This means that at the END of the 2 hours:

Andrew is at the end of the running track having covered 10x Dist.

Phil only covered 4x Dist. from the Starting Point.

GAP Distance between them = 10x - 4x = 6x "Gap" Distance that Each must cover as they Run towards each other in Opposite Directions.


However, now Andrew's Speed is (1/5)th of his Prior Speed: (5x) * (1/5) = 1x Speed of Andrew (in Ratio Units)


Time = ("Gap" Distance between them) / (Speed of Phil + NEW Speed of Andrew) = 6x / (2x + 1x) = 2 hours



(Lastly) Calculate the Distance Phil traveled and the Time it took him to do so


From above, it is explained how Phil only traveled 10 km away from the Starting Point when Andrew turns around to meet him.

Phil covered this 10 km in:

(1st) he ran for 2 hours when Andrew made it to the end of the running track

+

(2nd) he ran for ANOTHER 2 hours when he met Andrew again when Andrew turned around
_______________________________

4 hours



Actual Speed of Phil = (10 km) / (4 hours) = 2.5 km/h


-C-
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Two friends phil and andrew started running simultaneously from a poin 14 Jun 2021, 06:13
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these questions can never be pin down within 3 min it took me 4 +min i will lay put the plan total time = 2+t1 = 10/2x
= D/5x +D-10/x= 10/2x the time taken by phil which gives us the distance as 75/6 through which we find 2x as 2.5
imo C
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Two friends phil and andrew started running simultaneously from a poin 15 Jun 2021, 00:48
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The ratio of their speed is 2:5
Phil has reached 10 km from point P in 2 hours, so 2x*2=10
x=5/2
Speed of Andrew is 1/5 of its original speed: 5x*1/5=x
Speed of Andrew is 5/2=2.5 (C)
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Two friends phil and andrew started running simultaneously from a poin 15 Jul 2021, 11:50
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Bunuel
Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?

(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr

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With all due respect Question stem contains incorrect usage Bunuel

Lets say he is running towards East - Question Stem says - he turned back - Now he faces West - Now he is running backwards - Which means he runs towards EAST !!!!

A simple case of double negative !

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Two friends phil and andrew started running simultaneously from a poin 15 Jul 2021, 20:20
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Bunuel
Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?

(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr

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Let the initial speed of P and A be 2x and 5x. So, A gains 3x units of distance every hour.
After two hours, he would have gained 3x*2 or 6x.

He turns back and starts running (backwards) at a rate of 5x/5 or x. But P is running at 2x.
Now p is gaining 2x-x or x units of distance every hour.

P would gain 6x in 6 hours. So after 2+6 hours, they are at the same point, that is 10 km from start.

So the CONSTANT speed of P is 10/8=1.25 km per hour.
Incidentally, this will also be the AVERAGE speed of A.

hD13, you are correct the question does mean that A is running in the same direction
A=> U…>>5x>>>……V…….>>x>>……..W
P=> U…………………..>>2x>>………………..W
U is the start point. V is where A turns backwards and W is the point 10km away.
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Two friends phil and andrew started running simultaneously from a poin 15 Jul 2021, 23:25
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Bunuel
Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?

(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr

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Say their speeds are 2s and 5s.
In 2 hrs, Phil covers 4s distance and Andrew covers 10s distance. Now distance between them is 6s.

______(4s)__________P-->_____________(6s)_________________<--A


The two of them together need to cover this 6s distance. Now their speeds are in the ratio 2: 1 (because Andrew's speed becomes 1/5th his initial speed). So Phil will cover 4s of the 6s distance and Andrew will confer 2s of the 6s distance (in the ratio 2:1)

So total distance covered by Phil in all this time = 4s + 4s = 8s = 10 km
s = 10/8 km/hr
Phil's speed = 2s = 2*10/8 = 2.5 km/hr
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Two friends phil and andrew started running simultaneously from a poin 21 Apr 2024, 14:41
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Bunuel
Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?

(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr

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­Speed of Phill/Speed of Andrew=2/5=2x/5x.
Andrew covers the total distance in 2 hours therefore, the total distance one-way is 2*5x=10x km. Phill goes in 2 hours 2x*2=4x km. The remaining distance to be covered by Phill is 10x-4x=6x km.

Now, Andrew's new speed is 5x/5=x km/hr. Therefore, the relative speed of Phill and Andrew is x+2x=3x km/hr. The time taken by them to meet is 6x/3x=2 hours.

In this 2 hours, Phill goes 2x*2=4x km.

Given, 4x+4x=10 or, x=1.25. So, Phill's speed is 2*1.25=2.5 km/hr. Option (C) is correct.
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Two friends phil and andrew started running simultaneously from a poin 08 Aug 2025, 13:16
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