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Two friends, X and Y start from two points M and N and move towards 06 Jul 2017, 02:33
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55% (02:11) correct 45% (02:14) wrong based on 690 sessions

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Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.
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C
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Two friends, X and Y start from two points M and N and move towards 06 Jul 2017, 03:14
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Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.
Show more

\(Speed = \frac{Distance}{Time}\)

Question: Distance of Y = ?

Statement 1:X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively

The ratio of speeds = Ratio of their timings as Speed is directly proportional to Time

\(\frac{S_1}{S_2} = \frac{D_1}{D_2}\)
But since the total distance is unknown hence
NOT SUFFICIENT

Statement 2:Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P
i.e. @20% of speed of Y, the distance travelled in entire time of travel = 10 miles

but since we don't know the speed of Y hence
NOT SUFFICIENT

Combining the two statements
Speed of Y = 20 miles and hour
@20% of speed of Y, the distance travelled in entire time of travel = 10 miles
i.e. (0.2*20)*T = 10 where T is the total time of travel
i.e. T = 2.5 hours
So distance travelled by Y = 20*2.5 = 50 miles

answer: option C
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Two friends, X and Y start from two points M and N and move towards 07 Jul 2017, 03:35
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I originally thought that without any mention when they left, this wouldn't be possible. However that doesn't matter as using both statements allows you to set up 2 relevant distance equations for Person Y.
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Two friends, X and Y start from two points M and N and move towards 18 Jul 2017, 10:16
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Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.

\(Speed = \frac{Distance}{Time}\)

Question: Distance of Y = ?

Statement 1:X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively

The ratio of speeds = Ratio of their timings as Speed is directly proportional to Time

\(\frac{S_1}{S_2} = \frac{D_1}{D_2}\)
But since the total distance is unknown hence
NOT SUFFICIENT

Statement 2:Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P
i.e. @20% of speed of Y, the distance travelled in entire time of travel = 10 miles

but since we don't know the speed of Y hence
NOT SUFFICIENT

Combining the two statements
Speed of Y = 20 miles and hour
@20% of speed of Y, the distance travelled in entire time of travel = 10 miles
i.e. (0.2*20)*T = 10 where T is the total time of travel
i.e. T = 2.5 hours
So distance travelled by Y = 20*2.5 = 50 miles

answer: option C
Show more


When it is told that if Y had travelled 20% faster then they both would have met 10 mile away from point P

My doubt is
1) How do u know if that point is left to point P or right to point P? are we assuming it on left to P becoz Y would have travelled faster?
2) How can the distance between P and the new meeting point tells us that Y travelled only 10 miles??



Please i need urgent help
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Two friends, X and Y start from two points M and N and move towards 19 Jul 2017, 00:31
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Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.

(1) says-
the distance travelled by Y - 2/5 d.
the distance travelled by X - 3/5 d.

Not sure of d ( total distance). Clearly insuff.

(2)- No mention of speed of Y and what is the ratio of speed of X and Y.
Clearly insuff.

(1) + (2)-

YP = distance Y travelled = 2/5*d.

Now YP' = new distance travelled by Y = (20*1.20 / ( 20*1.2 + 30))d = 0.44d

YP' - YP = 0.04d

0.04d = 10; d = 250.
YP = 2/5*d = 100 miles.

Ans C
Hope it clears doubt.
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Two friends, X and Y start from two points M and N and move towards 19 Jul 2017, 01:10
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shekyonline
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.

(1) says-
the distance travelled by Y - 2/5 d.
the distance travelled by X - 3/5 d.

Not sure of d ( total distance). Clearly insuff.

(2)- No mention of speed of Y and what is the ratio of speed of X and Y.
Clearly insuff.

(1) + (2)-

YP = distance Y travelled = 2/5*d.

Now YP' = new distance travelled by Y = (20*1.20 / ( 20*1.2 + 30))d = 0.44d

YP' - YP = 0.04d

0.04d = 10; d = 250.
YP = 2/5*d = 100 miles.

Ans C
Hope it clears doubt.
Show more





the distance travelled by Y - 2/5 d.
the distance travelled by X - 3/5 d.

How?
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Two friends, X and Y start from two points M and N and move towards 19 Jul 2017, 01:39
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Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.

\(Speed = \frac{Distance}{Time}\)

Question: Distance of Y = ?

Statement 1:X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively

The ratio of speeds = Ratio of their timings as Speed is directly proportional to Time

\(\frac{S_1}{S_2} = \frac{D_1}{D_2}\)


But since the total distance is unknown hence
NOT SUFFICIENT

Statement 2:Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P
i.e. @20% of speed of Y, the distance travelled in entire time of travel = 10 miles

but since we don't know the speed of Y hence
NOT SUFFICIENT

Combining the two statements
Speed of Y = 20 miles and hour
@20% of speed of Y, the distance travelled in entire time of travel = 10 miles
i.e. (0.2*20)*T = 10 where T is the total time of travel
i.e. T = 2.5 hours
So distance travelled by Y = 20*2.5 = 50 miles

answer: option C
Show more


hey

I got your equation very right ....

please, however, tell me the reason why the below equation is not right for...

1.2y(d + 10) - yd = 10 miles
where y = y's speed and d= distance traveled by y

thanks in advance ...
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Two friends, X and Y start from two points M and N and move towards 19 Jul 2017, 02:30
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rocko911

Distance travelled by X = d(x) = speed(x)* time(x)
Distance travelled by Y = d(y) = speed(y)* time(y)

time(x) = time(y) as X, Y meet at a point.

So d(x)/(d(y) = 3/2

distance traveled by X will be d(x)/(d(x)+d(y)) of total distance= (3/ (3+2)) of d = 3/5d

distance traveled by Y will be d - 3/5d = 2/5d.

Hope its clear.
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Two friends, X and Y start from two points M and N and move towards 04 Oct 2017, 23:49
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Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.
Show more


Is statement B alone not sufficient to solve this? If speed of Y is "Y" and Time for them to meet at point P is "T" then we effectively need to calculate Y*T.
Distance when travelling at speed Y = Y*T
Distance when travelling at speed 1.2Y = Y*T + 10 (this distance will be more than Y*T because Y is walking faster)
So we get Y*T + 10 = 1.2 Y*T
which gives Y*T = 50 [isn't this sufficient by itself?]
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Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2)

Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.



Is statement B alone not sufficient to solve this? If speed of Y is "Y" and Time for them to meet at point P is "T" then we effectively need to calculate Y*T.
Distance when travelling at speed Y = Y*T

Distance when travelling at speed 1.2Y = Y*T + 10 (this distance will be more than Y*T because Y is walking faster)


So we get Y*T + 10 = 1.2 Y*T
which gives Y*T = 50 [isn't this sufficient by itself?]
Show more

Shiprasingh1100
I was thinking on the same line, but my friend we are wrong.

Check the above highlighted texts again and try to find a flaw in your reasoning. First try if you are still unable to find the flaw check the below spoilers.

Hint:
Show Spoiler
Statement 2 says Y travels faster and nothing about the speed of X.

Answer:
Show Spoiler
Statement 2 says 10miles away, and nothing about the speed of X.
Case 1. speed of X is constant then, your equation below is correct.
1.2Y = Y*T + 10

Case2. Speed of X is increased by a greater number such that speed of Y relative to its speed earlier is less. In this case Y would travek 10miles less than what Y travelled earlier. So the equation would be
1.2Y = Y*T - 10
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Two friends, X and Y start from two points M and N and move towards 06 Oct 2017, 22:44
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Shiprasingh1100
Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.


Is statement B alone not sufficient to solve this? If speed of Y is "Y" and Time for them to meet at point P is "T" then we effectively need to calculate Y*T.
Distance when travelling at speed Y = Y*T
Distance when travelling at speed 1.2Y = Y*T + 10 (this distance will be more than Y*T because Y is walking faster)
So we get Y*T + 10 = 1.2 Y*T
which gives Y*T = 50 [isn't this sufficient by itself?]
Show more

Can any expert respond to this. Taking simple numbers also, if the original speed is say 100 mph and the speed is increased by 20 percent. The distance traveled will also be increased by 20%? If so, statement B is sufficient.
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Bunuel

In the second case, What if the time is same.

So t1= t2
I.e Increase in speed will compensate by decreasing the distance. Time being same.

p/y = p+10/1.2y

Y get cancelled. We have p=50.
So statement 2 is sufficient.

Where I am making the error? Please suggest.
Thanks in advance.
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Two friends, X and Y start from two points M and N and move towards Updated on: 06 Apr 2018, 23:12
Originally posted by nahid78 on 02 Apr 2018, 09:55.
Last edited by nahid78 on 06 Apr 2018, 23:12, edited 1 time in total.
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Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.
Show more


Let assume that Y's speed= S
Time= T
Then Distance Y traveled up to the meeting point = D
We are supposed to find S*T or D

S*T = D

as per (2) 1.2S*T = D + 10
So, D = 1.2S*T -10


SO S*T= 1.2S*T - 10
1.2S*T - S*T = 10
ST (1.2-1) = 10
S*T= 10/.2= 50

So, (2) alone can answer this question. If i am wrong, what is wrong with my calculation?

chetan2u, Can you please help me?
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Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.

\(Speed = \frac{Distance}{Time}\)

Question: Distance of Y = ?

Statement 1:X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively

The ratio of speeds = Ratio of their timings as Speed is directly proportional to Time

\(\frac{S_1}{S_2} = \frac{D_1}{D_2}\)


But since the total distance is unknown hence
NOT SUFFICIENT

Statement 2:Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P
i.e. @20% of speed of Y, the distance travelled in entire time of travel = 10 miles

but since we don't know the speed of Y hence
NOT SUFFICIENT

Combining the two statements
Speed of Y = 20 miles and hour
@20% of speed of Y, the distance travelled in entire time of travel = 10 miles
i.e. (0.2*20)*T = 10 where T is the total time of travel
i.e. T = 2.5 hours
So distance travelled by Y = 20*2.5 = 50 miles

answer: option C


hey

I got your equation very right ....

please, however, tell me the reason why the below equation is not right for...

1.2y(d + 10) - yd = 10 miles
where y = y's speed and d= distance traveled by y

thanks in advance ...
Show more

I think i have got it

distance = speed * time, NOT speed * distance

thnaks
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Two friends, X and Y start from two points M and N and move towards 06 Apr 2018, 23:34
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Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.


Let assume that Y's speed= S
Time= T
Then Distance Y traveled up to the meeting point = D
We are supposed to find S*T or D

S*T = D

as per (2) 1.2S*T = D + 10
So, D = 1.2S*T -10


SO S*T= 1.2S*T - 10
1.2S*T - S*T = 10
ST (1.2-1) = 10
S*T= 10/.2= 50

So, (2) alone can answer this question. If i am wrong, what is wrong with my calculation?

chetan2u, Can you please help me?
Show more


Hi....
Where you are going wrong is taking same time T in both cases..
When you increase speed of one, they will NOT meet in same time as they were meeting earlier. It will now be lesser than T
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Two friends, X and Y start from two points M and N and move towards 11 Apr 2019, 08:31
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Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y traveled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.
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Hi Bunuel.

I thought it is E because it didn't state that X and Y started at the same time. Am I exaggerating?
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Two friends, X and Y start from two points M and N and move towards 10 Jul 2019, 23:56
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S1 & S2 are insufficient.
When combined: (refer to image)

The essence of the question is realising that even after the increase in speed, the time taken to meet is the same.
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Hi,

Unfortunately, I believe that most/all answers mentioned are wrong except for the fact that chetan2u has stated, which is true. If the speed of Y increases, and the speed of X remains the same, the total time taken for X and Y to meet will reduce. Why is this so? The answer is simple: consider 'd' to be the distance between M and N. When X and Y meet at P, they together cover a distance d. Now, at original speeds of x and y, total time taken to cover d is d/(x+y), where x and y is the combined rate of X and Y. If Y's speed increases by 20% and X's speed remains the same, then total time to cover d will be d/(x + 1.2y) which is less than d/(x+y). Therefore Statement II is insufficient, because all we have from St. II is that .2y*z = 10, where z is a time less than the some time t taken by Y, at its original speed, to cover distance until.

St I + St II:
We realise that X and Y travel for the same time until reaching point P. We'll call this time t. Let the total distance between M and N be represented by d.
Now, from St I, 30t + 20t = d - - > (1)
from St II, 30z + 24z = d -- >(2) (z is a time less than t, as per the explanation above)
we also know that 24z - 20t = 10 -->(3).

3 variables, 3 equations - solvable. Hence the answer is (C).

Bunuel chetan2u
GMATinsight - I believe your calculations are incorrect.

Thank you
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Two friends, X and Y start from two points M and N and move towards 05 Jun 2020, 19:36
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sidship21
Hi,
Unfortunately, I believe that most/all answers mentioned are wrong except for the fact that chetan2u has stated, which is true. If the speed of Y increases, and the speed of X remains the same, the total time taken for X and Y to meet will reduce. Why is this so? The answer is simple: consider 'd' to be the distance between M and N.

3 variables, 3 equations - solvable. Hence the answer is (C).

Bunuel chetan2u
GMATinsight - I believe your calculations are incorrect.

Thank you
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Hi,

You are correct with your observations, and answer can be marked C without solving.

But if we were to solve it fully, one easy way would be..
Quote:
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?
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No info given except both are travelling towards each other.

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
SO they have covered the distance in ratio X:Y = 30:20 or 3:2 or 3a:2a when they have traveled for equal time. We require to know the total distance to answer this question.

(2) Had Y traveled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.
So speed of \( Y = Y*1.2=24. \)

Combined
statement II tells us the new speed of Y = 20*1.2=24, and the distance covered now is \(X:Y=24:30=4:5=4b:5b\). Total distance = 4a+5a=9a.
Statement I told us they traveled a combined distance of 2a+3a=5a, but the distance has to be same in both instances, so \(5a=9b\)...(i)
Also, when speed is increased by 20% Y traveled 10 miles more. So, 4b-2a=10...a=2b-5...(ii)

Substitute the value of a in (i)
\(5(2b-5)=9b......10b-25=9b....b=25\), and Y travels = 4b=4*25=100
Total distance = 25*9=225= 5a...a=45
So 2a=2*45=90

C

NOTE : Y travels 2a or 90 when he travels with original speed and 4b or 100 with increased speed. difference = 100-90=10.
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Two friends, X and Y start from two points M and N and move towards 29 Jul 2020, 01:49
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Bunuel
Two friends, X and Y start from two points M and N and move towards each other and meet at a point P on the way. How much distance has Y travelled up to the meeting point?

(1) X and Y travel at speeds of 30 miles per hour and 20 miles per hour, respectively.
(2) Had Y travelled at 20 percent higher speed, X and Y would have met at a point which is 10 miles away from P.
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Responding to a pm:


M ......................................... N
(X) --->---------P----<--(Y)

We need the actual distance PN.

1. X:Y = 3:2
Speeds are 30 mph and 20 mph respectively.
So Y will cover 2/5th of the total distance. But we need to know the distance.


2. If Y were 1.2Y

M ......................................... N
(X) --->------P'---P----<--(Y)
Distance between P and P' is 10 miles. But we don't know how much time was taken in each case.

Using both,
Speed of Y is 24 mph in second case. So distance it will cover in second case is 24/54 of the total distance

(24/54)D - (2/5)D = 10 (= P'N - PN)

[(120 - 108)/54*5 ] * D = 10

D = 225

Since Y travels (2/5)th of this distance, Y travels 90 miles.

Answer (C)
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