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Math Expert V
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Two friends, X and Y start from two points M and N and move towards  [#permalink]

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18 00:00

Difficulty:   75% (hard)

Question Stats: 54% (02:13) correct 46% (02:11) wrong based on 248 sessions

### HideShow timer Statistics 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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Two friends, X and Y start from two points M and N and move towards  [#permalink]

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2
3
Bunuel wrote:
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

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##### General Discussion
Intern  B
Joined: 30 Mar 2017
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Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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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.
Manager  B
Joined: 11 Feb 2017
Posts: 187
Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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GMATinsight wrote:
Bunuel wrote:
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

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??

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Joined: 10 Apr 2015
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Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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2
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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Posts: 187
Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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shekyonline wrote:
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.

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

How?
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Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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GMATinsight wrote:
Bunuel wrote:
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

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

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Two friends, X and Y start from two points M and N and move towards  [#permalink]

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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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Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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Bunuel wrote:
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?]
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Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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1
Shiprasingh1100 wrote:
Bunuel wrote:
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?]

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

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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Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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Shiprasingh1100 wrote:
Bunuel wrote:
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?]

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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Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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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.
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Two friends, X and Y start from two points M and N and move towards  [#permalink]

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Bunuel wrote:
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?

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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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Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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gmatcracker2018 wrote:
GMATinsight wrote:
Bunuel wrote:
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

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

I think i have got it

distance = speed * time, NOT speed * distance

thnaks
Math Expert V
Joined: 02 Aug 2009
Posts: 7764
Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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nahid78 wrote:
Bunuel wrote:
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?

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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Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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Bunuel wrote:
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.

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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Re: Two friends, X and Y start from two points M and N and move towards  [#permalink]

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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.
Attachments IMG_20190711_122212__01.jpg [ 2.15 MiB | Viewed 120 times ] Re: Two friends, X and Y start from two points M and N and move towards   [#permalink] 10 Jul 2019, 23:56
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