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Two friends, X and Y start from two points M and N and move towards
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06 Jul 2017, 02:33
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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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Two friends, X and Y start from two points M and N and move towards
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06 Jul 2017, 03:14
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, respectivelyThe 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 Pi.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 statementsSpeed 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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Re: Two friends, X and Y start from two points M and N and move towards
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07 Jul 2017, 03:35
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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Re: Two friends, X and Y start from two points M and N and move towards
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18 Jul 2017, 10:16
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, respectivelyThe 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 Pi.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 statementsSpeed 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 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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Re: Two friends, X and Y start from two points M and N and move towards
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19 Jul 2017, 00:31
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 CHope it clears doubt.
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Re: Two friends, X and Y start from two points M and N and move towards
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19 Jul 2017, 01:10
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
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19 Jul 2017, 01:39
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, respectivelyThe 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 Pi.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 statementsSpeed 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 ...



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Two friends, X and Y start from two points M and N and move towards
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19 Jul 2017, 02:30
rocko911Distance 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 = 3/5ddistance 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
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04 Oct 2017, 23:49
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
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05 Oct 2017, 17:03
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?] Shiprasingh1100I 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. Answer: 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
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06 Oct 2017, 22:44
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
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09 Oct 2017, 01:57
BunuelIn 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
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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 (1.21) = 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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Re: Two friends, X and Y start from two points M and N and move towards
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04 Apr 2018, 01:14
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, respectivelyThe 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 Pi.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 statementsSpeed 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 ... I think i have got it distance = speed * time, NOT speed * distance thnaks



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Re: Two friends, X and Y start from two points M and N and move towards
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06 Apr 2018, 23:34
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 (1.21) = 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? 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
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11 Apr 2019, 08:31
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
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10 Jul 2019, 23:56
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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Re: Two friends, X and Y start from two points M and N and move towards
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