sidship21 wrote:
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 chetan2uGMATinsight - I believe your calculations are incorrect.
Thank you
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?
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=90C
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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