Two friends, X and Y start from two points M and N and move towards

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

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?

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 ...

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.

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:


Answer:

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.

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.

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?

I think i have got it

distance = speed * time, NOT speed * distance

thnaks

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

Hi Bunuel.

I thought it is E because it didn't state that X and Y started at the same time. Am I exaggerating?

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.

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

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..


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.

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)