While on a straight road, car X and car Y are traveling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y?
(1) Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour
(2) Three minutes ago car X was 1/2 mile ahead of car X.
Answer is D
Here is my explanation:
Let's say now is at time t = T mins
Cax X's speed = S1
Car Y's speed = S2
Distance travelled by X = d1
Distance travelled by Y = d2
We know, at t = T mins d1 = d2 + 1
We want to find after how many mins d1 = d2 + 2
Cax X's speed = 50 miles/hr
Car Y's speed = 40 miles/hr
Therefore at t = T mins Car X travels T5/6 miles and Car Y travels T4/6 miles.
we know d1 = d2 + 1 at time t = Tmins
Hence, T5/6 = T4/6 + 1 this give T = 6mins
Now we need to find t = M mins when d1 = d2 + 2
Therefore at t = M mins Car X travels M5/6 miles and Car Y travels M4/6 miles.
we know d1 = d2 + 2 at t = Mmins
Hence M5/6 = M4/6 + 2 this give M = 12 so after 6 min Car X will be 2 miles ahead of Car Y
Condition 1 is sufficient.
at time t = T mins we know d1 = d2 + 1
Car X will travel d1 = TS1/60 and Car Y will travel d2 = d1 + 1
Therefore, d2 = TS1/60 + 1 ---I
From condition 2 we know,
at time t = T-3 min d1 = d2 + 1/2
at time t= T-3
Car X will travel d1 = (T-3)*S1/60 and Car Y will travel d2 = d1 + 1/2
Therefore, d2 = (T-3)*S1/60 + 1/2 ---II
From I and II we can find out S1 = 10 miles/hr
We need find t = T + x when d1 = d2 + 2
at time t= T+x
Car X will travel d1 = (T+x)*S1/60 and Car Y will travel d2 = d1 + 2
Therefore, d2 = (T+x)*S1/60 + 2 ---III
From II and III we can find x and hence sufficient.
Answer is D
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