12:00 am, Mary passed a certain gas station on a highway while traveli

Question

12:00 am, Mary passed a certain gas station on a highway while traveling west at a constant speed of r miles per hour. Then, 1 hour later, Paul passed the same gas station while traveling west at a constant speed 3r/2 miles per hour. If both drivers maintained their speeds, how many miles from the station did Paul catch up with Mary?

(A) r/2

(B) r

(C) 2r

(D) 5r/2

(E) 3r

(A) r/2

(B) r

(C) 2r

(D) 5r/2

(E) 3r

ThatDudeKnows · Answer

If we see a variable in the question that is repeated in the answer choices, there's a good chance we can use Plugging In. Let's make r=2. So Mary travels at 2 mph and Paul travels at 3 mph.
At 12:00, Mary passes the station.
At 1:00, Mary is 2 miles down the road and Paul passes the station.
At 2:00, Mary is 4 miles down the road and Paul is 3 miles down the road.
At 3:00, Mary is 6 miles down the road and Paul is 6 miles down the road. Aha, he caught up! At 6 miles.
Check the answer choices.

(A) 2/2 Wrong
(B) 2 Wrong
(C) 2*2 Wrong
(D) (5*2)/2 Wrong
(E) 3*2

Answer choice E

ThatDudeKnowsPluggingIn

FazliddinKD · Answer

S * T = D

Mary = r * t= rt

Paul => 3r/2* t-1= 3r/2*(t-1)

rt=3r/2*(t-1) => 2rt=3rt-3r => rt=3r => t=3

Paul will catch up Mary at distance r*t = 3r Ans. (E)

azamatboden · Answer

Shrinking gap kind of problem here...
relative speed = 3r/2 - r = 1/2 r
in 1 hour Mary does r miles -- so the distance between her and Paul is r miles.

to find time when they catch up: r mile / (r/2) = 2

so Paul will catch up with mary in 2 * (3r/2) = 3 r

youcouldsailaway · Answer

Paul has to catch up to Mary (meet her where she is at), so we know that the distance travelled by both is going to be the same even if their individual time and speed varies.

Let's say Paul travelled the distance in t hours. Since Mary got a head start, she travelled the same distance in t+1 hours.

We know, Distance = Speed * Time.

So, Paul's equation looks like this:  \frac{3r}{2}  * t = d
and Mary's equation looks like this: r * (t+1) = d

Thus, (t+1) * r =  \frac{3r}{2}  * t
On solving we get t = 2

Now, since d = 3r/2 * t for Paul and t = 2, we get that d = 3r.

I was solving this problem quite well until I made one mistake, which was assigning t+1 to Paul instead of to Mary. Seems intuitive now, but basically whoever gets the head start, i.e, whoever starts first, should get the + n hrs / mins.

Paras96 · Answer

12:00 am, Mary passed a certain gas station on a highway while traveling west at a constant speed of r miles per hour. Then, 1 hour later, Paul passed the same gas station while traveling west at a constant speed 3r/2 miles per hour. If both drivers maintained their speeds, how many miles from the station did Paul catch up with Mary?

(A) r/2

(B) r

(C) 2r

(D) 5r/2

(E) 3r

Distance traveled by Mary in 1 hour = Speed of Mary*1 = r*1=r
The relative velocity of Paul w.r.t Mary = 3r/2 - r = r/2

Time taken by Paul to Catch Mary = Distance traveled by Mary in 1 hour/The relative velocity of Paul w.r.t Mary = r/(r/2)=2 hours

Distance travelled by Paul in 2 hours = (3r/2)*2 = 3r

Hence E

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12:00 am, Mary passed a certain gas station on a highway while traveli

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