Speed

[Mo2men]

In a race, Dan gave Mary a head start of 'x' m and took 20 sec to get ahead of Mary by 'x' m. Given that the ratio of speeds of Dan to Mary is 5:4 and Mary finished the race in 1 min. What fraction of race had Dan completed when he overtook Mary ?

a. 1/6

b. 5/24

c. 1/4

d. 7/24

e. 11/24

OA: B

[elias.latour.apex]

Just assign a number for x because it doesn't matter. I chose 10 meters, as it was a convenient number. Since Dan covered the 10 meters he was behind and got ahead by 10 meters in 20 seconds, he is traveling at 1 m/s faster than Mary is. Their ratio of speeds is 5:4 so Mary goes 4 m/s and Dan goes 5 m/s. If Mary finished the race in 60 seconds, then the total race is 240 meters. So when Dan overtook Mary he had been traveling for 10 seconds at 5 m/s so he had covered 50 meters, which is 50/240 of the race.

That reduces to 5/24. Answer choice (B).

[GMATGuruNY]

In a race, Dan gave Mary a head start of 'x' m and took 20 sec to get ahead of Mary by 'x' m. Given that the ratio of speeds of Dan to Mary is 5:4 and Mary finished the race in 1 min. What fraction of race had Dan completed when he overtook Mary ?

a. 1/6

b. 5/24

c. 1/4

d. 7/24

e. 11/24

Let Dan's rate = 5 meters per second and Mary's rate = 4 meters per second.
Since Mary finishes the race in 60 seconds, the total distance = rt = 4*60 = 240 meters.

Mary is given a head start of x meters.
Since Dan takes 20 SECONDS to catch up to Mary by x meters and then travel ahead another x meters -- for a total of 2x meters -- Dan must take 10 SECONDS to catch up to Mary by x meters and overtake her.
Since Dan's rate = 5 meters per second, the distance traveled by Dan in 10 seconds = rt = 5*10 = 50 meters.
Implication:
After traveling 10 seconds, Dan overtakes Mary at the 50-meter mark.
Resulting fraction:
(distance traveled by Dan when he overtakes Mary)/(total distance) = 50/240 = 5/24.

The correct answer is B.

[Mo2men]

In a race, Dan gave Mary a head start of 'x' m and took 20 sec to get ahead of Mary by 'x' m. Given that the ratio of speeds of Dan to Mary is 5:4 and Mary finished the race in 1 min. What fraction of race had Dan completed when he overtook Mary ?

a. 1/6

b. 5/24

c. 1/4

d. 7/24

e. 11/24

Let Dan's rate = 5 meters per second and Mary's rate = 4 meters per second.
Since Mary finishes the race in 60 seconds, the total distance = rt = 4*60 = 240 meters.

Dan takes 20 SECONDS to accomplish the following:
-- travel Mary's head start of x meters and overtake her
-- travel x meters ahead of Mary
Since the two distances in blue are EQUAL, it must take Dan 10 SECONDS to travel the first x meters and overtake Mary and then ANOTHER 10 SECONDS to travel x meters ahead of Mary.
Thus:
In 10 seconds, Dan overtakes Mary.

Since Dan's rate = 5 meters per second, the distance traveled by Dan in 10 seconds = rt = 5*10 = 50 meters.
Resulting fraction:
(distance traveled by Dan when he overtakes Mary)/(total distance) = 50/240 = 5/24.

The correct answer is B.

Dear Mitch,

I have stumbled in one issue in the above solution. From Mary's time of finishing the race, you got the total race 240 m but she was already given head start by x. Should not the total race (240 + x).

thanks in advance

[GMATGuruNY]

Dear Mitch,

I have stumbled in one issue in the above solution. From Mary's time of finishing the race, you got the total race 240 m but she was already given head start by x. Should not the total race (240 + x).

thanks in advance

In my post above, Dan takes 10 seconds to catch up to Mary.
Since Dan's rate = 5 meters per second, the distance traveled by Dan in 10 seconds = rt = 5*10 = 50 meters.
Implication:
Dan overtakes Mary at the 50-meter mark.
Put another way:
When Dan overtakes Mary, BOTH runners have traveled 50 meters of the 240-meter race.
Thus, Mary must travel another 190 meters to complete the 240-meter race.

[Mo2men]

Dear Mitch,

I have stumbled in one issue in the above solution. From Mary's time of finishing the race, you got the total race 240 m but she was already given head start by x. Should not the total race (240 + x).

thanks in advance

In my post above, Dan takes 10 seconds to travel Mary's head start of x meters.
Since Dan's rate = 5 meters per second, the distance traveled by Dan in 10 seconds = rt = 5*10 = 50 meters.
Implication:
Dan gives Mary a head start of x=50 meters.
In other words:
When Dan starts to run, Mary has already traveled 50 meters of the 240-meter race.
She must travel another 190 meters to complete the 240-meter race.

Thanks Mitch for you response.

According to you answer, I may have real problem with meaning "head start". I understand it as follows: every one of them is stationed in his/her position with distance X, then the race is triggered but the meaning seems something another. Does it mean that: Dan let Mary stars till distance x then he starts to run???

I'm in big confusion.

Thanks

[GMATGuruNY]

the meaning seems something another. Does it mean that: Dan let Mary stars till distance x then he starts to run???

Your understanding seems correct.
To give a person a head start = to allow that person to start EARLY.

In my solution, Dan takes 10 seconds to catch-up to Mary, with the result that Dan overtakes Mary at the 50-meter mark.
Since Mary's rate = 4 meters per second, the time for Mary to reach the 50-meter mark = d/r = 50/4 = 12.5 seconds.
Implication:
Since Mary takes 12.5 seconds to reach the 50-meter mark, whereas Dan takes only 10 seconds, Mary must begin the race 2.5 SECONDS EARLY.

Since Mary's rate = 4 meters per second, the distance traveled by Mary in 2.5 seconds = rt = (4)(2.5) = 10 meters.
Thus, Mary is given a head start of x=10 meters.

[GMATGuruNY]

Alternate approach:

Time and rate have a RECIPROCAL RELATIONSHIP:
If Amy's rate is 2 times Bob's rate, then Amy's time is 1/2 Bob's time.
If Amy's rate is 3 times Bob's rate, then Amy's time is 1/3 Bob's time.

Since the rate ratio for Dan and Mary = 5:4, Dan's rate is 5/4 Mary's rate.
Implication:
Dan's total time to complete the race will be 4/5 Mary's total time of 60 seconds:
(4/5)(60) = 48 seconds.
As noted in my earlier posts, Dan takes 10 seconds to catch-up to Mary and overtake her.
Thus:
(Dan's time to overtake Mary)/(Dan's total time) = 10/48 = 5/24.
Since Dan takes 5/24 of his total time to overtake Mary, he must travel 5/4 of the total distance to overtake her.

The correct answer is B.

[Mo2men]

the meaning seems something another. Does it mean that: Dan let Mary stars till distance x then he starts to run???

Your understanding seems correct.
To give a person a head start = to allow that person to start EARLY.

In my solution, Dan takes 10 seconds to catch-up to Mary, with the result that Dan overtakes Mary at the 50-meter mark.
Since Mary's rate = 4 meters per second, the time for Mary to reach the 50-meter mark = d/r = 50/4 = 12.5 seconds.
Implication:
Since Mary takes 12.5 seconds to reach the 50-meter mark, whereas Dan takes only 10 seconds, Mary must begin the race 2.5 SECONDS EARLY.

Since Mary's rate = 4 meters per second, the distance traveled by Mary in 2.5 seconds = rt = (4)(2.5) = 10 meters.
Thus, Mary is given a head start of x=10 meters.

Thanks Mitch for your excellent explanation
I have question.

Let Dan speed =5 m/s & Mary = 4 m/s

Dan takes 20 SECONDS to accomplish the following:
-- travel Mary's head start of x meters and overtake her
-- travel x meters ahead of Mary
Then
2X = (5) (20)..............then x =50 meter. However, you got x =10 meter.

Where does the discrepancy comes from?

Thanks

[GMATGuruNY]

Thanks Mitch for your excellent explanation
I have question.

Let Dan speed =5 m/s & Mary = 4 m/s

Dan takes 20 SECONDS to accomplish the following:
-- travel Mary's head start of x meters and overtake her
-- travel x meters ahead of Mary
Then
2X = (5)(20)..............then x =50 meter. However, you got x =10 meter.

Where does the discrepancy comes from?

Thanks

The value in red is incorrect.

x = Mary's head start.
When Dan starts to run, he is x meters behind Mary.
As Dan races to close the x-meter gap between him and Mary, Mary CONTINUES TO MOVE FORWARD.
Implication:
To close the x-meter gap and overtake Mary in 10 seconds, Dan must travel in 10 seconds X MORE METERS THAN MARY TRAVELS in 10 seconds.
By extension:
To get x meters ahead of Mary in the following 10 seconds, Dan must travel in the following 10 seconds X MORE METERS THAN MARY TRAVELS in the following 10 seconds.
Thus:
To close the x-meter gap and get ahead another x meters in a total of 20 seconds, Dan must travel in 20 seconds 2X MORE METERS THAN MARY TRAVELS in 20 seconds.

Since Dan's distance in 20 seconds (5*20) must be 2x more meters than Mary's distance in 20 seconds (4*20), we get:
5*20 = (4*20) + 2x
100 = 80 + 2x
20 = 2x
x = 10.