OA [spoiler]25%[/spoiler]

let the total distance = 120
speed of j = 30 , K = 60
j stops when k trevelled 25% distance means 30, it means oth J and K travelled half an hour before J stops

J travelled 15 , hence K travelled 105 ----------- 1

now the case when j would not have been stopped, let they meet at x distance from J

x/30 = 120-x/60, x= 40, it means K travelled 120-40 = 80

now time differance 105/60 - 80/60 = .4
.4/1.75(105/60)= approx 25%

Not clear at all!!

vipulgoyal wrote:OA [spoiler]25%[/spoiler]

let the total distance = 120
speed of j = 30 , K = 60
j stops when k trevelled 25% distance means 30, it means oth J and K travelled half an hour before J stops

J travelled 15 , hence K travelled 105 ----------- 1

now the case when j would not have been stopped, let they meet at x distance from J

x/30 = 120-x/60, x= 40, it means K travelled 120-40 = 80

now time differance 105/60 - 80/60 = .4
.4/1.75(105/60)= approx 25%

vipulgoyal wrote:OA [spoiler]25%[/spoiler]

let the total distance = 120
speed of j = 30 , K = 60
j stops when k trevelled 25% distance means 30, it means oth J and K travelled half an hour before J stops

J travelled 15 , hence K travelled 105 ----------- 1

now the case when j would not have been stopped, let they meet at x distance from J

x/30 = 120-x/60, x= 40, it means K travelled 120-40 = 80

now time differance 105/60 - 80/60 = .4
.4/1.75(105/60)= approx 25%

Hi!

Picking numbers is a great way to approach this question. However, you got a bit lost partway through your solution.

Let's go back a few steps!

First, let's state our task: We need to solve for the % greater Karen runs with gimpy John than with healthy John. In other words, percent change. So, we're solving for:

(Karen longer time - Karen shorter time)/(Karen shorter time) * 100%

We're told that Karen runs 50% faster than John - but 60 isn't 50% more than 30, it's 100% more (i.e. your numbers didn't actually match the info in the Q stem). So, let's set John's rate at 20 and Karen's rate at 30.

Now that we've chosen rates, let's pick a distance that works well with those rates. We know that John is going to stop after 25% of the distance, so let's pick a number that, when divided by 4, will be easily divisible by 20. 80 looks like a good fit.

Now that we finally have our numbers (spending a bit of extra time picking manageable numbers almost always saves you even more down in the long run), let's solve!

Total distance = 80
John only runs 25%, which is 20. At 20mph, John runs for 1 hour.
Karen runs the remaining 60. At 30mph, Karen runs for 2 hours. Hey, numbers are working out nicely so far!

Now we need to figure out how long Karen would have run if John hadn't pulled up lame. When solving for time of two objects moving TOWARD each other, we ADD their rates:

total time = (total distance)/(sum of rates)
t = (80)/(50) = 8/5

So, if John hadn't stopped, each would have run for 8/5 hours.

Finally, we plug into the originally noted formula:

changing 2hours to 10/5 hours, we get:

(10/5 - 8/5)/(8/5)
= (2/5)/(8/5)
= 2/8
= 25%

Pick (A)!

One important note: the question doesn't say approximately what percent, so the answer will be exact. Since you didn't end up with an exact match for one of the choices, you should have known that there was a mistake somewhere in your methodology.

Thanks Stuart for simplifying it!

Stuart Kovinsky wrote:

vipulgoyal wrote:OA [spoiler]25%[/spoiler]

let the total distance = 120
speed of j = 30 , K = 60
j stops when k trevelled 25% distance means 30, it means oth J and K travelled half an hour before J stops

J travelled 15 , hence K travelled 105 ----------- 1

now the case when j would not have been stopped, let they meet at x distance from J

x/30 = 120-x/60, x= 40, it means K travelled 120-40 = 80

now time differance 105/60 - 80/60 = .4
.4/1.75(105/60)= approx 25%

Hi!

Picking numbers is a great way to approach this question. However, you got a bit lost partway through your solution.

Let's go back a few steps!

First, let's state our task: We need to solve for the % greater Karen runs with gimpy John than with healthy John. In other words, percent change. So, we're solving for:

(Karen longer time - Karen shorter time)/(Karen shorter time) * 100%

We're told that Karen runs 50% faster than John - but 60 isn't 50% more than 30, it's 100% more (i.e. your numbers didn't actually match the info in the Q stem). So, let's set John's rate at 20 and Karen's rate at 30.

Now that we've chosen rates, let's pick a distance that works well with those rates. We know that John is going to stop after 25% of the distance, so let's pick a number that, when divided by 4, will be easily divisible by 20. 80 looks like a good fit.

Now that we finally have our numbers (spending a bit of extra time picking manageable numbers almost always saves you even more down in the long run), let's solve!

Total distance = 80
John only runs 25%, which is 20. At 20mph, John runs for 1 hour.
Karen runs the remaining 60. At 30mph, Karen runs for 2 hours. Hey, numbers are working out nicely so far!

Now we need to figure out how long Karen would have run if John hadn't pulled up lame. When solving for time of two objects moving TOWARD each other, we ADD their rates:

total time = (total distance)/(sum of rates)
t = (80)/(50) = 8/5

So, if John hadn't stopped, each would have run for 8/5 hours.

Finally, we plug into the originally noted formula:

changing 2hours to 10/5 hours, we get:

(10/5 - 8/5)/(8/5)
= (2/5)/(8/5)
= 2/8
= 25%

Pick (A)!

One important note: the question doesn't say approximately what percent, so the answer will be exact. Since you didn't end up with an exact match for one of the choices, you should have known that there was a mistake somewhere in your methodology.

Since, the question is of %.. so lets solve using assumption

Let Distance = 100
Initial Speed of John = 5
Intital Speed of Karen = 7.5

Time to meet = 100/(5+7.5) = 8

Distance traveled by Karen = 8*7.5 = 60

Distance Karen traveled due to John's cramp = (100 - 25) = 75

So, extra distance ran = 75 - 60 = 15

Percent = 15/60 * 100 = 25%

[spoiler]{A}[/spoiler]

thanks Stuart, Rahul you need have another look on solution, qustion is asking % change in terms of time

vipulgoyal wrote:thanks Stuart, Rahul you need have another look on solution, qustion is asking % change in terms of time

Vipul, it doesn't matter.. as percentage is unit free... and distance varies with time .. Also, the same is clear from my solution's answer

John and Karen begin running at opposite ends of a trail until they meet somewhere in between their starting points. They each run at their respective constant rates until John gets a cramp and stops. If Karen runs 50% faster than John, who is only able to cover 25% of the distance before he stops, what percent longer did Karen run than she would have had John been able to maintain his constant rate until they met?

25%

50%

75%

100%

200%

Karen's TIME will increase in proportion to how much her DISTANCE increases.

Since Karen's rate is 50% faster than John's rate, for every 2 miles that John travels, the distance traveled by Karen = 2 + .5(2) = 3 miles.
Implication:
When Karen and John travel toward each other, of every 5 miles, 2 miles are traveled by John, while 3 miles are traveled by Karen.
Thus, of the total distance between Karen and John, the portion normally traveled by Karen = 3/5 = 60%.
When John gets a cramp, he travels only 25% of the total distance, implying that the portion traveled by Karen = 75%.
Since Karen's portion of the DISTANCE increases from 60% to 75%, her TIME will increase by the same ratio:
Percent increase from 60% to 75% = Difference/Original * 100 = 15/60 * 100 = 25%.

The correct answer is A.