Problem Solving

Jane gives Karen a 5m head start in a 100 m race. Jane is beaten by Karen by 0.25 m.
In how many meters more would Jane have overtaken Karen?

The explanation is provided in the toughest 100 quant problems resource on this site, but its hard to follow.

Can someone pls help with not just the numerical solution, but also with some of the reasoning behind how you got there? Thanks!

Is the answer 5 more meters?

Hi Kancz:

This is how I did it.

First we need to set up the problem.
Let's look at the given data.
1. The total distance for the race is 100m.
2. Jane gives Karen a 5m head start.Also, Karen beats Jane. This means that Karen had run 100m-5m = 95m at the end of the race.
3. Jane is beaten by Karen by 0.25m. Therefore Jane ran 99.75m.

Now, here's the reasoning.
1. Relative distance between them at the start = 5m. Relative distance at the end = 4.75m.
2. They both run at a constant speed. Therefore, their relative speed must be constant as well.

Now we know that, in the time that Karen ran 95m, their relative distance decreased by 4.75m.
Also, since the relative speed, and Karen's own speed must be constants in their own right, The distances run in the same time (because distance = speed * time) must be proportional.

Now, for 4.75m of relative distance, Karen runs = 95m, therefore
for each 1m of relative distance, Karen runs = 95/4.75 = 20m.
Therefore for 0.25m of relative distance left at the end, Karen runs = 0.25*20m = 5m.

Now let's see Jane's perspective:
for 4.75m of relative distance, Jane runs = 100-0.25 = 99.75m
for 1m of relative distance; Jane runs = 99.75m/4.75 = 21m
Therefore, for 0.25m of relative left, Jane runs = .25*21m = 5.25m. (Which makes sense, since Jane has to cover Jane's 5m and the extra 0.25m left).

Clearly, if the race was 99.75+5.25 = 105m long or longer, Jane would have overtaken Karen.
The language of the question is a little ambiguous.
So, from Jane's perspective, she has to run 5.25m more to overtake Karen.
From Karen's perspective, she has to run 5m more before she is overtaken.

Let me know if this helps

Kancz44 wrote:Jane gives Karen a 5m head start in a 100 m race. Jane is beaten by Karen by 0.25 m.
In how many meters more would Jane have overtaken Karen?

The explanation is provided in the toughest 100 quant problems resource on this site, but its hard to follow.

Can someone pls help with not just the numerical solution, but also with some of the reasoning behind how you got there? Thanks!

Let the time for the race = 1 hour.

Since Jane loses by .25 meters, she travels 99.75 meters, implying that her rate = 99.75 meters per hour.
Since Karen gets a 5-meter head start, she travels 95 meters, implying that her rate = 95 meters per hour.

Jane's rate - Karen's rate = 99.75 - 95 = 4.75 meters per hour.
This is the rate at which Jane will catch up to Karen.

Since Jane must catch up by .25 meters, the time needed = d/r = .25/4.75 = 1/19 of an hour.
In 1/19 of an hour, the additional distance traveled by Karen = r*t = 95(1/19) = 5 meters.
Since Jane was .25 meters behind, the additional distance traveled by Jane = 5+25 = 5.25 meters.