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neoreaves
Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

We are asked: In how many meters more would Jane have overtaken Karen? So I think we are asked to calculate the distance Jane should cover to overtake Karen but all solutions above are calculating the distance Karen should cover. Can you please post answer choices to clear my doubts.

Anyway below is my solution:

Jane covered \(100m-0.25m=99.75m\) and Karen covered \(100-5=95m\) (in the same time interval).

Initial distance between them was \(5m\) and final distance between Jane and Karen was \(0.25m\). Thus Jane gained \(99.75m-95m=4.75m\) over Karen in \(99.75m\), hence Jane is gaining \(1m\) over Karen in every \(\frac{99.75}{4.75}=21m\).

Hence, Jane in order to gain remaining \(0.25m\) over Karen should cover \(21*0.25=5.25m\).

Answer: 5.25.

That is an excellent explanation
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ratio of distance covered = ratio of their speed in time t

hence \(\frac{95}{99.75} = \frac{vK}{vJ}\) --1

now Jane is 0.25 meter behind, suppose it overtakes him at x from 100m mark.

again using
ratio of distance covered = ratio of their speed in time t

\(\frac{(x+0.25)}{x} = \frac{vJ}{vK}\)---2

multiply equation 1 and 2
\(\frac{95}{99.75} = \frac{x}{(x+0.25)}\) => x =5

hence Jane needs to cover x+0.25 = 5.25
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Bunuel,

art thou a kind of math GOD?

how can I raise unto this magnitude?
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neoreaves
Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

We are asked: In how many meters more would Jane have overtaken Karen? So I think we are asked to calculate the distance Jane should cover to overtake Karen but all solutions above are calculating the distance Karen should cover. Can you please post answer choices to clear my doubts.

Anyway below is my solution:

Jane covered \(100m-0.25m=99.75m\) and Karen covered \(100-5=95m\) (in the same time interval).

Initial distance between them was \(5m\) and final distance between Jane and Karen was \(0.25m\). Thus Jane gained \(99.75m-95m=4.75m\) over Karen in \(99.75m\), hence Jane is gaining \(1m\) over Karen in every \(\frac{99.75}{4.75}=21m\).

Hence, Jane in order to gain remaining \(0.25m\) over Karen should cover \(21*0.25=5.25m\).

Answer: 5.25.

Excellent explanation!!! Much simpler than all the others i've read.

Thanks Bunuel
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1. The initial gap between Jane and Karen was 5m. It was reduced to 0.25 m as Jane covers 99.75 m. Jane gains 4.75m.
2. So the gap of 0.25m would be reduced to 0 or Jane would gain 0.25m, if Jane further covers (99.75/4.75) * 0.25 = 5.25 m.
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my approach is-

Jane covered 4.75m when she ran 99.75m
So for 1m gain 99.75/4.75 = 21
and so .25 meter gain .25*21 = 5.25
D
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How do you calculate 99.75/4.75 so easily ? I solved the problem quite quickly but got completely stuck in that calculation... Do you have a more straightforward way to solve that than doing the division as in primary school ?
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How do you calculate 99.75/4.75 so easily ? I solved the problem quite quickly but got completely stuck in that calculation... Do you have a more straightforward way to solve that than doing the division as in primary school ?

Even if you actually had 99.75/4.75, calculating this is also quite simple.

99.75/4.75 = 9975/475 (get rid of the decimals by multiplying and dividing by 100)

Now, you need to find common factors. Any number ending in 25/50/75/00 will be divisible by 25. Now the question is 9975 is which multiple of 25. Note that if 9975 had another 25, it would have been 10,000 and that is 25*400. So 9975 must be 25*399.
Similarly if 475 had another 25, it would have been 500 and 25*20 = 500. So 25*19 = 475

We get 9975/475 = 399/19 = 21
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neoreaves
Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

A 5m
B. 7m
C. 4.5m
D. 5.25m
E. 6m

Find my answer in the attached image.
Hope this helps.

The trick is to remember to add 0.25m after calculating the remaining distance.
Attachment:
Screenshot 2021-06-30 at 2.22.33 PM.png
Screenshot 2021-06-30 at 2.22.33 PM.png [ 1.85 MiB | Viewed 7427 times ]
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