Jane gave Karen a 5 m head start in a 100 race and Jane was

Question

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

Bunuel · Accepted Answer

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.

KarishmaB · Answer

Since speeds of the two are constant, we can directly use variation here.

Since Jane covers a gap of 4.75m by running 99.75m, she will cover a gap of 0.25 m by running another (99.75) x 0.25/4.75 = 5.25 m.

In races questions, making a diagram can give you a clear picture.

Attachment:

doc1.jpg [ 9.2 KiB | Viewed 25800 times ]

neoreaves · Answer

That is an excellent explanation

gurpreetsingh · Answer

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

mxgms · Answer

Bunuel,

art thou a kind of math GOD?

how can I raise unto this magnitude?

anon321 · Answer

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

Thanks Bunuel

SVaidyaraman · Answer

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.

AMITAGARWAL2 · Answer

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

Lapetiteflo · Answer

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 ?

KarishmaB · Answer

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

arya251294 · Answer

Find my answer in the attached image. Hope this helps. The trick is to remember to add 0.25m after calculating the remaining distance. Screenshot 2021-06-30 at 2.22.33 PM.png

bumpbot · Answer

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Jane gave Karen a 5 m head start in a 100 race and Jane was

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