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Director
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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was
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12 Feb 2015, 09:14
gurpreetsingh wrote: 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 Very elegant solution !! doable in 2minutes . +1 Kudos thank you .
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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was
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23 Feb 2015, 01:09
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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Jane gave Karen a 5 m head start in a 100 race and Jane was
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23 Feb 2015, 01:45
Lapetiteflo wrote: 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 ? Your are missing something , it should be X=95*0.25/4.75 and you do not need to solve all fraction numbers . 0.25 = 1/4 4.75 = 4 3/4 95*1/4* 4/19 = 5. so 5+0.25 = 5.25 ans.
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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was
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23 Feb 2015, 02:24
Lapetiteflo wrote: 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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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was
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05 Mar 2015, 07:44
I solved it using easy unitary method and got it right..



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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was
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09 Apr 2015, 14:33
This is what I did, but need to admit it took a while. So i first start to find teh gain of Jane when Karen passes one meter, so I get 99.75/95=1.05. this means that for every meter that Karen goes Jane will have traveled 1.05meter. so now is streight foreward. originaly JAne needs to pass 5 meters to overtake KAren, so JAne will need 5*1.05=5.25 meters.



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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was
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22 May 2015, 08:15
Bunuel wrote: neoreaves wrote: 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 \(100m0.25m=99.75m\) and Karen covered \(1005=95m\) (in the same time interval). Initial distance between them was \(5m\) and final distance between Jane and Karen was \(0.25m\). Thus Jane gained \(99.75m95m=4.75m\) 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. Bunuel, how would someone go about calculating that 99.75/4.75 = 21 on the test with everything else in 2 min? Is there an easy way to do these decimal divisions?



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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was
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22 May 2015, 08:21
torontoclub15 wrote: Bunuel wrote: neoreaves wrote: 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 \(100m0.25m=99.75m\) and Karen covered \(1005=95m\) (in the same time interval). Initial distance between them was \(5m\) and final distance between Jane and Karen was \(0.25m\). Thus Jane gained \(99.75m95m=4.75m\) 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. Bunuel, how would someone go about calculating that 99.75/4.75 = 21 on the test with everything else in 2 min? Is there an easy way to do these decimal divisions? Please read the whole thread: janegavekarena5mheadstartina100raceandjanewas9345120.html#p1488810 Chances are that your questions has already been addressed.
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Jane gave Karen a 5 m head start in a 100 race and Jane was
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Updated on: 28 Apr 2018, 08:28
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
let x=meters J would have needed to overtake K in running 99.75m, J gains 5.25=4.75m on K 99.75/4.75=.25/x x=5.25m D
J
Originally posted by gracie on 14 Feb 2016, 12:39.
Last edited by gracie on 28 Apr 2018, 08:28, edited 1 time in total.



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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was
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28 Apr 2018, 07:07
VeritasPrepKarishma wrote: 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 Hi Karishma, Can you please check where is my understanding going wrong. I got 5 as my answer. And here is how I arrived to this number Let SJ > speed of jane SK> speed of Karen Now, since the time in which karen covers 95m is equal to time in which jane covers 99.75m, we get, 95/sk = 99.75/sj > I Also, let x be the distance from the 100m finish line in which jane overtakes karen. So, (100+x)/sj = (95+x)/sk >II solving I & II, we get, x = 5, which is the desired answer



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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was
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28 Jun 2018, 07:29
torontoclub15 wrote: Bunuel wrote: neoreaves wrote: 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 \(100m0.25m=99.75m\) and Karen covered \(1005=95m\) (in the same time interval). Initial distance between them was \(5m\) and final distance between Jane and Karen was \(0.25m\). Thus Jane gained \(99.75m95m=4.75m\) 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. Bunuel, how would someone go about calculating that 99.75/4.75 = 21 on the test with everything else in 2 min? Is there an easy way to do these decimal divisions? One easy way which i did : we can calculate 4.75x20=95 easily. Then just need to add 4.75 to get 99.75. So it is 4.75x21=99.75




Re: Jane gave Karen a 5 m head start in a 100 race and Jane was &nbs
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