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rate problem

by The Jock » Tue Oct 27, 2009 10:30 pm
Jane gave Karen a 5 m head start in a 100 m race and 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

Please let me knoe the answer.

Thanks,
Varun
https://varunsaini06.blocked/
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Re: rate problem

by regor60 » Wed Oct 28, 2009 11:16 am
The Jock wrote:Jane gave Karen a 5 m head start in a 100 m race and 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

Please let me knoe the answer.

Thanks,
Varun
https://varunsaini06.blocked/

D1, R1, T1 = Karen
D2,R2,T2= jane

From race start to end, a fixed time, Karen travels 95m, Jane 99.75, so equating times....

D1/R1=D2/R2, so R2/R1=99.75/95

When Jane catches Karen, she will have traveled .25 m farther than Karen in the same amount of time
So equating times again...

D1/R1=(D1+.25)/R2

Substitute R2/R1 from above yields (D1+.25)/D1=99.75/95 > solve for D1 = 5
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Re: rate problem

by chipbmk » Wed Oct 28, 2009 1:48 pm
The Jock wrote:Jane gave Karen a 5 m head start in a 100 m race and 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

Please let me knoe the answer.

Thanks,
Varun
https://varunsaini06.blocked/
I got the answer more in a different way from the above poster.

I figured that Jane managed to make up 4.75m of the headstart after running 100m (take the 5m headstart minus the .25m remaining at the end of the race).

From there, I took 4.75/100 (the distance she made up divided by the distance it took her to make up that distance) to get how many meters she was gaining PER meter run. This equals .0475 meters.

Now you just need to know how many meters would it take before she reaches .25m. You can set that up like this.

.0475x=.25
x= .25/.0475
x=5.26
So, in 5.26m more she would have caught Karen, meaning she needs to run any distance further than this to overtake her.

Answer is E I think. I believe she would equal Karen at answer D and overtake her at E. (I may be over thinking this and it could be D)

Can you please post the OA?
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by NikolayZ » Wed Oct 28, 2009 2:33 pm
Hey guys !
Let the speed of Karen and Jane be K and J respectively.
After the ahead start
Jane ran 100-0.25=99,75
Karen ran 100-5=95
so, let's assume that X is the time when Jane will overtake Karen
97,5/J=95/K
and
K*X-J*X=0,25. (from the stem)
Solving this one for J*X we get JX=21/4=5.25
Hence D.
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by chipbmk » Wed Oct 28, 2009 4:36 pm
NikolayZ wrote:Hey guys !
Let the speed of Karen and Jane be K and J respectively.
After the ahead start
Jane ran 100-0.25=99,75
Karen ran 100-5=95
so, let's assume that X is the time when Jane will overtake Karen
97,5/J=95/K
and
K*X-J*X=0,25. (from the stem)
Solving this one for J*X we get JX=21/4=5.25
Hence D.
That is basically the same response I got, but doesnt 5.25 represent when she will be exactly EVEN with her? Maybe I am reading too much into this question.
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by NikolayZ » Wed Oct 28, 2009 4:57 pm
Oh yeah. I thought that it would be illogical, as you did.
Indeed, in 5.25 meter more than Jane ran, she will not overtake Karen, but they will arrive at the same time.
Though, i think that the answer choices would be different in that case : like 5.26 or 5.205 =)
6 is too much for Jane to overtake Karen.
So, consistently, 5.25 is the only answer that fits the "logic" test =)
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by regor60 » Thu Oct 29, 2009 12:02 pm
My answer of D1=5 of course means Jane's additional distance, but Karen will need to travel an extra .25m, so the answer is indeed 5.25 as the question is posed
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