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Car and time

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Car and time

by Viper83 » Sun Jun 19, 2011 2:43 am
how much time did it take a certain car to travel 400 km?

1. the car traveled the first 200 km in 2.5 hrs
2. if the car's average speed had been 20 km/hr greater than it was, it would have traveled the 400 km in 1 hr less than it did.

From the prompt we get 400=rt

Is this setup correct?
1. is obviously insufficient because we dont know anything about the additional 200 km.
2. we have 400=rt from the prompt. From statement 2 we get 400=(r+20)(t-1). Now we have 2 equations and can solve for t. Is this correct? Answer is B
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by amit2k9 » Sun Jun 19, 2011 2:52 am
in b

400/(s) - 400/(s+20) = 1
so you get s(s+20) = 8000
where s can be = 20 or so. Hence sufficient.

in your case
st = (s+20)(t-1) = st-s+20t-20
so, s+20 = 20t

as st=400 hence t=400/s
so substituting for s will give t too.
sufficient.
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by Anurag@Gurome » Sun Jun 19, 2011 3:03 am
Viper83 wrote:Is this setup correct?
1. is obviously insufficient because we dont know anything about the additional 200 km.
2. we have 400=rt from the prompt. From statement 2 we get 400=(r+20)(t-1). Now we have 2 equations and can solve for t. Is this correct?
Yes, it's correct but careless.
There is more to solving the equations than just saying "we have two equations in two unknown, hence we can solve..". That statement only holds strong when the equations are linear and independent of each other.

In this case, rt = 400 ........ (1)
And, (r + 20)(t - 1) = 400 .... (2)

Now replacing 400 in (2) by rt from (1), we have
  • --> (r + 20)(t - 1) = rt
    --> (rt + 20t - r - 20) = rt
    --> (20t - r - 20) = 0 ......... (3)
Now, replacing r = 400/t from (1) in (3),
  • --> (20t - 400/r - 20) = 0
    --> (20t² - 20t - 400) = 0
    --> (t² - t - 20) = 0 .......... (4)
Solving (4) we get, t = 5 or t = -4

Now in this case t = -4 is not possible, hence t = 5.
BUT what if the equation gave two possible solution for t? In that case statement 2 wouldn't had been sufficient.
Last edited by Anurag@Gurome on Tue Jun 21, 2011 10:47 pm, edited 1 time in total.
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by Viper83 » Sun Jun 19, 2011 3:08 am
Thanks Anurag, so in this kind of question one should actually solve to get the answer since it is a quadratic equation?
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by Anurag@Gurome » Sun Jun 19, 2011 3:11 am
Viper83 wrote:Thanks Anurag, so in this kind of question one should actually solve to get the answer since it is a quadratic equation?
In my opinion, yes.
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by anuragvaid » Tue Jun 21, 2011 10:35 pm
Anurag_Gurome,

Thanks for the fundoo explanation and the absolutely correct way to approach these quesns...

Just a quickie - I think the correct value of T will be 5 not 4. Thats cos, T = 5 or T=-4 will be the roots here.....

Anurag
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by anuragvaid » Wed Jun 22, 2011 7:30 am
Anurag,

Its my mistake.....U had indeed correctly mentioned the solution to be 5...dunno, Y I read 4....

Nevertheless, my two cents !!!

Anurag
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by Anurag@Gurome » Wed Jun 22, 2011 8:51 am
anuragvaid wrote:Anurag,

Its my mistake.....U had indeed correctly mentioned the solution to be 5...dunno, Y I read 4....
No no. Not yours.
You're correct. I edited that after seeing your post. I thought I had posted a message acknowledging that! :(
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