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# Quant Q Distance Q followup

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Intern
Joined: 09 Jan 2009
Posts: 10

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Schools: harvard
Quant Q Distance Q followup [#permalink]

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28 Jan 2009, 10:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Question: How much time did it take a certain car to travel 400km?
If the car's avg speed had been 20km/hr greater than it was, it would have traveled the 400km in 1hr less time than it did. How is this sentence sufficient?

Answer: let v be the original speed and t be the original time.
(v+20)*(t-1) = 400
vt=400 =>v=400/t
Thus, you can solve for t.

Followup: But the solution t still includes another variable v? So you can not find a finite answer for t, the answer for t will always include the unknow variable v, wont it?

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Senior Manager
Joined: 30 Nov 2008
Posts: 484

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Schools: Fuqua
Re: Quant Q Distance Q followup [#permalink]

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28 Jan 2009, 12:34
Here is another way to elaborate the solution.

Actual scenario -
Let r be the rate, t be the time and d be the distance covered.
Given that distand d = 400.

By Rate forumyla, rt = d ==> rt = 400 -----> Eq 1

Hypothetical scenario -
if rate is 20 more than the actual rate, then new rate will be r + 20.
and time will be 1 hr less the actual time, then new time will be t - 1.
Distance is the same = 400.

By Rate forumyla, rate * time = distance, we have (r+20)(t-1) = 400. ---> Eq 2.

We have two unknowns r and t and two eq. Try solving the two equations to get the unknown values.
From eq1, rt= 400 ==> r = 400/t------->Eq 1a derived from Eq 1.

Substitute r from eq 1a in eq 2.

(400/t +20) (t-1) = 400.

Try soving for t from the above eq, and you should be getting the value of t as 5.

Question is asking for the actual time and hence our ans will be 5.

Hope this is clear.

Kudos [?]: 361 [0], given: 15

Re: Quant Q Distance Q followup   [#permalink] 28 Jan 2009, 12:34
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