M04 Q21 - soccer coach bike ride

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Joined: 02 Jul 2012

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Re: M04 Q21 - soccer coach bike ride [#permalink]

18 Oct 2012, 00:59

samiam7 wrote:

A soccer coach riding his bike at 24 km/h reaches his office 5 minutes late. Had he traveled 25% faster, he would have reached the office 4 minutes earlier than the scheduled time. How far is his office from his house?

(A) 18 km
(B) 24 km
(C) 36 km
(D) 40 km
(E) 72 km

[Reveal] Spoiler: OA

Source: GMAT Club Tests - hardest GMAT questions

I used equations to solve and it took me longer than I would have liked.

The OA says "The best choice is backsolving." Can someone please walk me though the way they would backsolve?

Thanks

Finding the answer algebraically is simple enough.. Wont take more than a minute. Since the answer here is A, backsolving might be quicker. But that might not be the case everytime.

Let "x" be the distance and "m" be the time taken when 5 minutes late

\frac{x}{24}=m

1.25 times the speed is 30 km/hr. So
\frac{x}{30}= m - \frac{9}{60} = \frac{x}{24} - \frac{9}{60}

Solving the equation is basic math which gives 18 as the answer.
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Joined: 15 May 2012

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Re: M04 Q21 - soccer coach bike ride [#permalink]

14 Dec 2012, 12:23

There was a post by Sriharimurthy and I really got a lot benefitted by that post. It gives idea about how to approach rate problems.
Here is how I solved it:

speed distance time

late 24km/h d t+5

early 30km/h d t-4

25% faster is 24*25/100=6 Hence 24+6=30

Now, d/t+5=24 and d/t-4=30
Solving this equation, we get t=40 min. Apply this value ion the above equation and you get 18 kilometers.

Re: M04 Q21 - soccer coach bike ride [#permalink] 14 Dec 2012, 12:23

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M04 Q21 - soccer coach bike ride

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M04 Q21 - soccer coach bike ride

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