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M04 #1

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Re: M04 #1 [#permalink] New post 17 Jan 2011, 11:46
well a is the only one that give u time. otherwise - u cannot know how long is the drive...

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Re: M04 #1 [#permalink] New post 17 Jan 2011, 20:01
A. B has nothing to do with solving the question and with A you can quickly back into his normal time which is 30 mins.
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Re: M04 #1 [#permalink] New post 11 Jun 2011, 17:45
lets say the normal speed is v => v= d/t1

where t1 is the time taken to cover distance d.

when speed is
v/1.5 = d/t2

1. Sufficient

v/1.5 = d/t1+15

v = d/t1

solving these two equations t1 can found out.

2. Not sufficient

d = 15

no info on v or there is no way to find out t1.

Answer is A.
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Re: M04 #1 [#permalink] New post 18 Jan 2012, 22:36
Rate and Work problems are handled like this.

Rate x Time = Distance (Work) :idea:

Now in this question below is given:
Time2 = 1.5 Time1 .... (1)And question is asking Time1 =?

Statement1 : Time2 = 15 + Time1 ... (2).
Using both we get 1.5Time1 = 15 + Time1 2 Equations 1 variable so solvable and Time1 comes out to be 30 mins.

Statement 2: Distance is given so if we use the main equation
Rate x Time = Distance (Work)
Rate1 x Time1 = 15

So Answer is A.

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Re: M04 #1 [#permalink] New post 14 Sep 2012, 11:06
Here we go, since the question is Data Sufficiency we do not need exact answer, just get the gist.

You know that Time = Distance/Speed, and from here we can find either value, yep?

Let the Speed be X, Distance D and Time Spent D/X - all the usual indicators

However, Today occurred some changes as follow:

TODAY: - Speed Today X/1.5 (1.5 times slower) and Time Spent 1.5D/X and the Distance remain the same, lets now move to options what we got:

1. From Statement 1 we get that 1.5D/X - D/X = 15/60 --->>>> a little bit manipulation gives us that D/X = 1/2, OR usual time (Before) it took Bill 1/2 hour or 30 minutes to drive to school, hence Sufficient.

2. The distance between home and school is irrelevant here, it can not help us that much, thus Insufficient.

Therefore, answer would be (A)

Please, correct me if I went awry.

dczuchta wrote:
Driving 1.5 times slower, Bill was late for school today. What is the usual time it takes Bill to drive to school? (Assume that each day Bill takes the same route).

1) It took Bill 15 more minutes to drive to school today than usually
2) The distance between home and school is 15 miles

[Reveal] Spoiler: OA

Source: GMAT Club Tests - hardest GMAT questions

S1 tells us that on average Bill drives for 15 minutes less. Since he drove 1.5 times slower, the 15 minutes account for the difference. Therefore, it is possible to find the difference. (Can Someone Please Explain to Me Here In an Equation HOw we Can come up with that Answer if We Needed To. Thank you.

S2 does not tell us much; we don't really need to know the distance and by itself it is insufficeint.


God loves the steadfast.

Last edited by Javoni on 14 Sep 2012, 11:17, edited 1 time in total.
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Re: M04 #1 [#permalink] New post 18 Jan 2013, 05:05
Expert's post
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Re: M04 #1 [#permalink] New post 18 Jan 2013, 08:12
Bunuel wrote:
REVISED VERSION OF THIS QUESTION IS HERE: m04-70604-20.html#p1121614

Yep the revised question makes complete sense. Thanks Bunuel.
Re: M04 #1   [#permalink] 18 Jan 2013, 08:12

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M04 #1

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