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If Bill drove 1.5 times slower than normal and was late for
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12 Jul 2010, 03:46
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I came across this qs in M04. Please guide me to the link if this qs has already been asked. If Bill drove 1.5 times slower than normal and was late for school today, how long does it normally take Bill to drive to school? (Assume that each day Bill takes the same route). 1. It took Bill 15 minutes longer to drive to school today than normal. 2. The distance between home and school is 15 miles. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifbilldrove15timesslowerthannormalandwaslatefor129323.html== Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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Re: If Bill drove 1.5 times slower than normal and was late for
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12 Jul 2010, 04:52
Normal rate = x today his rate is x/1.5
Let's look on 1. normal time is t, today's time is t+15
We have: xt=(x/1.5)*(t+15) Remove x and solve for t.
2. gives nothing.
The answer is (A).



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Re: If Bill drove 1.5 times slower than normal and was late for
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12 Jul 2010, 07:16
I would go also for be, because if we have the length of the route, we still do not have the time and cannot answer the question. I will go for A as well.
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Re: If Bill drove 1.5 times slower than normal and was late for
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12 Jul 2010, 10:11
A Nicely explained by "ulm"
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Re: If Bill drove 1.5 times slower than normal and was late for
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30 Aug 2010, 10:06
I don’t understand the explanation below. Can someone please help.
If Bill drove 1.5 times slower than normal and was late for school today, how long does it normally take Bill to drive to school? (Assume that each day Bill takes the same route).
1. It took Bill 15 minutes longer to drive to school today than normal. 2. The distance between home and school is 15 miles.
Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient Statements (1) and (2) TOGETHER are NOT sufficient
Explanation:  What is ‘s’ and ‘v’ in the below equation.
Statement (1) by itself is sufficient. Let denote the time it normally takes Bill to drive to school. Decreasing Bill's speed by 1.5 times results in increasing the normal time he spends to drive to school:
\(t = s/v/1.5 = 1.5s/v = 1.5t\) (sorry if the equation looks awkward but I dont know how to put the division sign)
\(1.5t = t + 15\)
We can solve the equation and find the normal time.



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Re: If Bill drove 1.5 times slower than normal and was late for
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30 Aug 2010, 10:41
"s" is the distance of school from his home, "v" is normal speed.



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Re: If Bill drove 1.5 times slower than normal and was late for
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31 Aug 2010, 03:41
Try it this way.. D = t1 * S1 = t2*s2, In addition, s2=S1/1.5
From statement1, t2=t1+1/4 Solve this equations and you get the answer from st1



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Re: If Bill drove 1.5 times slower than normal and was late for
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31 Aug 2010, 03:49
seekmba wrote: I don’t understand the explanation below. Can someone please help.
If Bill drove 1.5 times slower than normal and was late for school today, how long does it normally take Bill to drive to school? (Assume that each day Bill takes the same route).
1. It took Bill 15 minutes longer to drive to school today than normal. 2. The distance between home and school is 15 miles.
Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient Statements (1) and (2) TOGETHER are NOT sufficient
Explanation:  What is ‘s’ and ‘v’ in the below equation.
Statement (1) by itself is sufficient. Let denote the time it normally takes Bill to drive to school. Decreasing Bill's speed by 1.5 times results in increasing the normal time he spends to drive to school:
\(t = s/v/1.5 = 1.5s/v = 1.5t\) (sorry if the equation looks awkward but I dont know how to put the division sign)
\(1.5t = t + 15\)
We can solve the equation and find the normal time. Put statement A in simple words... Speed has declined to 1.5 times to result in a 15 minute delay Had bill drove at normal speed then there would have been no delay so clearly this 0.5 times takes 15 minutes (extra)..........now we can obviously calculate 1 time from that..... Hence OA ......... A Hope this helps !!!
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Re: If Bill drove 1.5 times slower than normal and was late for
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31 Aug 2010, 04:16
It doesn't make any sense to say someone drove "1.5 times slower" than someone else; that's not something that you can say in English. If you ever see a sentence that says "X is 1.5 times Y", you can always write an equation that begins "X = 1.5*Y". You can't do that here. We're left to guess the meaning of the phrase, and I imagine the question designer intended to say that Bill's time was 1.5 times longer than usual. Then Statement 1 is sufficient, since we get a simple equation in one unknown. But as it stands, it's a badly written question.
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Re: If Bill drove 1.5 times slower than normal and was late for
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23 Jan 2011, 12:00
a similar question. question8inm10105223.html#p856317
1) \(\frac{d}{s} =\) the time that Bill usually takes
\(\frac{d*1.5}{s}  \frac{d}{s} = 15\)
\(\frac{d}{s}(1.5  1) = 15\)
\(\frac{d}{s} = 30 mins\) ... SUFFICIENT
2) NOT SUFFICIENT



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Re: If Bill drove 1.5 times slower than normal and was late for
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23 May 2011, 23:11
agree with Ian, also could we rephrase the question as " the speed decreased by a factor 1.5 ? " then we could write v = v/1.5 i dont think so
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Re: If Bill drove 1.5 times slower than normal and was late for
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24 May 2011, 01:36
3/2 times slower means 2/3 times more time.
hence t2= t1+ 2/3 * t1 = 5t1/3
thus t2t1 = 15 and t2= 5t1/3 will give the values.
Hence A
b nothing is given for t1 and t2. Not sufficient.



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Re: If Bill drove 1.5 times slower than normal and was late for
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24 May 2011, 05:42
Please explain how did you solve it
I did not understand
Thanks



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Re: If Bill drove 1.5 times slower than normal and was late for
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24 May 2011, 07:36
seekmba wrote: \(t = s/v/1.5 = 1.5s/v = 1.5t\) (sorry if the equation looks awkward but I dont know how to put the division sign)
\(1.5t = t + 15\)
We can solve the equation and find the normal time.
You can see that \(time = \frac{distance}{speed}\) > This is the fundamental equation used to Calculate Bill's normal time. Since the question states that Bill drove 1.5 times slower, this means that New Speed = \(\frac{Normal Speed}{1.5}\) Now if we take the slower speed (V/1.5) and plug it in the fundamental formula \(time = \frac{distance}{speed}\) we get: \(time =\frac{distance}{speed/1.5}\) \(time = \frac{1.5 * distance}{speed}\) The above equation also equals to: \(1.5 * t2 = \frac{distance}{speed}\) where t2 is the new time and t1 is the original time. Therefore, the equation boils down to: 1.5 * New Time = Original Time + 15 which equals to 1.5 t2 = t1 + 15 > Delta T = 1.5t  t = 15 > 0.5 t = 15 therefore the new time it took for him to drive to school with a speed 1.5 times slower is 30 minutes. Therefore (A) is SUFFICIENT.
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Re: If Bill drove 1.5 times slower than normal and was late for
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24 May 2011, 19:05
v = d/t New speed v2 = d/t2 = v/1.5 1. Sufficient given t2 = 15+t => \((v/1.5) = d/(15+t)\) => \((d/(1.5t)) = d/(15+t)\) enough to find t 2. Not sufficient we only know d. nothing about v. Answer is A. AmmarAlsultan wrote: Please explain how did you solve it
I did not understand
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Re: If Bill drove 1.5 times slower than normal and was late for
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27 Jan 2018, 06:46
If Bill drove 1.5 times slower than normal and was late for school today, how long does it normally take Bill to drive to school? (Assume that each day Bill takes the same route).Suppose regular time in minutes is t, then the time for today would be 1.5t. Question: t=? (1) It took Bill 15 minutes longer to drive to school today than normal > 1.5t=t+15 > t=30 minutes. Sufficient. (2) The distance between home and school is 15 miles. Useless info. Not sufficient. Answer: A. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifbilldrove15timesslowerthannormalandwaslatefor129323.html== Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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