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Driving 1.5 times slower, Bill was late for school today.

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Manager
Joined: 11 Apr 2008
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Location: Chicago
Driving 1.5 times slower, Bill was late for school today. [#permalink]

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18 Sep 2008, 20:53
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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

This question may be the end of me, but if you go 1.5 times slower aren't you going a negative speed? For example, if you are going 10 mph, and you drive 1.5 times slower (10 * 1.5 = 15) ---> 10 - 15 = -5, doesn't this mean you are going -5 mph?
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Joined: 04 Aug 2008
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19 Sep 2008, 00:02
Hi rate cannot be negative. it can be direction. Let's say it 10 mph then it's 1.5 times down is 10 /1.5 .

I think if we combine both stats then u have 1.5 x 15/x = 15/x + 15/60 where x is normal speed mph.

so u get x and then usual time for 15 miles

so Ans is C
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19 Sep 2008, 00:19
1.5 times is supposed to be multiplied and not divided right ? How can something be 1.5 times slow ? do we have to multiply by -1.5 ? What am I missing here ?
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19 Sep 2008, 01:19
1.5 times low doesnt mean multiply by -1.5 times.

if rate is 1.5 less means (x * 1.5) = 10 so x = 10 /1.5

Hope it clarifies..

cheers
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Joined: 19 Sep 2008
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19 Sep 2008, 01:42
The rate is absolutely possible, it just meens that the speed was 1.5 time less than Bill's everyday speed. Find below my logic:
Actually we don't need to know the distance to find how much time it takes to Bill to get to the school, so we can assume that distance is "1", then we get the following equation: 1/x=1/x/1.5-15/60, where x - his normal speed, 1/x - usual time he spends everyday to get to the school, 1/x/1.5 - this particular day time, 15/60 - minutes in hours, thus we have:
1/x=3/2x-0.25 => x = 2
The second statement doesn't give us anything, so I think the asnwer is "A".

PS: Sorry for my English
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Joined: 11 Apr 2008
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19 Sep 2008, 08:56
rakeshmrana wrote:
1.5 times low doesnt mean multiply by -1.5 times.

if rate is 1.5 less means (x * 1.5) = 10 so x = 10 /1.5

Hope it clarifies..

cheers

But according to your formula, which is (old rate)/(factor) = new rate ---> new rate = 10/1.5, if the question were to state that "if you are going 10 mph, and you drive 50% slower" then 10/0.5 = 20 --> you would be going faster not slower. How does this make sense?
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19 Sep 2008, 09:17
Basically the distance is the same, so:

Distance = rate x time
rate1 x time1 = rate1 x 1.5 x time2

We have an equation with three variables, but rate1 can be cancelled out. Using Statement 1 you can get an equation with only one variable. Answer should be A.
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19 Sep 2008, 09:35
Nerdboy wrote:
Basically the distance is the same, so:

Distance = rate x time
rate1 x time1 = rate1 x 1.5 x time2

We have an equation with three variables, but rate1 can be cancelled out. Using Statement 1 you can get an equation with only one variable. Answer should be A.

Ahh, that makes sense. So, in a way, the new time is 1.5 times as long. This is easier to think of then trying to manipulate the rate. Thanks.
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Factorials were someone's attempt to make math look exciting!!!

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19 Sep 2008, 10:28
Nerdboy wrote:
Basically the distance is the same, so:

Distance = rate x time
rate1 x time1 = rate1 x 1.5 x time2

We have an equation with three variables, but rate1 can be cancelled out. Using Statement 1 you can get an equation with only one variable. Answer should be A.

Get it now !!!!
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Joined: 05 Oct 2008
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05 Nov 2008, 01:27
Nerdboy - you still have 3 variables left as t1 and t2 are not the same? Would you please explain by solving the equation completely, considering it a problem solving question. Thanks
Re: possibly impossible rate   [#permalink] 05 Nov 2008, 01:27
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Driving 1.5 times slower, Bill was late for school today.

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