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# If Bill drove 1.5 times slower than normal and was late for

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If Bill drove 1.5 times slower than normal and was late for [#permalink]  19 Mar 2012, 09:42
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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.

Can someone please explain step by step solution to this? This is how I am trying to do this:

Let T be the time taken normally by Bill. S be the speed and D be the distance
Speed = \frac{Distance}{Time}

Or Distance = Time * Speed = T * S ---------------------------------------(1)

Now, considering Bill drove 1.5 times slower then speed becomes 1.5 s and t becomes t +15

So, 1.5 S = \frac{D}{T+15}

D = 1.5 S(T+15) --------------------------------------------------------(2)

Equating 1 & 2

TS=1.5S(T+15)

T = 1.5T+15 and we can find T. Is my approach correct?
[Reveal] Spoiler: OA

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Re: Time taken by Bill [#permalink]  19 Mar 2012, 10:08
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.

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Re: If Bill drove 1.5 times slower than normal and was late for [#permalink]  23 Mar 2012, 15:31
I set up normal speed as: 1.5R
and slower speed as: R

and its fine.

BUT how can I properly express "1.5 times slower than normal" algebraically???

Many thanks.
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Re: If Bill drove 1.5 times slower than normal and was late for [#permalink]  24 Mar 2012, 03:44
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LA2DC wrote:
I set up normal speed as: 1.5R
and slower speed as: R

and its fine.

BUT how can I properly express "1.5 times slower than normal" algebraically???

Many thanks.

The wording of the question is not perfect (to say the least).

"Bill drove 1.5 times slower than normal" is intended to be translated algebraically as t(today)=1.5*t(normal), where t stands for time.

Hope it helps.
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Re: If Bill drove 1.5 times slower than normal and was late for   [#permalink] 24 Mar 2012, 03:44
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