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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]

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New post 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?

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Re: Time taken by Bill  [#permalink]

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New post 19 Mar 2012, 10:08
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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).

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.
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Re: If Bill drove 1.5 times slower than normal and was late for  [#permalink]

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New post 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]

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New post 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]

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New post 01 Aug 2013, 15:31
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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).

Time = Distance/Rate

(1) It took Bill 15 minutes longer to drive to school today than normal.
1.5t (the normal time multiplied by 1.5)
t+15 = the minutes more it took him to get to school.

1.5 times longer than his normal time = his normal time + 15
1.5t = t+15
.5t=15
t=30

SUFFICIENT

(2) The distance between home and school is 15 miles.
We don't know the typical rate of travel.
INSUFFICIENT

(A)
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Re: If Bill drove 1.5 times slower than normal and was late for  [#permalink]

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New post 10 Sep 2013, 03:58
I also understood that "1.5 times slower" refers that the rate, and computed as follows

today normal
T t+15 t
r r 1.5r
d 15 15

and computed using 2 equations.
If I had understood correctly that it was a reference for rate, would my above solution be correct?
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Re: If Bill drove 1.5 times slower than normal and was late for  [#permalink]

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New post 20 Jan 2015, 07:17
Bunuel wrote:
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.


Hi,

Probably this does not matter to answer the data sufficiency of above question but for my understanding the question is:

For 1.5x "slower", how can we multiply T by 1.5? If T = 100, then 1.5T = 150

Please explain.

TO
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Re: If Bill drove 1.5 times slower than normal and was late for  [#permalink]

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New post 20 Jan 2015, 07:24
thorinoakenshield wrote:
Bunuel wrote:
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.


Hi,

Probably this does not matter to answer the data sufficiency of above question but for my understanding the question is:

For 1.5x "slower", how can we multiply T by 1.5? If T = 100, then 1.5T = 150

Please explain.

TO


Check here: if-bill-drove-1-5-times-slower-than-normal-and-was-late-for-129323.html#p1064476
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Re: If Bill drove 1.5 times slower than normal and was late for  [#permalink]

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New post 05 Apr 2017, 07:19
enigma123 wrote:
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?



IMO: A.
The wording of the question is not correct. No one can drive 1.5 times SLOWER. It should be "DROVE 0.5 TIMES SLOWER THAN NORMAL SPEED" or "NORMAL SPEED IS 1.5 TIMES TODAY'S SPPED".
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Re: If Bill drove 1.5 times slower than normal and was late for  [#permalink]

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Re: If Bill drove 1.5 times slower than normal and was late for   [#permalink] 12 Mar 2019, 02:05
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