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M04 #1

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M04 #1 [#permalink] New post 22 Sep 2008, 18:48
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

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A

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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.

REVISED VERSION OF THIS QUESTION IS HERE: m04-70604-20.html#p1121614
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Re: M04 #1 [#permalink] New post 23 Sep 2008, 14:33
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Let d be the distance, and v the uniform speed of travel. Then,

d/(v/1.5) - d/v = 15 min

You're looking for (d/v), which is 30 mins.
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Re: M04 #1 [#permalink] New post 02 Sep 2009, 00:59
Quote:
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


Guys I dont understand how stmt 1 is suff..don't we have two variables?Distance and speed?
Somebody please explain in detail..
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Re: M04 #1 [#permalink] New post 02 Sep 2009, 01:16
While d is distance and v is speed, \frac{d}{v} is time. So you end up with a "single variable" \frac{d}{v}.

Hope it helped.
tejal777 wrote:
Quote:
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


Guys I dont understand how stmt 1 is suff..don't we have two variables?Distance and speed?
Somebody please explain in detail..

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Re: M04 #1 [#permalink] New post 22 Sep 2009, 11:18
tejal777 wrote:
I get all the tougher questions of DTS but easy questions like these...ARRRGH..can somebody please form equations the OE is not clear:(

Will the one below suffice?
mmond4 wrote:
Let d be the distance, and v the uniform speed of travel. Then,

d/(v/1.5) - d/v = 15 min

You're looking for (d/v), which is 30 mins.

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Re: M04 #1 [#permalink] New post 11 Jan 2010, 09:04
A, Statement 1 only. You could figure out that 30 minutes is the regular time it takes to get to school.
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Re: M04 #1 [#permalink] New post 11 Jan 2010, 09:37
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Distance=Speed*time
Here distance is constant, so
Original Speed*Original time=New Speed*New time
Acc. to stmt 1
s*t=(s/1.5)*(t+15)
1.5t= t+15
0.5t=15
t=30 min.

Hope this helps.
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Re: M04 #1 [#permalink] New post 11 Jan 2010, 10:11
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Ouch!! It is the first time I am hearing the phrase "times slower", in this case "1.5 times slower". I am not sure if this automatically means Velocity/1.5 just as"1.5 times faster" would mean Velocity*1.5. Can someone clarify this ? This seems pretty basic but I have difficulty with it.

Thanks
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Re: M04 #1 [#permalink] New post 11 Jan 2010, 19:57
Almost the same solution we'll get if we denote the usual time by t and distance by d.
Then the equation looks like d/t=1.5d/(t+15), which doesn't depend on d and finally
t+15=1.5t. A is suff alone.
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Re: M04 #1 [#permalink] New post 12 Jan 2010, 05:55
Yes, you understood it correctly :).
kaptain wrote:
Ouch!! It is the first time I am hearing the phrase "times slower", in this case "1.5 times slower". I am not sure if this automatically means Velocity/1.5 just as"1.5 times faster" would mean Velocity*1.5. Can someone clarify this ? This seems pretty basic but I have difficulty with it.

Thanks

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Re: M04 #1 [#permalink] New post 12 Jan 2010, 23:00
dzyubam wrote:
Yes, you understood it correctly :).
kaptain wrote:
Ouch!! It is the first time I am hearing the phrase "times slower", in this case "1.5 times slower". I am not sure if this automatically means Velocity/1.5 just as"1.5 times faster" would mean Velocity*1.5. Can someone clarify this ? This seems pretty basic but I have difficulty with it.

Thanks


Thanks dzyubam.
I sometimes have problems with the language in the questions.Making an equation itself becomes hard. My english skills are't bad and I have gone through Manhattan Word Translations as well. Still there is a gap as long as interpreting the questions go. I get trapped in 600 level questions because of this. Any suggestions ?
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Re: M04 #1 [#permalink] New post 12 Jan 2010, 23:09
An example of the question type I just talked on is the one below. Took lots of time to formulate the equation. Picking numbers gets confusing as well:-

"I am twice as old as you were when I was as old as you are.The sum of the ages is 64.Find the ages"
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Re: M04 #1 [#permalink] New post 13 Jan 2010, 00:05
For a problem like this you might want to consider picking numbers instead of actually solving the problem. The wording of this problem is indeed tricky.

As for advice, please check out this thread if you haven't yet:
word-problems-made-easy-87346.html

It has a lot of good stuff there. Hope it helps :).
kaptain wrote:
An example of the question type I just talked on is the one below. Took lots of time to formulate the equation. Picking numbers gets confusing as well:-

"I am twice as old as you were when I was as old as you are.The sum of the ages is 64.Find the ages"

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Re: M04 #1 [#permalink] New post 13 Jan 2010, 12:55
kaptain wrote:
An example of the question type I just talked on is the one below. Took lots of time to formulate the equation. Picking numbers gets confusing as well:-

"I am twice as old as you were when I was as old as you are.The sum of the ages is 64.Find the ages"


First, the sum S of the ages must divisible by 7.
If S=63 (instead of the given 64), the ages are 36 and 27 (9 years ago: 27 and 18).
if S=70, then 40 and 30 (30 and 20), if S=7 then 4 and 3.
If x and y are the ages, then till the difference between the ages x-y=y-x/2, or 3x=4y.
In general, the ages are 4S/7 and 3S/7. And the difference between the ages will be S/7. If S is not a multiple of 7 you'll get confusing fractions as ages.
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Re: M04 #1 [#permalink] New post 14 Jan 2011, 06:30
Let T(1) be the usual time
Let T(2) be todays time
then the problem reads:
T(2)=1,5 * T(1) Solve for T(1)
Data 1 reads:
T(2)=T (1)+15 min ==> 2 equations with 2 variables ==> sufficient
Data 2 reads:
Distance = 15 miles ==> no information about T(1) ==> not necessary and insufficient
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Re: M04 #1 [#permalink] New post 14 Jan 2011, 07:16
Yups, Answer is A

d = distance
s = speed
t = time

d=st

We have following information on normal days Bill goes with '1.5s' and today he went with speed 's'

From 1st Clause we get additional info that on normal days Bill takes time 't' mins and today he took 't+15' mins

So we get,

d = (1.5s)(t)
d = (s)(t+15)

That gives ..

1.5st = st + 15s
0.5st = 15s
0.5t = 15 (cancelling out s, As s<>0)
t = 30 mins

From 1st Clause we can derive all things

From 2nd Clause,

We dont have any information about 'time taken' so cannot complete the formula..

So answer is 'A'
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Re: M04 #1 [#permalink] New post 15 Jan 2011, 09:37
C

Let speed be B. New speed is B'
B= 1.5B'
t = ?
s1 x/B = x/B' + 15
=> x/1.5B' = x/B' + 15
two unkowns. Not suff
s2 x = 15
insuff

combine
15/1.5B' = 15/B' +15
suff

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Re: M04 #1 [#permalink] New post 16 Jan 2011, 00:16
k..1.5 times slower means =
if original speed is 6x
then 1.5 times slower speed is =4x

a) t=d/s
so for original thing t= d/6x and with reduced speed t+15=d/4x
so one can calculate frm here two equations , two unknowns , and no quadratic so u have the ans!

B) how late u cannot say

ans A!
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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.
Re: M04 #1   [#permalink] 17 Jan 2011, 20:01
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