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# Bob bikes to school every day at a steady rate of x miles

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Director
Joined: 23 May 2008
Posts: 800

Kudos [?]: 86 [0], given: 0

Bob bikes to school every day at a steady rate of x miles [#permalink]

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12 Sep 2008, 15:48
00:00

Difficulty:

25% (medium)

Question Stats:

84% (01:23) correct 16% (02:07) wrong based on 199 sessions

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Bob bikes to school every day at a steady rate of x miles per hour. On a particular day, Bob had a flat tire exactly halfway to school. He immediately started walking to school at a steady pace of y miles per hour. He arrived at school exactly t hours after leaving his home. How many miles is it from the school to Bob's home?

A. (x + y) / t

B. 2(x + t) / xy

C. 2xyt / (x + y)

D. 2(x + y + t) / xy

E. x(y + t) + y(x + t)

OPEN DISCUSSION OF THIS QUESTION IS HERE: bob-bikes-to-school-every-day-at-a-steady-rate-of-x-miles-57035.html
[Reveal] Spoiler: OA

Kudos [?]: 86 [0], given: 0

Manager
Joined: 09 Jul 2007
Posts: 244

Kudos [?]: 281 [1], given: 0

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12 Sep 2008, 17:46
1
KUDOS
IMO C

Let distance be S ;

so , bob cycled S/2 => S/2= x*t1 ( t1= time taken to cycle ) => t1=S/2x
and S/2 = y*t2 ( t2 is time taken to walk ) => t2 = S/2y

now t1 + t2 = t
=> S/2x + S/2y= t => S* ( x+y)/2xy = t=> S= 2xyt/( x+y)

Kudos [?]: 281 [1], given: 0

Intern
Joined: 14 Feb 2012
Posts: 27

Kudos [?]: 5 [0], given: 6

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09 Feb 2013, 08:47
ssandeepan wrote:
IMO C

Let distance be S ;

so , bob cycled S/2 => S/2= x*t1 ( t1= time taken to cycle ) => t1=S/2x
and S/2 = y*t2 ( t2 is time taken to walk ) => t2 = S/2y

now t1 + t2 = t
=> S/2x + S/2y= t => S* ( x+y)/2xy = t=> S= 2xyt/( x+y)

I don't get this question.
Normally I just use the pick-a-number-approach.

If we choose for d (distance): 10 miles, for x 10 and for y 5.
t would be 90 minutes or 1,5 hours.

If I try this for answer choice A it fits.

Can anyone explain what I am doing wrong?

Many thanks.

Kudos [?]: 5 [0], given: 6

Math Expert
Joined: 02 Sep 2009
Posts: 42583

Kudos [?]: 135540 [0], given: 12697

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10 Feb 2013, 01:46
waltiebikkiebal wrote:
ssandeepan wrote:
IMO C

Let distance be S ;

so , bob cycled S/2 => S/2= x*t1 ( t1= time taken to cycle ) => t1=S/2x
and S/2 = y*t2 ( t2 is time taken to walk ) => t2 = S/2y

now t1 + t2 = t
=> S/2x + S/2y= t => S* ( x+y)/2xy = t=> S= 2xyt/( x+y)

I don't get this question.
Normally I just use the pick-a-number-approach.

If we choose for d (distance): 10 miles, for x 10 and for y 5.
t would be 90 minutes or 1,5 hours.

If I try this for answer choice A it fits.

Can anyone explain what I am doing wrong?

Many thanks.

For plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

OPEN DISCUSSION OF THIS QUESTION IS HERE: bob-bikes-to-school-every-day-at-a-steady-rate-of-x-miles-57035.html
_________________

Kudos [?]: 135540 [0], given: 12697

Re: mgmat   [#permalink] 10 Feb 2013, 01:46
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