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Bob bikes to school every day at a steady rate of x miles per hour. On
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24 May 2021, 03:59
24 May 2021, 03:59
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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)
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)
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Re: Bob bikes to school every day at a steady rate of x miles per hour. On
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24 May 2021, 05:38
24 May 2021, 05:38
1
Carcass wrote:
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)
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)
Let d = the TOTAL distance to school.
Bob had a flat tire exactly halfway to school
So, d/2 = distance spent biking
and d/2 = distance spent walking
We can write: (time spent biking) + (time spent walking) = t
time = distance/speed
We get: (d/2)/x + (d/2)/y = t
Simplify: d/2x + d/2y = t
Find a common denominator of 2yx to get: dy/2yx + dx/2yx = t
Combine terms: (dy + dx)/2yx = t
Multiply both sides by 2yx to get: dy + dx = 2xyt
Factor: d(y + x) = 2xyt
Divide both sides by (x + y) to get: d = 2xyt/(x+y)
Answer: C
gmatclubot
Re: Bob bikes to school every day at a steady rate of x miles per hour. On [#permalink]
24 May 2021, 05:38
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