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Bob bikes to school

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Bob bikes to school

by BTGmoderatorDC » Fri Jan 26, 2018 10:15 pm
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)

How will i solve this kind of problem? Can some experts teach me?

OA C
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by GMATGuruNY » Sat Jan 27, 2018 3:57 am
lheiannie07 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)
Let the distance to school = 20 miles.
Let x = 5 miles per hour.
Time spent biking = 10/5 = 2 hours.
Let y = 2 miles per hour.
Time spent walking = 10/2 = 5 hours.
Total time = t = 2+5 = 7.
The questions asks for the total distance: 20 miles. This is our target.

New we plug x=5, y=2, and t=7 into the answers to see which yields our target of 20.

Only answer choice C works:
2xyt/(x+y) = (2*5*2*7)/(5+2) = 140/7 = 20.

The correct answer is C.
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by Brent@GMATPrepNow » Sat Jan 27, 2018 7:23 am
lheiannie07 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)

How will i solve this kind of problem? Can some experts teach me?

OA C
Here's an algebraic solution:

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
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