Problem Solving

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BTGmoderatorRO: Moderator; Posts: 772; Joined: Wed Aug 30, 2017 6:29 pmFollowed by: 6 members

by BTGmoderatorRO » Sat Feb 10, 2018 5:20 am

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)

OA is C

What do I need to solve this, I need the help of an expert here. Thanks

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by GMATGuruNY » Sat Feb 10, 2018 5:26 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Feb 10, 2018 7:09 am

Roland2rule 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)

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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8157; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sun Jul 28, 2019 9:22 am

BTGmoderatorRO 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)

OA is C

We are given that Bob bikes at a rate of x miles per hour and walks at a rate of y miles per hour. If we let the distance between his home and school be d, then the distance he bikes is d/2, and the distance he walks is d/2.

Thus, his biking time is (d/2)/x = d/(2x), and his walking time is (d/2)/y = d/(2y). Since he spent t total hours biking and walking:

d/(2x) + d/(2y) = t

Multiplying the entire equation by 2xy, we have:

dy + dx = 2xyt

d(y + x) = 2xyt

d = 2xyt/(x + y)

Answer: C

What do I need to solve this, I need the help of an expert here. Thanks

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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