Bob bikes to school every day

rsarashi 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)

These kinds of questions (Variables in the Answer Choices - VIACs) can be answered algebraically or using the INPUT-OUTPUT approach.

I was about to solving it using the INPUT-OUTPUT approach, but then I saw Mitch post his INPUT-OUTPUT solution.
So, here's an algebraic solution:

We know the TOTAL travel time = t hours
Let's let B = the time spent BIKING
So, t - B = time spent walking.

Bob had a flat tire exactly halfway to school.
Let's start with a word equation.
Distance traveled on bike = Distance traveled by foot
Distance = (speed)(time)
So, we get: (x)(B) = (y)(t - B)
Expand: xB = yt - yB
Add yB to both sides: xB + yB = yt
Factor: B(x + y) = yt
Divide both sides by (x+y) to get: B = yt/(x+y)

So, the TIME spent biking = yt/(x+y)
We can now use this to find the DISTANCE spent biking.
distance = (speed)(time)
So, DISTANCE spent biking = (x)yt/(x+y)
= xyt/(x+y)

Since Bob had a flat tire exactly halfway to school, we know that xyt/(x+y) represents HALF the distance to school.

So, the ENTIRE distance to school = 2xyt/(x+y)

Answer: C

rsarashi 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 a different 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

rsarashi 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)

''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