Carmen rides her bicycle at a constant rate to the market. When she rides her bicycle back home along the same route, she bikes at three-quarters the rate she biked to the market. At any given time, t, the distance biked can be calculated using the formula d = rt, where d represents distance and r represents rate. If the trip home takes 12 minutes longer than the trip to the market, how many minutes does it take Carmen to bike home?

Question

Carmen rides her bicycle at a constant rate to the market. When she rides her bicycle back home along the same route, she bikes at three-quarters the rate she biked to the market. At any given time, t, the distance biked can be calculated using the formula d = rt, where d represents distance and r represents rate. If the trip home takes 12 minutes longer than the trip to the market, how many minutes does it take Carmen to bike home?

Answer 1

distance = d

rt = (3/4 r)(t+12)
rt = 3/4 rt + 9r
1/4 rt = 9r
t = 36

So, it takes 36+12=48 minutes to bike home

Answer 2

Let's assume that it takes Carmen t minutes to bike to the market.

Given that she rides her bicycle at a constant rate to the market, we can write the equation for the distance biked to the market as: d1 = r*t

Since she bikes at three-quarters the rate when she bikes back home, we can write the equation for the distance biked back home as: d2 = (3/4)*r*t

We know that the trip back home takes 12 minutes longer than the trip to the market. Therefore, we can write the equation: d2 = d1 + 12

Now, let's substitute the equations for d1 and d2: (3/4)*r*t = r*t + 12

Let's simplify this equation: 3r*t/4 = rt + 12

Now, let's solve for t:

3rt/4 - rt = 12
(3rt - 4rt)/4 = 12
-rt = 12 * 4
rt = -48

Since time cannot be negative, there is no valid solution for t. Therefore, there must be an error in the given information or equations. Can you please double-check the problem?

Answer 3

To solve this problem, let's define some variables.

Let r be the rate Carmen bikes to the market (in distance per minute).
Carmen's rate back home is three-quarters (3/4) of her rate to the market, so her rate back home is (3/4)r.

Let d be the distance from Carmen's home to the market.

We know that the formula to calculate distance is d = rt, where d is the distance, r is the rate, and t is the time.

For the trip to the market, the time taken is t1, and for the trip back home, the time taken is t2.

We can set up two equations to represent the time for each trip:

For the trip to the market: d = rt1
For the trip back home: d = (3/4)r*t2

We also know that the trip back home takes 12 minutes longer than the trip to the market. So we can set up another equation:

t2 = t1 + 12

Now, we have three equations:

d = rt1
d = (3/4)r*t2
t2 = t1 + 12

We need to find the value of t2, which represents the time Carmen takes to bike back home.

To solve this system of equations, we can substitute the values from the first and third equations into the second equation.

Substituting d = rt1 into the second equation gives:
rt1 = (3/4)r*t2

We can cancel out the r from both sides of the equation:
t1 = (3/4)t2

Now we can substitute this value of t1 into the third equation:
(3/4)t2 = t2 + 12

To get rid of the fraction, we can multiply both sides of the equation by 4:
3t2 = 4t2 + 48

Now, rearrange the equation:
4t2 - 3t2 = 48
t2 = 48

Therefore, it takes Carmen 48 minutes to bike back home.