carmen rides her bicycle at a constant rate to the market. when she rides her bicycle back home along the…

Explanation:

Step1: Let the rate to the market be $r_1$ and time be $t_1$, rate back home be $r_2$ and time be $t_2$.

Since distance $d = r\times t$ and the distance to market and back home is the same, so $r_1t_1=r_2t_2$. Given $r_2=\frac{3}{4}r_1$.

Step2: Substitute $r_2$ into the distance - equality equation.

We get $r_1t_1=\frac{3}{4}r_1t_2$, then $t_1=\frac{3}{4}t_2$.

Step3: Use the time - difference condition.

We know that $t_2=t_1 + 12$. Substitute $t_1=\frac{3}{4}t_2$ into it: $t_2=\frac{3}{4}t_2+12$.

Step4: Solve the equation for $t_2$.

Subtract $\frac{3}{4}t_2$ from both sides: $t_2-\frac{3}{4}t_2 = 12$, $\frac{1}{4}t_2=12$. Multiply both sides by 4, we get $t_2 = 48$.

Answer:

D. 48 min