**Problem**

A man covers the distance *A* to *B* in 5 hours and 30 minutes. During the second half of the distance he traveled 1/2 kilometer less per hour than during the first half. In the return trip from *B* to *A*, by traveling 1 km/hr faster than during the first half of his trip from *A* to *B*, he consumed 3-3/4 hours. Determine the distance *AB*.

**Answer Key**

15 km

**Solution**

Let

*x*= distance between*A*and*B*in kilometer*v*= original velocity (velocity during the 1st half of distance from*A*to*B*)From *A* to *B*

$\dfrac{0.5x}{v} + \dfrac{0.5x}{v - 0.5} = 5.5$ ← Equation (1)

From *B* to *A*

$\dfrac{x}{v + 1} = 3.75$

$v = \dfrac{x}{3.75} - 1$

From Equation (1)

$\dfrac{0.5x}{\dfrac{x}{3.75} - 1} + \dfrac{0.5x}{\left( \dfrac{x}{3.75} - 1 \right) - 0.5} = 5.5$

$x = 15 ~ \text{km}$ *answer*