1. The next day, 2400×( 1/4)m was completed, and there were 2400-2400× 20%-2400× (1/4) m left on this road.

2. 15%×60÷5%-60

3. There is X, [(1/4) x+140]: [x-(1/4) x-140] = 3: 2.

4.35÷(28÷20%)

5. Let the total length of this highway be xm, (7/20) x-32% x = 270.