Solution Naval Numerical 4

Numerical 4: The speed of a ship is increased to 18% above normal for 7.5 hours, and then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption for the day is the normal amount. Find the percentage difference between the distance traveled in that day and the normal distance traveled per day.
Solution Let, V = normal speed, V1 is increased speed for 7.5 hrs, V2 is reduced speed for 10 hrs and V3 is speed for remaining hrs.
c = normal consumption per hour
Then, 24c = normal consumption per day.
Now,
V1 = V + 0.18V = 1.8V
V2 = V - 0.09V = 0.91V 
V3 = ?
(i) For the first 7.5hours, speed is v1.
cons/h = c x
=C x =1.643c
thus, consumption for 7.5hrs = 7.5 x 1.643c =12.32c
similarly, Consumption for 10hrs with V2 = 7.536c
and for remaining 6.5hrs = 24c - 12.32c - 7.536c = 4.144c
Cons/h = 4.144c/6.5 = 0.637c
Now,
thus, V3 = 0.86V
Normal distance travelled/days = 24V
New distance travelled/day
=(1.18V x 7.5) + (0.91V x 10) + (.86V x 6.5) = 23.54V
percentage reduction in distance/day =
Percentage reduction in distance/day = 1.92%