39.NA March 2019 Q.8

The speed of a ship is increased to 18 % above normal for 7.5 hours , then reduced to 9 % below normal for 10 hours. The speed is then reduced for remainder of the day so that the consumption for the day is normal amount. Find the percentage difference between the distance traveled in that day and the normal distance traveled per day.

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Given

The speed of a ship is increased to 18 % above normal for 7.5 hours
The speed of a ship is decreased to 9 % above normal for 10 hours

To find

1. The percentage difference between the distance traveled in that day
2. The normal distance traveled per day.

Solution

Let us consider
Normal speed of ship = V
Normal Distance travel / day = 24 V
Normal consumption / hr = C
Consumption for 24 hours = 24 C

We know that C ∝ V³

As given above the speed of a ship is increased to 18 % above normal for 7.5 hours

Consumption for 7.5 hrs
Let us consider C1 = Consumption / Hr
                         V1 = 1.18 x Normal speed
                               = 1.18 V

We know that C1 / C = (V1/ V)³

                             C1 =  C x (1.18 V / V)³

                             C1 = 1.643 C per hour

For 7.5 hour Consumption = 7.5 x 1.643 C
For 7.5 hour Consumption = 12.37 C -----------------------------------(1)

As given above the speed of a ship is decreased to 9 % above normal for 10 hours

Consumption for 10 hrs
Let us consider C2 = Consumption / Hr
                         V2 = 0.91 x Normal speed
                               = 0.91 V

We know that C2 / C = (V2/ V)³

                             C2 =  C x (0.91 V / V)³

                             C2 = 0.753 C per hour

For 10 hour Consumption = 10 x 0.753 C
For 10 hour Consumption = 7.53 C --------------------------------------(2)

Considering (1) and (2)
Total consumption for 17.5 hours = 12.37 C + 7.53 C
                                                       = 19.9 C

As given above the consumption is normal for remaining day. So for remaining (24 - 17.5) = 6.5 hours Consumption is (24 C - 19.9 C) = 4.1 C

So consumption / hr, C3 = 4.1 C / 6.5 = 0.63 C
And speed for 6.5 hours = V3

We know that C3 / C = (V3/ V)³

                 0.63 C / C = (V3/ V)³

                             V3 = 0.86 V


1. Actual distance traveled / day = (V1 x 7.5) + (V2 x 10) + (V3 x 6.5)
                                                 = (1.18 V x 7.5) + (0.91 V x 10) + (0.86 V x 6.5)
                                                 = 23.54 V

2. % Reduction in distance traveled = (Normal distance - Actual distance) / Normal distance
                                                      = (24 V - 23.54 V) / 24 V
                                                      = 1.92 %