Let normal speed be V and normal daily consumption be C. Consumption varies as the cube of speed, so C = k.V^3.

Step 1 - Find the speed for the remainder of the day.

For 7.5 hours the speed is 1.18V. Consumption in that period = k.(1.18V)^3 x 7.5/24 = k.V^3 x 1.643 x 0.3125 = 0.5134 k.V^3.

For 10 hours the speed is 0.91V. Consumption = k.(0.91V)^3 x 10/24 = k.V^3 x 0.7536 x 0.4167 = 0.3140 k.V^3.

Total consumption in first 17.5 hours = 0.5134 + 0.3140 = 0.8274 k.V^3.

Remaining time in the day = 24 - 17.5 = 6.5 hours.

For the day's total consumption to equal the normal amount k.V^3, the remaining consumption must be k.V^3 - 0.8274 k.V^3 = 0.1726 k.V^3.

If the reduced speed is Vr, then k.Vr^3 x 6.5/24 = 0.1726 k.V^3.

So Vr^3 = 0.1726 x 24/6.5 x V^3 = 0.6373 V^3.

Vr = (0.6373)^(1/3) V = 0.8606 V.

So the ship travels at 86.06% of normal speed for the last 6.5 hours.

Step 2 - Find the distance travelled that day.

Distance = speed x time.

Normal distance per day = V x 24 = 24V.

Actual distance = 1.18V x 7.5 + 0.91V x 10 + 0.8606V x 6.5

= 8.85V + 9.10V + 5.594V = 23.544V.

Step 3 - Percentage difference.

Difference = 24V - 23.544V = 0.456V.

Percentage difference = 0.456/24 x 100 = 1.9%.

Answer: The distance travelled that day is 1.9% less than the normal distance per day.