AQA AS level physics 2021 paper 2(MECHANICS)

It looks like you're given values to 2sf so calculating results to more than 2sf is spurious precision, you should round to 2sf.

The motor is connected to a 110V dc supply that has negligible internal resistance. The maximum operating current in the motor is 5.0A.
The efficiency of the motor and drive system of the conveyor is 28%. The belt travels at 0.32ms^-1 and is 8.0m long. Deduce the maximum number of blocks that can be moved on the belt at one time.

In the question before this, it was saying the total friction force needed for the block to move in 0.32m/s is 82 N, therefore I used this to find the power needed to move 1 block:
P=Fv
P=0.32x82=26.24
I got the useful power input by doing P=VI, P=110x5=550, 550x0.28=154W
Then I did 154/26.24 which got me 5.8333, but the mark scheme says the answer is 6. Could any of you tell me what I've done wrong?
The friction force I got in the question before was correct.