A rectangular container 120 cm by 60 cm by 40 cm was 20% filled with water. A tap was turned on to fill it up with water at a rate of 8 ℓ/min. Every 30 seconds after the tap was turned on, an iron ball of volume 300 cm3 was added to the container. How many iron balls of the same volume would there be in the container when the container is 100% filled with water?
Percentage of volume of space to be filled
= 100% - 20%
= 80%
Volume of space in the container
= 80% x 120 x 60 x 40
=
80100 x 120 x 60 x 40
= 230400 cm
3 1 ℓ = 1000 mℓ
8 ℓ = 8 x 1000 = 8000 mℓ
Number of sets of 30 seconds in 1 min
= 60 ÷ 30
= 2
Water flowed in every 30 seconds
= 8000 ÷ 2
= 4000 mℓ
Total volume added for every 30 seconds
= 4000 + 300
= 4300 cm
3 Number of sets of 4300 cm
3= 230400 ÷ 4300
= 53 r 2500
After the last iron ball is added, 2500 cm
3 of water will flow into the tank so that the container is 100% filled with water.
Number of iron balls added = 53
Answer(s): 53