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Volume · filling and draining
P6 · Volume · PSLE Maths
Question
The school Eco Club has two rectangular tanks. Tank A has a base of 40 cm × 25 cm and a water depth of 18 cm. It is being filled at 1.5 L per minute. Tank B has a base of 50 cm × 30 cm and a water depth of 42 cm. It is being drained at 3 L per minute. Both start at the same time. After how many minutes will the water levels in the two tanks be the same?
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Hint
First convert each volume rate into a change in water level per minute by dividing by the base area.
Show the answer and step-by-step explanation
Answer: Time taken 6 6/7 min
- Draw heights — The question asks when the water levels are the same, so we draw a model for the starting depths.
- Gap is 24 cm — At first, Tank B is 42 − 18 = 24 cm higher than Tank A. This gap closes because A rises while B falls.
- Convert to height — Be careful: the tanks have different base areas, so we cannot compare L/min directly. Convert each rate to change in water level per minute.
- Closing each min — A rises 1.5 cm each minute, and B falls 2 cm each minute. So the gap decreases by 1.5 + 2 = 3.5 cm each minute.
- Find the time — The starting gap is 24 cm, and it closes by 3.5 cm per minute. Time = 24 ÷ 3.5 = 6 6/7 min.
- Check — Check the water levels: A becomes 28 2/7 cm, and B also becomes 28 2/7 cm. They are equal. 📌 Key idea: when one tank is filling and the other is draining, convert volume rates to height rates first. Then use starting gap ÷ gap closed per minute.
Method: To compare water levels, compare changes in height. Convert L/min to cm³/min, then use volume ÷ base area = height to find the change in water level per minute.
Practice: 3 similar questions
- Tank A has a base of 30 cm × 30 cm and a water depth of 10 cm. It is filled at 1.8 L per minute. Tank B has a base of 40 cm × 30 cm and a water depth of 34 cm. It is drained at 2.4 L per minute. Both start at the same time. After how many minutes will the water levels be the same?
Answer
6 min — Tank A base area = 30 × 30 = 900 cm². 1.8 L = 1800 cm³, so A rises 1800 ÷ 900 = 2 cm per minute. Tank B base area = 40 × 30 = 1200 cm². 2.4 L = 2400 cm³, so B falls 2400 ÷ 1200 = 2 cm per minute. The starting gap is 34 − 10 = 24 cm; it closes by 2 + 2 = 4 cm per minute. Time taken = 24 ÷ 4 = 6 min.
- Mei Ling has two rectangular fish tanks. Tank X has a base of 30 cm × 40 cm and a water depth of 20 cm. It is filled at 2.4 L per minute. Tank Y has a base of 60 cm × 40 cm and a water depth of 50 cm. It is drained at 3.6 L per minute. Both start at the same time. After how many minutes will the water levels be the same?
Answer
8 4/7 min — X base area = 30 × 40 = 1200 cm². 2.4 L = 2400 cm³, so X rises 2400 ÷ 1200 = 2 cm per minute. Y base area = 60 × 40 = 2400 cm². 3.6 L = 3600 cm³, so Y falls 3600 ÷ 2400 = 1.5 cm per minute. The starting gap is 50 − 20 = 30 cm; it closes by 2 + 1.5 = 3.5 cm per minute. Time taken = 30 ÷ 3.5 = 60/7 = 8 4/7 min.
- Raju has two rectangular tanks for a science experiment. Tank A has a base area of 800 cm² and a water depth of 16 cm. It is filled at 2 L per minute. Tank B has a base area of 1250 cm² and a water depth of 56 cm. It is drained at 1.875 L per minute. Both start at the same time. After how many minutes will the water level in Tank A be 8 cm higher than that in Tank B?
Answer
12 min — A rises each minute: 2 L = 2000 cm³, 2000 ÷ 800 = 2.5 cm. B falls each minute: 1.875 L = 1875 cm³, 1875 ÷ 1250 = 1.5 cm. The difference in their water levels changes by 2.5 + 1.5 = 4 cm per minute. At first, B is 56 − 16 = 40 cm higher than A. For A to be 8 cm higher than B, the total change needed is 40 + 8 = 48 cm. Time taken = 48 ÷ 4 = 12 min.
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