Two open rectangular watering trays, Tray A and Tray B, are placed side by side. Tray A has a base measuring 6 cm by 4 cm and initially contains 480 cm³ of water. Tray B is empty and has a base measuring 8 cm by 5 cm. Water is transferred through a pipe from Tray A to Tray B at 5 cm³ per second. Neither tray overflows.
- What is the initial depth of water in Tray A?
- How many seconds does it take for the water level in Tray B to be 2 cm higher than the water level in Tray A?
- How much water is in Tray B when this happens?
Need a hint?
When Tray B's water is 2 cm higher, imagine taking away the top 2 cm of water in Tray B. Then the water left in both trays is at the same depth.
Worked answer
- (a) Base area of Tray A = 6 × 4 = 24 cm².
- Depth of water in Tray A at first = 480 ÷ 24 = 20 cm.
- (b) Base area of Tray B = 8 × 5 = 40 cm².
- When Tray B's water is 2 cm higher, imagine taking away its top 2 cm of water: 40 × 2 = 80 cm³.
- The other 480 − 80 = 400 cm³ fills both trays to the same depth. Together the bases are 24 + 40 = 64 cm², so that depth = 400 ÷ 64 = 6.25 cm.
- Depth in Tray B = 6.25 + 2 = 8.25 cm, so Tray B holds 40 × 8.25 = 330 cm³ (check: Tray A holds 24 × 6.25 = 150 cm³, and 150 + 330 = 480).
- Time = 330 ÷ 5 = 66 seconds.
- (c) Water in Tray B = 330 cm³.
The mistake this question tests
Fraction of a remainder
A common mistake is to find only one water depth and compare it with 2 cm. The depths of both trays must be found separately using their different base areas.
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