At a community sports day, water was stored in two rectangular coolers, Cooler R and Cooler S. The base of Cooler R measured 24 cm by 15 cm. The base of Cooler S measured 20 cm by 18 cm, and Cooler S was 30 cm high. At first, the two coolers contained 14 400 cm³ of water altogether. The water level in Cooler R was 4 cm higher than that in Cooler S.
- Find the volume of water in Cooler R at first.
- Arif poured water from Cooler R into Cooler S until Cooler S was full. What was the percentage decrease in the volume of water in Cooler R? Round your answer to 1 decimal place.
Need a hint?
Find the base area of each cooler first. Use the total volume and the difference in water levels to find both initial water levels before calculating the percentage decrease.
Worked answer
- The base area of each cooler is 360 cm²: 24 × 15 = 360 and 20 × 18 = 360.
- Let the water level in Cooler S be h cm. Then 360h + 360(h + 4) = 14 400, so h = 18 cm and the level in Cooler R is 22 cm.
- The initial volume of water in Cooler R is 360 × 22 = 7 920 cm³.
- The water transferred is 360 × (30 − 18) = 4 320 cm³.
- The percentage decrease is 4 320 ÷ 7 920 × 100% = 54.5%.
The mistake this question tests
Percentage of a new total
A common mistake is to divide the water transferred by the volume left in Cooler R. A percentage decrease must be calculated using the original volume in Cooler R as the denominator.
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See the answer →Practise PSLE area & volume until the mistakes stop.
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