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Question
primary 6 | Maths
| Measurement
2 Answers Below
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Pls help
Date Posted: 4 years ago
Views: 250
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A)
Time difference :
1900-0900=10 h
Vol of water lost
= 8000cm2 x 3cm
=24000=24 litres
Rate of water leakage= 24/10= 2.4 h/l
B) time difference :
2300h -1900h = 4h
Water leakage rate = 2.4 h/l
In 4 hrs, volume of water lost = 2.4 x4 = 9.6 litres
Ht of water level in tank at 1900h
= (80% x50cm) -3cm= 37cm
Vol of water in tank at 1900= 8000x37= 296000cm3 = 296 litres
Remaining amount of water in tank
= 296 - 9.6 = 286.4 litres
Time difference :
1900-0900=10 h
Vol of water lost
= 8000cm2 x 3cm
=24000=24 litres
Rate of water leakage= 24/10= 2.4 h/l
B) time difference :
2300h -1900h = 4h
Water leakage rate = 2.4 h/l
In 4 hrs, volume of water lost = 2.4 x4 = 9.6 litres
Ht of water level in tank at 1900h
= (80% x50cm) -3cm= 37cm
Vol of water in tank at 1900= 8000x37= 296000cm3 = 296 litres
Remaining amount of water in tank
= 296 - 9.6 = 286.4 litres
done
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Hope this helps
Date Posted:
4 years ago
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