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primary 6 | Maths
| Fractions
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Date Posted: 5 years ago
Views: 307
We can let tank A's capacity be 7u and B's be 3u.
If pour from B to A until reaches A's brim,
Total volume of water in both tanks = 7u + 9 litres
If the other way round,
Total volume of water in both tanks = 3u + 33 litres
Notice that the total volume of water in both tanks is the same for both cases.
So 7u + 9 = 3u + 33 litres
4u = 24 litres
1u = 24 ÷ 4 = 6 litres
Total volume when both containers are full = 7u + 3u
= 10u
= 10 x 6 litres
= 60 litres
Volume currently in the tanks
= 7 x 6 litres + 9 litres (or 3 x 6 litres + 33 litres)
= 51 litres
60 litres - 51 litres = 9 litres
9 more litres of water are needed
If pour from B to A until reaches A's brim,
Total volume of water in both tanks = 7u + 9 litres
If the other way round,
Total volume of water in both tanks = 3u + 33 litres
Notice that the total volume of water in both tanks is the same for both cases.
So 7u + 9 = 3u + 33 litres
4u = 24 litres
1u = 24 ÷ 4 = 6 litres
Total volume when both containers are full = 7u + 3u
= 10u
= 10 x 6 litres
= 60 litres
Volume currently in the tanks
= 7 x 6 litres + 9 litres (or 3 x 6 litres + 33 litres)
= 51 litres
60 litres - 51 litres = 9 litres
9 more litres of water are needed
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