The ratio of the capacity of Tank A to that of Tank B is 7 : 3. Each tank is filled with some water. If the water from Tank B is poured into Tank A until it reaches the brim, there will be 9 liters of water left in Tank B. If the water from Tank A is poured into Tank B until it reaches the brim, there will be 33 liters of water left in Tank A. How much more water are needed to fill both tanks completely ?
We can go hard way with algebraic method and some illustrations to clarify the scenario above, or do it in much simpler way with some twisted idea.
First thing first, it is a total unchanged problem, just internal transfer. All the if statement in the problem shows there is no additional water involved.
Method 1
Capacity Tank A : Tank B = 7 : 3
=> CA : CB = 7 : 3
Original water volume in tank A = va
Original water volume in tank B = vb
Illustration
If the water from Tank B is poured into Tank A until it reaches the brim, there will be 9 liters of water left in Tank B
==> CA = va + (vb - 9)
If the water from Tank A is poured into Tank B until it reaches the brim, there will be 33 liters of water left in Tank A
==> CB = vb + (va - 33)
7 : 3 = va + vb - 9 : va + vb - 33
==> 7 (va + vb - 33) = 3 (va + vb - 9)
4 va + 4 vb = 204 or va + vb = 51
From two illustrations above:
To fill up tank A needs (vb - 9)
To fill up tank B needs (va - 33)
Method 2
This is internal transfer where total of water exchanged is unchanged.
Try to understand that left and right part of the equation are same.
7u + 9 = 3u + 33 --> 4u = 24 --> u = 6
So CA = 7u = 42 L and CB = 3u = 18 L.
In order to fill both tanks up to the brim, we need
(1) Let' s try if we fill up tank A, then 9 L remained in tank B. So (18 - 9) = 9 L water needed to fill up tank B, and both tanks were full.
(2) Another scenario, if we fill up tank B, then 33 L remained in tank A. So (42 - 33) = 9 L water needed to fill up tank A, and both tanks were full.
Both scenarios show that 9 L is needed !
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