A water tank A contains 3 times as much water as another water tank B. When 25 liters are poured from tank A into B, there is now twice as much water in B as in tank A. How much water did each tank contain originally? A water tank A contains 3 times as much water as another water tank B. Let the water be in tank A 3w and that in tank B be w. > When 25 liters are poured from tank A into B, there is now twice as much water in B as in tank A. 3w - 25 = 2w How much water did each tank contain originally? I can solve the problem if only I understand the English or words of the problem correctly. Please can any one explain it to me?
A starts with 3 times that of B so let B's amount be W and therefore A's amount 3W. Pouring 25 litres from A to B means A now has 3W-25 and B has W+25. The question says now B has twice as much as A so W+25 = 2(3W-25). Expanding this out gives W+25 = 6W-50. Therefore 5W = 75 and so W = 15. Let's check, so now B starts with 15 and A starts with 45. Pouring 25 from A to B gives B being 40 and A being 20, so B is twice A now. So W=25 is the answer.
Wait, A starts with 3W and B with W We transfer 25 liters form A to B, so A now has 3W-25 and B has W+25. Tank A now has two times as much water as B, so 3W-25 = 2(W+25) Expanding and solving we get W= 75. Thus A starts with 225 liters and B 75. Moving 25 liters from A to B makes 200 liters in A and 100 liters in B, or A has twice the water as B.
Yes, W=15, not W=25, that was a typo. The question says "there is now twice as much water in B as in tank A", so B is twice A, W+25 = 2*(3W-25), rather than A is twice B, 3W-25 = 2*(W+25), as you have. Of course we could do this in a more general manner. A starts with k times as much as B, so B=W and A = kW. We pour amount P from A into B so now we have A=kW-P and B = W+P. B is now m times as much as A so W+P = m(kW-P). We now rearrange to get (mk-1)W = (1+m)P so \(W = \frac{1+m}{mk-1}P\). A quick check from the original question has k=3, P=25 and m=2 so \(W = \frac{3}{5}25 = 15\) Please Register or Log in to view the hidden image! In your case we'd have \(m = \frac{1}{2}\) since you've got A and B the other way around in the second part so \(W = \frac{(\frac{3}{2})}{(\frac{1}{2})}25 = \frac{3}{1}25 = 75\). If we poured from B to A then P would be negative but there'd be a sign change in mk-1 so the amount is still positive, as A/B would go up, not down. You could do it even more generally by saying initially if B=W then \(A = aW+b\) and then after pouring amount P we have \(B=W+P\) and \(A = aW+b-P\) and then we can say \(B = xA+y\) so \(W+P = x(aW+b-P)+y\) and so \((ax-1)W = (1+x)P - (bx+y)\) so \(W = \frac{1+x}{ax-1}P - \frac{bx+y}{ax-1}\). Checking this with the example means a=3, b=0, P=25, x=2 and y=0 so \(W = \frac{1+2}{6-1}25 - \frac{0}{6-1} = 15\) and for your example we change x to x=1/2 and repeat.
looks like simultaneous linear equations 1. b/a=1/3 2. (b+25)/(a-25)=2 b=a/3=2a-50-25 5/3a-75=0 a=45, b=15
