Mary has two tanks of fish. If she transfers 15 fish from the larger tank to the smaller tank then the number of fish in the smaller tank
will be 5/7 of the number of fish in the large, tank. Given that there are 35 fish in the smaller tank originally, find the number of fish in the larger tank before the transfer of fish.
Let's say the number of fish in the larger tank before the transfer is x.
After transferring 15 fish from the larger tank to the smaller tank, the number of fish in the smaller tank becomes 35+15 = 50.
According to the given information, the number of fish in the smaller tank (50) is 5/7 of the number of fish in the larger tank (x).
So we can set up the following equation:
50 = (5/7)x
To find the number of fish in the larger tank before the transfer (x), we can isolate x by multiplying both sides of the equation by 7/5:
(7/5) * 50 = x
Simplifying this equation gives:
70 = x
Therefore, the number of fish in the larger tank before the transfer is 70.
To find the number of fish in the larger tank before the transfer, we can set up an equation based on the given information.
Let's assume the number of fish in the larger tank before the transfer is "x".
According to the given condition, if Mary transfers 15 fish from the larger tank to the smaller tank, the number of fish in the smaller tank becomes 5/7 of the number of fish in the larger tank.
We can express this condition using the equation:
35 + 15 = (5/7) * x
First, we add 35 to represent the original number of fish in the smaller tank. Then, we add the 15 fish transferred from the larger tank.
Simplifying, we get:
50 = (5/7) * x
To isolate "x", we multiply both sides of the equation by 7/5:
(7/5) * 50 = x
Now, we calculate:
(7/5) * 50 = 70
Therefore, the number of fish in the larger tank before the transfer is 70.
35 fish in smaller tank.
X fish in larger tank.
5/7(x-15) = 35+15 = 50.
5(x-15) = 350, 5x - 75 = 350, X = 85.