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.
To solve this problem, we can set up an equation based on the given information.
Let's say the number of fish in the larger tank before the transfer is x.
According to the problem, if Mary transfers 15 fish from the larger tank to the smaller tank, the number of fish in the smaller tank will be 5/7 of the number of fish in the larger tank.
So, the equation can be written as:
35 + 15 = (5/7)x
Let's solve for x to find the number of fish in the larger tank before the transfer.
35 + 15 = (5/7)x
50 = (5/7)x
To isolate x, we can multiply both sides of the equation by (7/5):
(7/5) * 50 = (7/5) * (5/7)x
70 = x
Therefore, the number of fish in the larger tank before the transfer is 70.
Let's denote the original number of fish in the larger tank as "x".
According to the given information, if Mary transfers 15 fish from the larger tank to the smaller tank, the number of fish in the smaller tank will be 5/7 of the number of fish in the larger tank.
So, after the transfer, the number of fish in the smaller tank will be (x - 15), and it will be equal to 5/7 of the number of fish in the larger tank (x):
(x - 15) = 5/7 * x
To find the value of x, let's solve this equation step-by-step:
Multiply both sides of the equation by 7 to eliminate the fraction:
7 * (x - 15) = 5 * x
Simplify the equation:
7x - 105 = 5x
Combine like terms:
7x - 5x = 105
2x = 105
Divide both sides of the equation by 2:
2x/2 = 105/2
x = 52.5
Therefore, the original number of fish in the larger tank before the transfer is 52.5. Since we cannot have a fraction of a fish, we can round the answer to the nearest whole number.
Hence, Mary originally had 53 fish in the larger tank.