Answer:
S = 20
F = 40
T = 80
L = 100
Explanation:
Let 1st fish= F
2nd fish = S
3rd fish = T
4th fish = last fish = L
From the question,
The 1st fish weighs twice as much as the 2nd fish. I.e
F = 2S ..... (1)
and the 3rd fish weighs twice as much as the 1st fish. I.e
T = 2F .......(2)
But F = 2S from (1) substitute for F
T = 2(2S)
T = 4S .......(3)
The weight of the last fish weighs 5 times as much as the 2nd fish. I.e
L = 5S .... (4)
We have,
1st + 2nd + 3rd + 4th = 240 I.e
F + S + T + L = 240 .......(5)
Substitute F, T and L into (5)
2S + S + 4S + 5S = 240
12S = 240
S = 20
From equation (1)
F = 2S = 40
From equation ( 3)
T = 4S = 80
From equation (4)
L = 5S = 100