Two worded maths questions. even if you know one answer it would ! also give working out if possible.

Step-by-step explanation:

1. The total amount of oil in the two tanks at the start is ...

  total oil = n + (n +150) + 0 = 2n+150 . . . . litres

The amount of oil in tank C at the end is 1/3 this amount, or

  oil in tank C = (total oil)/3 = (2n+150)/3 . . . . litres

The amount pumped into tank C was 500 litres, so we have ...

  (2n +150)/3 = 500

  2n +150 = 1500 . . . . . . multiply by 3

  2n = 1350 . . . . . . . . . . . subtract 150

  1350/2 = n = 675 . . . . . divide by the coefficient of n

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2. Let x represent the amount remaining in container A. Then 4x is the amount in tank B. Before the transfer, the amount in each container was the same, so ...

  x +120 = 4x -120 . . . . . A had 120 more than remains; B had 120 less

  240 = 3x . . . . . . . . add 120-x

  240/3 = x = 80 . . . divide by the coefficient of x

The amount of liquid left in container A is 80 mL.

207/8 is the right

sine

step-by-step explanation: