Let "x" be the amount/the mass of the original mixture which should be combined (in kilograms).

Let "y" be the amount/the mass of the (outside) sand which should be combined (in kilograms)


Then you have the system of these two equations for x and y:


 = 0.16     (1)   (to provide the cement 16% content)

  = 0.18     (2)   (to provide the sand 18% contents)


Simplify by multiplying both sides of each equation by the respective denominator. You will get

0.05x + 255 = 0.16*(x + y + 255),     (3)
0.07x  +  y = 0.18*(x + y + 255).     (4)


Simplify the system one step further by multiplying both sides of the equations by 100. You will get

5x + 25500  = 16*(x + y + 255),      (5)
7x +   100y = 18*(x + y + 255).      (6)


Write the system in canonical/standard form

11x + 16y = 21420,                   (7)
11x - 82y = -4590.                   (8)


The setup is completed. The rest is arithmetic and the technique for solving this system.


First step in solving is to subtract equation (8) from the equation (7). You will get

98y = 21420 + 4590 = 26010  ====>  y =  = 265.408 kg of of the (outside) sand.


Having the found value of y, you can calculate x from the equation (7)

x =  =  = 1561.225 kg.


Answer.  1561.225 kg of the original mixture and 265.408 kg of the (outside) sand should be added.


Check.  Eq(1):   = 0.16.

        Eq(2):   = 0.18.