Answer :
Answer:
[tex]1.6m^3[/tex] sand is to be added to make sand percentage [tex]60\%[/tex]
Step-by-step explanation:
From the question we are told that:
Total soil [tex]N=3.2m^3[/tex]
Initial sand percentage [tex]S_1=40\%[/tex]
Final sand percentage[tex]s_2=60\%[/tex]
Generally the sand and other particles division is mathematically given by
[tex]40\% of 3,2=1.28[/tex]
and
[tex]60\% of 3.2=1.92[/tex]
Generally what will make 60\% of 3.2 40\% is mathematically given by
[tex]x=1.92*0.4\\x=4.8[/tex]
Generally the sand to be added is mathematically given by
[tex]4.8sand-3.2sand =1.6m^3[/tex]
Therefore
[tex]1.6m^3[/tex] sand is to be added to make sand percentage [tex]60\%[/tex]
To achieve a 60% sand mixture from 3.2m³ of soil with 40% sand, you will need to add 1.6m³ of additional sand.
To determine how much sand to add to 3.2m³ of soil with 40% sand to achieve soil with 60% sand, we can set up the following equation:
Let x be the volume of sand to add. The total volume of the mixture would be 3.2m³ + x. Because we are dealing with percentages, we can write that 40% of 3.2m³ plus 100% of x will be equal to 60% of the total new volume (3.2m³ + x).
0.40(3.2) + 1(x) = 0.60(3.2 + x)
Now, let's solve for x:
1.28 + x = 1.92 + 0.60x
x - 0.60x = 1.92 - 1.28
0.40x = 0.64
x = 0.64 / 0.40
x = 1.6m³
Therefore, you would need to add 1.6m³ of sand to the existing soil to achieve a mixture with 60% sand content.
