An aluminum alloy containing 48% aluminum is to be made by containing 30% and 60% alloys. how many pounds of the 60% alloy must be added to 24 pounds of the 30% alloy to produce the desired aluminum alloy?

36 pounds

Step-by-step explanation:

48% = 48/100 = 0.48

30% = 30/100 = 0.3

60% = 60/100 = 0.6

Let pounds of 60% alloy added = x

So, from the statement, we can write:

0.6x + 0.3(24) = 0.48(x + 24)

Note "x + 24" in total because we are adding "x" from 60% and "24" from 30% alloys.

Now, we have to solve the above equation for x:

Thus,

We would need 36 pounds added from 60% alloy

C. π(2ft)²(10ft) + 1/3π(2ft)²(13ft-10ft)

Step-by-step explanation:

Given:

diameter = 4ft

length of the cylindrical part = 10ft

entire length of the silo = 13ft

Find the volume of the silo = volume of cylinder and volume of the cone

Formulas:

Volume of cylinder: πr²h

Volume of come: 1/3πr²h

2ft= radius

height =10

volume π×2²×10

radius 2ft

height 13-10=3ft

Volume= 1/3×π×2²×3 cube feet

Therefore the third answer is correct: C. π(2ft)²(10ft) + 1/3π(2ft)²(13ft-10ft)