How much candy at $1.20 a pound should be mixed with candy worth 95¢ a pound in order to obtain a mixture of 50 pounds of candy worth a dollar a pound?

10 pounds of candy costing $1.20 per pounds must be mixed with 40 pounds of candy costing $ 0.95 per pound to obtain a 50 pounds of candy costing $ 1 per pound

Solution:

Let "x" be the pounds of candy at $ 1.20 per pound

Then (50 - x) is the pounds of candy at $ 0.95 per pound ( since 95 cents is equal to 0.95 pound)

Therefore, x pounds of candy costing $1.20 per pounds must be mixed with (50 - x)  pounds of candy costing $ 0.95 per pound to obtain a 50 pounds of candy costing $ 1 per pound

Then the equation becomes,

1.20x + (50 - x)0.95 = 50 x 1

On expanding we get,

1.20x + 47.5 - 0.95x = 50

0.25x = 50 - 47.5

0.25x = 2.5

x = 10

Then (50 - x) = 50 - 10 = 40

So we need 10 pounds of candy worth $ 1.20 per pound and 40 pounds of candy worth $ 0.95 to obtain 50 pounds of candy worth a dollar a pound

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".

7,24

step-by-step explanation: