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A grocer mixes two kinds of nuts costing $6.13 per pound and $7.13 per pound to make 100 pounds of mixture costing $6,53 per pound. How many pounds of each kind of
nut are in the mixture?

Answer :

MHANIFAIN

Answer:

  • 60 pounds of $6.13 nuts and
  • 40 pounds of $7.13 nuts

Step-by-step explanation:

  • $6.13 nut = x pounds
  • $7.13 nut = y pounds

Equations:

  • 6.13x + 7.13y = 6.53*100
  • x + y = 100

Solve by substitution x = 100 - y:

  • 6.13(100 - y) + 7.13y = 653
  • 613 - 6.13y + 7.13y = 653
  • y = 653 - 613
  • y = 40

Find x:

  • x = 100 - 40 = 60

SEMSEE45IN

Answer:

60 lb of the nuts costing $6.13 per pound.

40 lb of the nuts costing $7.13 per pound.

Step-by-step explanation:

Given information:

  • Cost of nut A = $6.13 per pound
  • Cost of nut B = $7.13 per pound
  • Total = 100 lbs of nuts
  • Total cost = $6.53 per pound

Define the variables

Let a = number of pounds of nut A

Let b = number of pounds of nut B

From the given information and the defined variables, create a system of equations:

[tex]\begin{cases}\sf a + b = 100\\\sf 6.13a + 7.13b = 6.53 \times 100 \end{aligned}[/tex]

Rewrite the first equation to make b the subject:

[tex]\implies \sf b = 100-a[/tex]

Substitute the found expression for b into the second equation and solve for a:

[tex]\implies \sf 6.13a+7.13(100-a)=653[/tex]

[tex]\implies \sf 6.13a+713-7.13a=653[/tex]

[tex]\implies \sf -a+713=653[/tex]

[tex]\implies \sf -a=-60[/tex]

[tex]\implies \sf a=60[/tex]

Substitute the found value of a into the first equation and solve for b:

[tex]\implies \sf 60+b=100[/tex]

[tex]\implies \sf b=40[/tex]

Therefore, the number of pounds of each kind of nut in the mixture are:

  • 60 lb of the nuts costing $6.13 per pound.
  • 40 lb of the nuts costing $7.13 per pound.

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