A company produces 3 types of widgets. On a given day, the company was able to produce 9 total widgets (combining all 3 types). Widget A costs $3 to produce, Widget B costs #2 to produce, and Widget C costs $1 to produce. The cost to produce the widgets each day is $18. The company produces one more Widget B than the sum of the Widgets A & C. How many of each type of Widget is produced?

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DarcySea DarcySea · Answer 1 · 2019-10-30T19:34:09+01:00

Answer:

Number of widgets A and C are 2 each, widgets B are 5.

Step-by-step explanation:

Let [tex]x[/tex] be the number of widgets A, [tex]y[/tex] be the number of widgets B and [tex]z[/tex] be the number of widgets C.

As per the question,

On a given day, total 9 widgets are produced. So,

[tex]x+y+z=9[/tex]

Cost for 1 widget A is $3. So, cost of [tex]x[/tex] widgets A is 3x.

Cost for 1 widget B is $2. So, cost of [tex]y[/tex] widgets A is 2y.

Cost for 1 widget C is $1. So, cost of [tex]z[/tex] widgets A is z.

Now, total cost of the widgets is $18. So,

[tex]3x+2y+z=18[/tex]

Also, widget B is one more than the sum of widgets A and C. So,

[tex]y=x+z+1[/tex]

Now, we have 3 equations and 3 unknowns.

[tex]x+y+z=9[/tex]

[tex]3x+2y+z=18[/tex]

[tex]y=x+z+1[/tex]⇒[tex]x+z=y-1[/tex]

Now, using the first and third equation, we get

[tex]y+y-1=9\\2y=10\\y=5[/tex]

Plug in 5 for [tex]y[/tex] in the first 2 equations. This gives,

[tex]x+5+z=9[/tex]⇒[tex]x+z=4[/tex]

[tex]3x+2(5)+z=18[/tex]⇒[tex]3x+z=8[/tex]

Subtracting the above 2 equations, we get,

[tex]x-3x+z-z=4-8\\-2x=-4\\x=2[/tex]

Using the value of x in the above equation, we find z.

[tex]x+z=4\\2+z=4\\z=2[/tex]

Therefore, [tex]x=2,y=5,z=2[/tex]

Hence, number of widgets A and C are 2 each, widgets B are 5.