The Freeman family is barbecuing veggie burgers, corn cobs, and mushroom caps in their local park. If 3 8 of the items barbecued are veggie burgers, and 1 3 of the items barbecued are corn cobs, what fraction of barbecued items are mushroom caps?

The Freeman family barbecued veggie burgers, corn cobs, and mushroom caps. 3/8 of the items barbecued are veggie burgers, and 1/3 of the items barbecued are corn cobs.

Let the total number of berbecued items be x. Therefore:

x = barbecued veggie burgers + barbecued corn cobs + barbecued mushroom caps

Barbecued veggie burgers = (3/8)x, barbecued corn cobs = (1/3)x, Let barbecued mushroom caps be y

Substituting:

x = (3/8)x + (1/3)x + y

Multiply through by 24

24x = 9x + 8x + 24y

24x = 17x + 24y

24y = 24x - 17x

24y = 7x

y = (7/24)x

barbecued mushroom caps = (7/24) of items

7/24 of the items barbecued are mushroom caps

joaobezerra joaobezerra · Answer 2 · 2021-09-03T20:48:54+02:00

Using fractions, it is found that the fraction of barbecued items that are mushroom caps is of [tex]\frac{7}{24}[/tex].

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The total proportion of all products is 100% = 1.
The fraction corresponding to veggie burgers is [tex]\frac{3}{8}[/tex].
The fraction corresponding to corn cobs is [tex]\frac{1}{3}[/tex].
The fraction corresponding to mushroom caps is x.

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Thus:

[tex]\frac{3}{8} + \frac{1}{3} + x = 1[/tex]

Solving for x, we find the fraction of mushroom caps.
The least common multiple of 3 and 8 is 24.

Then:

[tex]\frac{3\times3 + 8\times1 + 24x}{24} = 1[/tex]

[tex]\frac{17 + 24x}{24} = 1[/tex]

[tex]17 + 24x = 24[/tex]

[tex]24x = 7[/tex]

[tex]x = \frac{7}{24}[/tex]

The fraction of barbecued items that are mushroom caps is of [tex]\frac{7}{24}[/tex].

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