A restaurant offers the possibility of 168 three-course dinners. Each dinner has an appetizer, an entrée, and a dessert. If the number of appetizers decreases from 7 to 5, how many fewer possible three-course dinners can the restaurant offers?

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A restaurant offers the possibility of 168 three-course dinners. Each dinner has an appetizer, an entrée, and a dessert. If the number of appetizers decreases from 7 to 5, how many fewer possible three-course dinners can the restaurant offers?

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Answer 1 · 2017-03-23T08:59:13

120

Explanation:

Let's look at it this way:

Before the change is made, there are 168 three-course dinners possible. How can we achieve that number? We have Appetizers, Entrees, and Desserts. We can reach 168 by saying:

$A \times E \times D = 168$ $AxxExxD=168$

We know there are 7 appetizers, so $A = 7$ $A=7$ , and so:

$7 \times E \times D = 168$ $7xxExxD=168$

$E \times D = 24$ $ExxD=24$

We can now multiply in the new number of appetizers, $A = 5$ $A=5$ :

$5 \times 24 = 120$ $5xx24=120$

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Notice that we could have simply said:

$\frac{168}{7} \times 5 = 120$ $168/7xx5=120$

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A restaurant offers the possibility of 168 three-course dinners. Each dinner has an appetizer, an entrée, and a dessert. If the number of appetizers decreases from 7 to 5, how many fewer possible three-course dinners can the restaurant offers?

1 Answer

Explanation:

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