Carcass wrote:

At a certain fruit stand, the price of an apple is twice the price of an orange. For which of the following combinations of apples and oranges is the total price equal to the total price of 20 oranges?

Indicate

all such combinations

A. 2 apples and 16 oranges

B. 3 apples and 14 oranges

C. 4 apples and 10 oranges

D. 6 apples and 8 oranges

E. 10 apples and 5 oranges

F. 12 apples and 4 oranges

Here's an algebraic solution.

Let x = the cost of ONE orange

So 2x = the cost of ONE appleFor which of the following combinations of apples and oranges is the total price equal to the total price of 20 oranges?The cost of 20 oranges =

20xSo, we're looking for answer choices that equal 20x

A. 2 apples and 16 oranges

Cost = 2(2x) + 16x = 4x + 16x = 20x. WORKS!

B. 3 apples and 14 oranges

Cost = 3(2x) + 14x = 6x + 14x = 20x. WORKS!

C. 4 apples and 10 oranges

Cost = 4(2x) + 10x = 8x + 10x = 18x. nope

D. 6 apples and 8 oranges

Cost = 6(2x) + 8x = 12x + 8x = 20x. WORKS!

E. 10 apples and 5 oranges

Cost = 10(2x) + 5x = 25x. nope

F. 12 apples and 4 oranges

Cost = 12(2x) + 4x = 28x. nope

Answer: A,B,D

Cheers,

Brent

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Brent Hanneson – Creator of greenlighttestprep.com

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