LucyNeedsMathHelp
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George just bought a new scooter for $8000. He plans on keeping it until it is worth a fourth of it's original price.

He believes that every year the scooter loses an eighth of its value. He can represent the value of the scooter by the equation V = 8000(7/8)t, where V is the value of the scooter and t is the number of years that have passed.

As he was leaving the dealership he was told that the scooter really loses a sixth of it's value every year. What should George change in the equation V = 8000(7/8)t to represent this new situation?


A) Replace the 7/8 with a 6.
B) Replace the 8000 with 5/6.
C) Switch the V and t
D) Replace the 7/8 with a 5/6.

Respuesta :

MadBulgarian
since it's 1/6th and not 1/8th then replace 7/8 with a 5/6

Answer:

D) Replace the 7/8 with a 5/6.

Step-by-step explanation:

Given,

The original value of the scooter = $ 8000,

If it loses an eighth of its value.

Then the decrement in the price of scooter = [tex]\frac{1}{8}[/tex] of the original price of the scooter

[tex]=\frac{1}{8}\times 8000[/tex]

[tex]=\frac{8000}{8}[/tex]

Now, the final price of the scooter = Original price of the scooter - Decrement in price

[tex]=8000-\frac{8000}{8}[/tex]

[tex]=8000(1-\frac{1}{8})[/tex]

[tex]=8000(\frac{8-1}{8}[/tex]

[tex]=8000(\frac{7}{8})[/tex]

Similarly, If it loses an sixth of its value.

Then, decrement in price = [tex]\frac{1}{6}[/tex] of 8000

[tex]=\frac{8000}{6}[/tex]

And, the final price of the scooter = [tex]8000-\frac{8000}{6}[/tex]

[tex]=8000(1-\frac{1}{6})[/tex]

[tex]=8000(\frac{6-1}{6})[/tex]

[tex]=8000(\frac{5}{6})[/tex]

Hence, Option D is correct.

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