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Algebra

by swerve » Wed Mar 25, 2020 12:14 pm

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A

B

C

D

E

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If a is a positive integer and \(y=3.2\times 10^a\), what is the value of y?

1) \(400 < y < 30,000\)
2) \(y^2=1.024 \times 10^7\)

The OA is D

Source: Princeton Review

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Re: Algebra

by deloitte247 » Fri Mar 27, 2020 6:38 pm

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B

C

D

E

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$$y=3.2\cdot10^9;\ where\ a\ is\ a\ positive\ integer$$
What is the value of y?
Statement 1: 400 < y < 30,000
Since a is a positive value, the possible value of y includes 32, 320, 3200, 32000, 320000, etc. But given that y is greater than 400 and less than 30,000, the only possible value of y is 3,200.
So, therefore, statement 1 is SUFFICIENT.

Statement 2:
$$y^2=1.024\cdot10^7$$
$$y=\pm\sqrt{1.024\cdot10^7}$$
Since y cannot be negative because a is positive from the question stem.
$$y=\sqrt{10,240,000}=3,200$$
Statement 2 is SUFFICIENT

Since each statement alone is sufficient, the correct option is option D