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In the formula w = \frac{p}{\sqrt[t]{v}} , integers p and t

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by BTGmoderatorDC » Fri Dec 13, 2019 4:48 pm
In the formula \(w = \frac{p}{\sqrt[t]{v}}\), integers p and t are positive constants. If w = 2 when v = 1 and if w = 1/2 when v = 64, then t =

(A) 1
(B) 2
(C) 3
(D) 4
(E) 16



OA C

Source: Official Guide
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by henilshaht » Sun Dec 15, 2019 9:02 am
We are given that w = 2 when v = 1. So,

\(2 = \frac{p}{\sqrt[t]{1}}\)

Since \(\sqrt[t]{1}\) is always going to be 1, we now know the value of p as 2.

Now, if w = 1/2 when v = 64

\(\frac{1}{2} = \frac{2}{\sqrt[t]{64}}\)

This gives us values of$$\frac{1}{\sqrt[t]{64}} = \frac{1}{4}$$ . So, we know that t is 3.

Answer C
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by Scott@TargetTestPrep » Mon Dec 23, 2019 5:01 pm
BTGmoderatorDC wrote:In the formula \(w = \frac{p}{\sqrt[t]{v}}\), integers p and t are positive constants. If w = 2 when v = 1 and if w = 1/2 when v = 64, then t =

(A) 1
(B) 2
(C) 3
(D) 4
(E) 16



OA C

Source: Official Guide

We can create the equations:

1/2 = p/(^t√64)

(^t√64)/2 = p

And

2 = p/(^t√1)

2(^t√1) = p

Substituting, we have:

(^t√64)/2 = 2(^t√1)

(^t√64) = 4(^t√1)

(^t√64)/(^t√1) = 4

(^t√64) = 4

Thus, t must be 3.

Alternate Solution:

First, we substitute w = 2 and v = 1 into the given equation to obtain:

2 = p / ^t√1

Since t is positive, we know that ^t√1 = 1 for any t. Thus, we have:

2 = p

Now, we substitute w = ½ and v = 64 and p = 2 into the given equation to obtain:

1/2 = 2 / ^t√64

1 = 4 / ^t√64

^t√64 = 4

Raising both sides to the t power, we have:

64 = 4^t

t = 3

Answer: C

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