The price of a diamond varies inversely with the square of the percentage of impurities. The cost of a diamond with

[Vincen]

The price of a diamond varies inversely with the square of the percentage of impurities. The cost of a diamond with \(0.02\%\) impurities is \(\$2500.\) What is the cost of a diamond with \(0.05\%\) impurities (keeping everything else constant)?

(A) \(\$400\)
(B) \(\$500\)
(C) \(\$1000\)
(D) \(\$4000\)
(E) \(\$8000\)

Answer: A

Source: Veritas Prep

[Scott@TargetTestPrep]

The price of a diamond varies inversely with the square of the percentage of impurities. The cost of a diamond with \(0.02\%\) impurities is \(\$2500.\) What is the cost of a diamond with \(0.05\%\) impurities (keeping everything else constant)?

(A) \(\$400\)
(B) \(\$500\)
(C) \(\$1000\)
(D) \(\$4000\)
(E) \(\$8000\)

Answer: A

Solution:

Recall that if A varies inversely with the square of B, then A_1 x (B_1)^2 = A_2 x (B_2)^2. Since the cost of a diamond varies inversely with the square of the percentage of impurities we have (letting x = the cost of the diamond with 0.05% impurities):

2500(0.02)^2 = x(0.05)^2

2500(0.0004) = x(0.0025)

2500(4) = x(25)

400 = x

Alternate Solution:

For this inverse proportionality question, we can first determine the value of k, the proportionality constant, by letting P = price of the diamond and m = the percentage of impurities in the diamond. We have:

P = k/m^2

2500 = k / 0.02^2

2500 = k / 0.0004

1 = k

To determine the price of the second diamond, we use the same formula, but we know k = 1 and we have a new value for m:

P = 1/0.05^2

P = 1/0.0025

P = 400

Answer: A

[swerve]

The price of a diamond varies inversely with the square of the percentage of impurities. The cost of a diamond with \(0.02\%\) impurities is \(\$2500.\) What is the cost of a diamond with \(0.05\%\) impurities (keeping everything else constant)?

(A) \(\$400\)
(B) \(\$500\)
(C) \(\$1000\)
(D) \(\$4000\)
(E) \(\$8000\)

Answer: A

Source: Veritas Prep

Can be solved with ratios.

Let the price be \(x\)
\begin{align*}
2500 &\Rightarrow \dfrac{1}{0.0004} \qquad (1) \\
x &\Rightarrow \dfrac{1}{0.0025} \qquad (2)
\end{align*}

Divide both the equations
\begin{align*}
\dfrac{2500}{x} &= \dfrac{25}{4} \\
x &= 2500 \cdot \left(\dfrac{4}{25}\right) \\
x &= 400
\end{align*}