Other Mathematical Relationships - SAT Subject Test in Math II

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Question

Add in modulo 9:

Answer

In modulo 9 arithmetic, a number is congruent to the remainder of its division by 9.

Since

and

$23 \div 9 = 2\textrm{ R }5$ ,

$5 + 6 + 5 + 4 + 3 \equiv 5 \mod 9$ ,

making "5" the correct response.

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Question

varies directly as and inversely as .

If and , then .

To the nearest whole number, evaluate if and .

Answer

varies directly as and inversely as . This means that for some constant of variation ,

$\frac{V P }{T } = k$

Alternatively,

$\frac{V_{1}P_{1}}{T_{1}} = \frac{V_{2}P_{2}}{T_{2}}$

We can substitute the initial conditions for thevariables on the left side and the final conditions for those on the right side, then solve for $V_{2}$ :

$\frac{1,000 \cdot 100}{273} = \frac{V_{2} \cdot 110}{293}$

$\frac{V_{2} \cdot 110}{293} \cdot \frac{293}{ 110} = \frac{1,000 \cdot 100}{273} \cdot \frac{293}{ 110}$

$V_{2} = \frac{1,000 \cdot 100 \cdot 293}{273\cdot 110} \approx 976$

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Question

varies directly as both and the square of .

If and , then .

Evaluate if and .

Answer

varies directly as both and the square of . This means that for some constant of variation ,

$\frac{E}{mv^{2}} = k$ .

Alternatively stated,

$\frac{E_{1}}{m_{1}v_{1}^{2}} = \frac{E_{2}}{m_{2}v_{2}^{2}}$ .

We can substitute the initial conditions for the variables on the left side and the final conditions for those on the right side, then solve for $E_{2}$ :

$\frac{8,000}{10 \cdot 40^{2}} = \frac{E_{2}}{15 \cdot 30^{2}}$

$\frac{8,000}{10 \cdot 40^{2}} \cdot 15 \cdot 30^{2} = \frac{E_{2}}{15 \cdot 30^{2}} \cdot 15 \cdot 30^{2}$

$E_{2} = \frac{8,000 \cdot 15 \cdot 30^{2}}{10 \cdot 40^{2}} = \frac{108,000,000}{16,000}= 6,750$

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Question

$4 \sqrt[4] {a}+ 7 \sqrt[4] {b} =18$

$3\sqrt[4] {a} -5 \sqrt[4] {b} = 34$

Evaluate .

Answer

Rewrite the two equations by setting $x = \sqrt [4] {a}$ and $y = \sqrt [4] {b}$ and substituting:

$4 \sqrt[4] {a}+ 7 \sqrt[4] {b} =18$

$3\sqrt[4] {a} -5 \sqrt[4] {b} = 34$

The result is a two-by-two linear system in terms of and :

This can be solved, among other ways, using Gaussian elimination on an augmented matrix:

$\begin{bmatrix} 4& 7\ 3& -5 \end{matrix} \begin{vmatrix} 18 \ 34 \ \end{bmatrix}$

Perform row operations until the left two columns show identity matrix $\begin{bmatrix} 1 & 0\ 0 &1 \end{bmatrix}$ . One possible sequence:

$R1-R2 \rightarrow R1$

$\begin{bmatrix} 4 - 3 & 7- (-5 )\ 3& -5 \end{matrix} \begin{vmatrix} 18 - 34 \ 34 \ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \ 3& -5 \end{matrix} \begin{vmatrix} -16 \ 34 \ \end{bmatrix}$

$R2 - 3R1 \rightarrow R2$

$\begin{bmatrix} 1 & 12 \ 3 - 3(1)& -5 - 3(12) \end{matrix} \begin{vmatrix} -16 \ 34 - 3(-16) \ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \ 0& -41\end{matrix} \begin{vmatrix} -16 \ 82\ \end{bmatrix}$

$R2 \div (-41 )\rightarrow R2$

$\begin{bmatrix} 1 & 12 \ 0\div (-41 )& -41\div (-41 )\end{matrix} \begin{vmatrix} -16 \ 82\div (-41 )\ \end{bmatrix}$

$\begin{bmatrix} 1 & 12 \ 0& 1 \end{matrix} \begin{vmatrix} -16 \-2\ \end{bmatrix}$

$R1 - 12 R2 \rightarrow R1$

$\begin{bmatrix} 1 - 12(0) & 12 - 12(1) \ 0& 1 \end{matrix} \begin{vmatrix} -16 - 12(-2) \ 2\ \end{bmatrix}$

$\begin{bmatrix} 1 & 0 \ 0& 1 \end{matrix} \begin{vmatrix} 8 \ 2\ \end{bmatrix}$

and . In the former equation, substitute $\sqrt [4] {a}$ back for , and raise both sides to the power of 4:

$\sqrt [4] {a} = 8$

$(\sqrt [4] {a}) ^{4}= 8^{4}$

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