Digital SAT Math Practice Test 4

Instead of solving a complex system for \(x\) and \(y\) individually, look for a linear combination by adding the two equations together vertically:

\[ \left(\dfrac{1}{2} + \dfrac{1}{5}\right)x + \left(\dfrac{1}{5} + \dfrac{1}{2}\right)y = \dfrac{1}{5} + \dfrac{1}{5} \]

Find a common denominator (10) for the fractions on the left side:

\[ \left(\dfrac{5}{10} + \dfrac{2}{10}\right)x + \left(\dfrac{2}{10} + \dfrac{5}{10}\right)y = \dfrac{2}{5} \]

\[ \dfrac{7}{10}x + \dfrac{7}{10}y = \dfrac{2}{5} \]

Factor out the common term \(\dfrac{7}{10}\):

\[ \dfrac{7}{10}(x + y) = \dfrac{2}{5} \]

To isolate \((x + y)\), multiply both sides by the reciprocal \(\dfrac{10}{7}\):

\[ (x + y) = \dfrac{2}{5} \times \dfrac{10}{7} = \dfrac{20}{35} = \dfrac{4}{7} \]

The question asks for the value of \( 2(x + y) \):

\[ 2\left(\dfrac{4}{7}\right) = \dfrac{8}{7} \]

Let the integer be \(k\), where \(k > 2\). We can set up the equation:

\[ \dfrac{x+2}{7} = k \Rightarrow x + 2 = 7k \]

Isolate \(x\): \[ x = 7k - 2 \]

To find the remainder when \(x\) is divided by 7, we need to express \(x\) in the form \(7m + r\), where \(r\) is a positive remainder (\(0 \le r < 7\)).

Rewrite \(7k - 2\) by "borrowing" 7 from the \(7k\) term:

\[ x = 7(k - 1) + 7 - 2 \]

\[ x = 7(k - 1) + 5 \]

Since \(7(k - 1)\) is a multiple of 7, the remaining term is the remainder. Therefore, the remainder is 5.

Start by isolating \(M\) in the equation to determine the constraints on \(N\):

\[ -3M = 10 - 4N \Rightarrow M = \dfrac{4N - 10}{3} \]

Because \(M\) must be a negative integer, the numerator \((4N - 10)\) must be a negative multiple of 3. Let's test negative integer values for \(N\):

• If \(N = -1\): \(M = \dfrac{4(-1) - 10}{3} = \dfrac{-14}{3}\) (Not an integer).
• If \(N = -2\): \(M = \dfrac{4(-2) - 10}{3} = \dfrac{-18}{3} = -6\).

Since both \(M = -6\) and \(N = -2\) satisfy the condition of being negative integers, -2 is a valid possible value for \(N\).

First, expand the equation and rewrite it in standard quadratic form (\(ax^2 + bx + c = 0\)):

\[ x^2 + kx + 4 = 0 \]

A quadratic equation has exactly one real solution when its discriminant (\(b^2 - 4ac\)) is equal to zero:

\[ k^2 - 4(1)(4) = 0 \]

\[ k^2 - 16 = 0 \Rightarrow k^2 = 16 \]

Taking the square root gives \(k = 4\) or \(k = -4\). Since the question specifies a positive value, the answer is 4.

To find \(x^8\), we need to find a way to combine the given \(x^6\) with another factor of \(x\). Notice that \(x^8 = x^6 \cdot x^2\).

Step 1: Find \(x^2\) from the given equation. Take the cube root of both sides:

\[ (x^6)^{1/3} = 20^{1/3} \Rightarrow x^2 = \sqrt[3]{20} \]

Step 2: Substitute both \(x^6 = 20\) and \(x^2 = \sqrt[3]{20}\) into our expression for \(x^8\):

\[ x^8 = 20 \cdot \sqrt[3]{20} \]

The exact value is \(20\sqrt[3]{20}\).

For a product to be zero, at least one of the individual factors must be zero. Let's analyze each factor:

1. \(x^2 + 3 = 0 \Rightarrow x^2 = -3\). No real number squared results in a negative value, so this factor provides no real solutions.
2. \(2|x| + 4 = 0 \Rightarrow 2|x| = -4 \Rightarrow |x| = -2\). Since an absolute value must be 0 or positive, this factor also provides no real solutions.
3. \(-x + 3 = 0 \Rightarrow x = 3\). This is a valid real solution.

The only real value that satisfies the equation is 3.

Calculate the area of both tables (\(A = s^2\)):

Smaller table: \(10^2 = 100 \text{ sq in}\)

Larger table: \(15^2 = 225 \text{ sq in}\)

To find the percent increase, use the formula: \(\dfrac{\text{Difference}}{\text{Original}} \times 100\).

\[ \text{Percent More} = \dfrac{225 - 100}{100} \times 100 = 125\% \]

The larger table's area is 125% more than the smaller table's area.

Warning: Average speed is not the average of the two speeds. It is \(\dfrac{\text{Total Distance}}{\text{Total Time}}\).

Leg 1 Distance: \(1.5 \text{ h} \times 60 \text{ km/h} = 90 \text{ km}\)

Leg 2 Distance: \(3.0 \text{ h} \times 70 \text{ km/h} = 210 \text{ km}\)

Total Distance = \(90 + 210 = 300 \text{ km}\)

Total Time = \(1.5 + 3.0 = 4.5 \text{ h}\)

Average Speed = \(300 / 4.5 \approx 66.666...\)

Rounded to one decimal place, the average speed is 66.7 km/h.

Let \(S\) and \(E\) be the counts in 2001. Then \(R = \dfrac{S}{E}\).

In 2002, the counts became \(1.4S\) and \(0.5E\). The new ratio is \(r = \dfrac{1.4S}{0.5E}\).

We are asked for \(\dfrac{r}{R}\):

\[ \dfrac{r}{R} = \dfrac{\dfrac{1.4S}{0.5E}}{\dfrac{S}{E}} = \dfrac{1.4}{0.5} = 2.8 \]

Complete the square for the \(y\) terms to find \(r^2\):

\[ x^2 + (y^2 - 8y + 16) = 48 + 16 \]

\[ x^2 + (y - 4)^2 = 64 \]

The standard form \((x-h)^2 + (y-k)^2 = r^2\) shows that \(r^2 = 64\).

Area = \(\pi r^2 = 64\pi\).

The empty space is the difference between the two volumes (\(V = \pi r^2 h\)):

Large Cylinder: \(\pi(10)^2(20) = 2000\pi\)

Small Cylinder: \(\pi(5)^2(20) = 500\pi\)

Difference: \(2000\pi - 500\pi = 1500\pi \text{ cm}^3\).

Volume is the product of three linear dimensions (\(L \times W \times H\)). Increasing a dimension by 50% is the same as multiplying by 1.5.

\[ V_{\text{new}} = (L \times 1.5) \times (W \times 1.5) \times (H \times 1.5) \]

\[ V_{\text{new}} = V_{\text{old}} \times (1.5)^3 \]

\[ V_{\text{new}} = 1000 \times 3.375 = 3375 \text{ m}^3 \]