Alberta's Provincial Achievement Test (PAT) - Free, HARD Math Questions + Answers + Explanations

To simplify: \(\left( \frac{5}{2} \right)^3 = \frac{125}{8}\), \(\sqrt{36} = 6\), \(2 \cdot 6 = 12\), \(\frac{7}{3} = \frac{14}{6}\). Adding: \(\frac{125}{8} - 12 + \frac{14}{6} = \frac{137}{6}\).

The perimeter is calculated as \(2(l + w)\). Simplify \(\sqrt{72} = 6 \sqrt{2}\) and \(3 \sqrt{18} = 9 \sqrt{2}\). Adding: \(2(6 \sqrt{2} + 9 \sqrt{2}) = 30 \sqrt{2}\).

After 10 hours, the population has doubled \(\frac{10}{2} = 5\) times. Using \(P = P_0 \cdot 2^n\): \(P = 1000 \cdot 2^5 = 1000 \cdot 32 = 32000\).

Combine the fractions: \(\frac{4(2x - 1) + 3x}{6} = 5 \implies \frac{8x - 4 + 3x}{6} = 5 \implies 11x - 4 = 30 \implies 11x = 34 \implies x = \frac{34}{11}\).

Find two numbers that multiply to \(6 \cdot 6 = 36\) and add to \(-13\): \(-9\) and \(-4\). Rewrite: \(6x^2 - 9x - 4x + 6\). Factor pairs: \(3x(2x - 3) - 2(2x - 3)\). Final: \((3x - 2)(2x - 3)\).

The triangle is a right triangle because \(8^2 + 15^2 = 17^2\). The area is \(\frac{1}{2} \cdot 8 \cdot 15 = 60\) cm\(^2\).

The surface area of a sphere is \(4\pi r^2\). Setting \(4\pi r^2 = 144\pi\), divide by \(\pi\): \(4r^2 = 144\). Simplify: \(r^2 = 36 \implies r = 6\) cm.

The original speed is \(\frac{120}{2} = 60\) km/h. After a \(20\%\) decrease, the new speed is \(60 \times 0.8 = 48\) km/h. The time to travel \(150\) km is \(\frac{150}{48} \approx 3.125\) hours

The combined rate is \(\frac{1}{3} + \frac{1}{5} = \frac{5}{15} + \frac{3}{15} = \frac{8}{15}\) of the tank per hour. We reverse it such that it takes \(\frac{15}{8} \) hours to fill up the whole tank. To fill \(75\%\) or \(\frac{3}{4}\) of the tank, time required therefore is 75% of the total time which is \(\frac{\frac{3}{4}}{\frac{15}{8}} = \frac{45}{32} \approx 1.4\) hours.

Let the width be \(x\). Then the length is \(x + 4\). The area is \(x(x + 4) = 96\). Solving, \(x^2 + 4x - 96 = 0\). Factoring, \((x - 8)(x + 12) = 0\), so \(x = 8\) (positive root). The dimensions are \(8\) cm by \(12\) cm.

Students who study only math are those who study math but not both subjects. The total studying math is \(18\), and those studying both are \(10\). Thus, students studying only math is \(18 - 10 = 8\).

To get a sum of \(9\), possible pairs are [(3,6), (4,5), (5,4), (6,3)], totaling \(4\) favorable outcomes. The total outcomes when rolling two dice are \(36\). The probability is \(\frac{4}{36} = \frac{1}{9}\).

The mean of five numbers is \(\frac{\text{sum of numbers}}{5}\). Solving, \(12 \times 5 = 60\). The sum of the four known numbers is \(9 + 10 + 15 + 14 = 48\). The fifth number is \(60 - 48 = 12\).