\begin{problembox}
    \subsection{Problems:}
    \begin{questions}
        \question[12] Solve these inequalities.
        \begin{parts}\begin{multicols}{2}
            \part \[x+7 \geq 17\]
        \begin{solutionordottedlines}[1in]
        Subtract 7 from both sides: \\
        \(x \geq 17 - 7\) \\
        \(x \geq 10\)
        \end{solutionordottedlines}
            \part \[x-7 \leq-4\]
        \begin{solutionordottedlines}[1in]
        Add 7 to both sides: \\
        \(x \leq -4 + 7\) \\
        \(x \leq 3\)
        \end{solutionordottedlines}
            \part \[-2 x \geq 16\]
        \begin{solutionordottedlines}[1in]
        Divide both sides by -2 (and reverse the inequality): \\
        \(x \leq \frac{16}{-2}\) \\
        \(x \leq -8\)
        \end{solutionordottedlines}
            \part \[\frac{x}{5}-\frac{1}{2} \geq 2\]
        \begin{solutionordottedlines}[1in]
        Add \(\frac{1}{2}\) to both sides: \\
        \(\frac{x}{5} \geq 2 + \frac{1}{2}\) \\
        \(\frac{x}{5} \geq \frac{5}{2}\) \\
        Multiply both sides by 5: \\
        \(x \geq \frac{5}{2} \times 5\) \\
        \(x \geq \frac{25}{2}\) \\
        \(x \geq 12.5\)
        \end{solutionordottedlines}
            \part \[6(x-3) \leq 15\]
        \begin{solutionordottedlines}[1in]
        Divide both sides by 6: \\
        \(x - 3 \leq \frac{15}{6}\) \\
        \(x - 3 \leq 2.5\) \\
        Add 3 to both sides: \\
        \(x \leq 2.5 + 3\) \\
        \(x \leq 5.5\)
        \end{solutionordottedlines}
            \part \[7-4 x \leq-12\]
        \begin{solutionordottedlines}[1in]
        Subtract 7 from both sides: \\
        \(-4x \leq -12 - 7\) \\
        \(-4x \leq -19\) \\
        Divide both sides by -4 (and reverse the inequality): \\
        \(x \geq \frac{-19}{-4}\) \\
        \(x \geq 4.75\)
        \end{solutionordottedlines}
            \part \[11-3 x>18\]
        \begin{solutionordottedlines}[1in]
        Subtract 11 from both sides: \\
        \(-3x > 18 - 11\) \\
        \(-3x > 7\) \\
        Divide both sides by -3 (and reverse the inequality): \\
        \(x < \frac{7}{-3}\) \\
        \(x < -\frac{7}{3}\) \\
        \(x < -2.\overline{3}\)
        \end{solutionordottedlines}
            \part \[6(1-x) \leq 2.4(15-12 x)\]
        \begin{solutionordottedlines}[1in]
        Expand both sides: \\
        \(6 - 6x \leq 36 - 28.8x\) \\
        Add \(6x\) to both sides and subtract 36: \\
        \(6 \leq -22.8x + 36\) \\
        \(-30 \leq -22.8x\) \\
        Divide both sides by -22.8 (and reverse the inequality): \\
        \(x \geq \frac{30}{22.8}\) \\
        \(x \geq \frac{5}{3.8}\) \\
        \(x \geq 1.\overline{315789473684210526}\)
        \end{solutionordottedlines}
            \part \[\frac{x-3}{3} \leq \frac{3+x}{2}\]
        \begin{solutionordottedlines}[1in]
        Cross multiply to compare: \\
        \(2(x - 3) \leq 3(3 + x)\) \\
        \(2x - 6 \leq 9 + 3x\) \\
        Subtract \(2x\) and add 6 to both sides: \\
        \(-6 \leq 9 + x\) \\
        \(-15 \leq x\) \\
        \(x \geq -15\)
        \end{solutionordottedlines}
            \part \[\frac{2 x+3}{2}-\frac{x-4}{3}>2\]
        \begin{solutionordottedlines}[1in]
        Find a common denominator and combine: \\
        \(\frac{3(2x + 3) - 2(x - 4)}{6} > 2\) \\
        \(\frac{6x + 9 - 2x + 8}{6} > 2\) \\
        \(\frac{4x + 17}{6} > 2\) \\
        Multiply both sides by 6: \\
        \(4x + 17 > 12\) \\
        Subtract 17 from both sides: \\
        \(4x > -5\) \\
        Divide both sides by 4: \\
        \(x > -\frac{5}{4}\) \\
        \(x > -1.25\)
        \end{solutionordottedlines}
            \part \[\frac{4-x}{2}+\frac{3-x}{4}<1\]
        \begin{solutionordottedlines}[1in]
        Find a common denominator and combine: \\
        \(\frac{2(4 - x) + (3 - x)}{4} < 1\) \\
        \(\frac{8 - 2x + 3 - x}{4} < 1\) \\
        \(\frac{11 - 3x}{4} < 1\) \\
        Multiply both sides by 4: \\
        \(11 - 3x < 4\) \\
        Subtract 11 from both sides: \\
        \(-3x < -7\) \\
        Divide both sides by -3 (and reverse the inequality): \\
        \(x > \frac{7}{3}\) \\
        \(x > 2.\overline{3}\)
        \end{solutionordottedlines}
            \part \[\frac{3-2 x}{2} \geq \frac{7-10 x}{5}\]
        \begin{solutionordottedlines}[1in]
        Cross multiply to compare: \\
        \(5(3 - 2x) \geq 2(7 - 10x)\) \\
        \(15 - 10x \geq 14 - 20x\) \\
        Add \(10x\) to both sides: \\
        \(15 \geq 14 - 10x + 10x\) \\
        \(15 \geq 14\) \\
        The inequality is always true, so \(x\) can be any real number.
        \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[4] At 2 p.m., two aeroplanes leave airports \(2880 \mathrm{~km}\) apart and fly towards one another. The average speed of one plane is twice that of the other. If they pass each other at 5 p.m., what is the average speed of each plane?
        \begin{solutionordottedlines}[1in]
        Let the speed of the slower plane be \(v\) km/h. \\
        Then the speed of the faster plane is \(2v\) km/h. \\
        They meet after 3 hours, so the total distance covered is: \\
        \(3v + 3(2v) = 2880\) \\
        \(3v + 6v = 2880\) \\
        \(9v = 2880\) \\
        \(v = 320\) km/h \\
        The speed of the slower plane is 320 km/h, \\
        and the speed of the faster plane is \(2 \times 320 = 640\) km/h.
        \end{solutionordottedlines}
        \question[3] When a mathematics teacher was asked her age she replied, 'One-fifth of my age three years ago, when added to half my age last year, gives my age 11 years ago.' How old is she?
        \begin{solutionordottedlines}[1in]
        Let her current age be \(x\) years. \\
        One-fifth of her age three years ago: \(\frac{x - 3}{5}\) \\
        Half her age last year: \(\frac{x - 1}{2}\) \\
        Her age 11 years ago: \(x - 11\) \\
        The equation is: \\
        \(\frac{x - 3}{5} + \frac{x - 1}{2} = x - 11\) \\
        Multiply through by 10 (LCM of 5 and 2): \\
        \(2(x - 3) + 5(x - 1) = 10(x - 11)\) \\
        \(2x - 6 + 5x - 5 = 10x - 110\) \\
        \(7x - 11 = 10x - 110\) \\
        Subtract \(7x\) from both sides: \\
        \(-11 = 3x - 110\) \\
        Add 110 to both sides: \\
        \(99 = 3x\) \\
        Divide by 3: \\
        \(x = 33\) \\
        The teacher is 33 years old.
        \end{solutionordottedlines}
        \question[3] \(3 \$ 420\) is divided between \(A, B\) and \(C . B\) receives \(\$ 20\) less than \(A\), and \(C\) receives half as much as \(A\) and \(B\) together. How much does each receive?
        \begin{solutionordottedlines}[1in]
        Let \(A\) receive \(x\) dollars. \\
        Then \(B\) receives \(x - 20\) dollars. \\
        \(C\) receives \(\frac{1}{2}(x + (x - 20))\) dollars. \\
        The total amount is: \\
        \(x + (x - 20) + \frac{1}{2}(x + (x - 20)) = 3420\) \\
        \(2x - 20 + x + (x - 20) = 3420\) \\
        \(4x - 40 = 3420\) \\
        Add 40 to both sides: \\
        \(4x = 3460\) \\
        Divide by 4: \\
        \(x = 865\) \\
        So, \(A\) receives \$865, \(B\) receives \(865 - 20 = \$845\), \\
        and \(C\) receives \(\frac{1}{2}(865 + 845) = \$855\).
        \end{solutionordottedlines}
        \question[3] In a printing works, 75000 leaflets are run off by two printing presses in \(18 \frac{3}{4}\) hours. One press delivers 200 more leaflets per hour than the other. Find the number of leaflets produced per hour by each of the machines.
        \begin{solutionordottedlines}[1in]
        Let the number of leaflets the slower press prints per hour be \(x\). \\
        Then the faster press prints \(x + 200\) leaflets per hour. \\
        Total time is \(18 \frac{3}{4}\) hours or \(18.75\) hours. \\
        The equation is: \\
        \(18.75x + 18.75(x + 200) = 75000\) \\
        \(18.75x + 18.75x + 3750 = 75000\) \\
        \(37.5x + 3750 = 75000\) \\
        Subtract 3750 from both sides: \\
        \(37.5x = 71250\) \\
        Divide by 37.5: \\
        \(x = 1900\) \\
        The slower press prints 1900 leaflets per hour, \\
        and the faster press prints \(1900 + 200 = 2100\) leaflets per hour.
        \end{solutionordottedlines}
        \question[4] A salesman makes a trip to visit a client. The traffic he encounters keeps his average speed to \(40 \mathrm{~km} / \mathrm{h}\). On the return trip, he takes a route \(6 \mathrm{~km}\) longer, but he averages \(50 \mathrm{~km} / \mathrm{h}\). If he takes the same time each way, how long was the total journey?
        \begin{solutionordottedlines}[1in]
        Let the distance of the initial trip be \(d\) km. \\
        Time for the initial trip: \(\frac{d}{40}\) hours. \\
        Time for the return trip: \(\frac{d + 6}{50}\) hours. \\
        Since the times are equal: \\
        \(\frac{d}{40} = \frac{d + 6}{50}\) \\
        Cross multiply: \\
        \(50d = 40(d + 6)\) \\
        \(50d = 40d + 240\) \\
        Subtract \(40d\) from both sides: \\
        \(10d = 240\) \\
        Divide by 10: \\
        \(d = 24\) km \\
        Total distance for both trips: \(24 + (24 + 6) = 54\) km \\
        Total time for both trips: \(\frac{24}{40} + \frac{30}{50} = 1.2\) hours
        \end{solutionordottedlines}
        \question[4] Solve these inequalities:
        \begin{parts}
            \part \[\frac{5-2 x}{3}-\frac{5+2 x}{4} \geq-1\]
        \begin{solutionordottedlines}[1in]
        Find a common denominator and combine: \\
        \(\frac{4(5 - 2x) - 3(5 + 2x)}{12} \geq -1\) \\
        \(\frac{20 - 8x - 15 - 6x}{12} \geq -1\) \\
        \(\frac{5 - 14x}{12} \geq -1\) \\
        Multiply both sides by 12: \\
        \(5 - 14x \geq -12\) \\
        Add 14x to both sides: \\
        \(5 + 14x \geq -12 + 14x\) \\
        Subtract 5 from both sides: \\
        \(14x \geq -17\) \\
        Divide by 14: \\
        \(x \geq -\frac{17}{14}\) \\
        \(x \geq -1.\overline{214285714285714285}\)
        \end{solutionordottedlines}
            \part \[\frac{2 x-1}{7} \geq \frac{2 x+3}{4}\]
        \begin{solutionordottedlines}[1in]
        Cross multiply to compare: \\
        \(4(2x - 1) \geq 7(2x + 3)\) \\
        \(8x - 4 \geq 14x + 21\) \\
        Subtract \(8x\) from both sides: \\
        \(-4 \geq 6x + 21\) \\
        Subtract 21 from both sides: \\
        \(-25 \geq 6x\) \\
        Divide by 6 (and reverse the inequality): \\
        \(-\frac{25}{6} \leq x\) \\
        \(x \geq -\frac{25}{6}\) \\
        \(x \geq -4.\overline{1}\)
        \end{solutionordottedlines}
            \part \[\frac{6 x+1}{3}>\frac{3 x-1}{2}+3\]
        \begin{solutionordottedlines}[1in]
        Find a common denominator and combine: \\
        \(\frac{2(6x + 1) - 3(3x - 1)}{6} > 3\) \\
        \(\frac{12x + 2 - 9x + 3}{6} > 3\) \\
        \(\frac{3x + 5}{6} > 3\) \\
        Multiply both sides by 6: \\
        \(3x + 5 > 18\) \\
        Subtract 5 from both sides: \\
        \(3x > 13\) \\
        Divide by 3: \\
        \(x > \frac{13}{3}\) \\
        \(x > 4.\overline{3}\)
        \end{solutionordottedlines}
            \part \[\frac{6-3 x}{2} \geq \frac{3 x-6}{5}\]
        \begin{solutionordottedlines}[1in]
        Cross multiply to compare: \\
        \(5(6 - 3x) \geq 2(3x - 6)\) \\
        \(30 - 15x \geq 6x - 12\) \\
        Add \(15x\) to both sides: \\
        \(30 \geq 21x - 12\) \\
        Add 12 to both sides: \\
        \(42 \geq 21x\) \\
        Divide by 21: \\
        \(2 \geq x\) \\
        \(x \leq 2\)
        \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{problembox}