A - Numerical Skills/Pre-Algebra
Problem 1:
Solution
| Use order of operation to evaluate multiplication and division first from left to right | \( 9 \div 3 \cdot 2 = 6 \) |
| Insert the result in the whole expression | \( 72 - 6 + 2 \) |
| Evaluate addition and subtraction from left to right | \( 72 - 6 + 2 = 68 \) |
Problem 2: In scientific notation, \( 3.0 \times 10^{-5} + 0.0000022 = \)
Solution
| Rewrite \(0.0000022\) using \(10^{-5}\) | \( 0.0000022 = 0.22 \times 10^{-5} \) |
| Add the terms | \( 3.0 \times 10^{-5} + 0.22 \times 10^{-5} = 10^{-5}(3.0 + 0.22) = 3.22 \times 10^{-5} \) |
Problem 3: If \[ \frac{2}{3} + \left(\frac{1}{2} - \frac{1}{6}\right) + \left(\frac{1}{3} - \frac{3}{4}\right) \] is calculated and reduced to simplest form, what is its numerator?
Solution
| Reduce all fractions to LCD 12 | \( \frac{8}{12} + \left(\frac{6}{12} - \frac{2}{12}\right) + \left(\frac{4}{12} - \frac{9}{12}\right) = \frac{8 + 6 - 2 + 4 - 9}{12} = \frac{7}{12} \) |
| Numerator after reduction | \( 7 \) |
Problem 4: \[ \frac{1}{4} + \left(\frac{4}{3} \times \frac{3}{5}\right) - \left(\frac{1}{4} \div \frac{4}{5}\right) = \]
Solution
| Evaluate brackets | \( \frac{4}{3} \times \frac{3}{5} = \frac{4}{5},\quad \frac{1}{4} \div \frac{4}{5} = \frac{1}{4} \times \frac{5}{4} = \frac{5}{16} \) |
| Substitute back | \( \frac{1}{4} + \frac{4}{5} - \frac{5}{16} \) |
| LCD 80 | \( \frac{20}{80} + \frac{64}{80} - \frac{25}{80} = \frac{59}{80} \) |
Problem 5: A shop owner increased the selling price of a shirt from $20 to $26. By what percentage was the price increased?
Solution
| Original price | \(20\) |
| Selling price | \(26\) |
| Difference | \(26 - 20 = 6\) |
| Relative change | \(\frac{6}{20} = 0.30\) |
| Percent increase | \(30\%\) |
Problem 6: \( \frac{3}{4} + 0.85 + 20\% = \)
Solution
| Convert to decimals | \( \frac{3}{4} = 0.75,\; 20\% = 0.2 \) |
| Sum | \( 0.75 + 0.85 + 0.2 = 1.8 \) |
Problem 7: Tom worked 6 hours at $5.50/hr, bought two magazines at $9.50 each and a pen at $8.25. How much money left?
Solution
| Earnings | \(6 \times 5.50 = 33\) |
| Magazines | \(2 \times 9.50 = 19\) |
| Pen | \(8.25\) |
| Total spent | \(19 + 8.25 = 27.25\) |
| Money left | \(33 - 27.25 = 5.75\) |
Problem 8: Bill bought \(1\frac{3}{4}\) lb cheddar, \(3.75\) lb blue, \(4\frac{1}{2}\) lb goat. Total pounds?
Solution
| Convert mixed numbers | \(1\frac{3}{4}=1.75,\; 4\frac{1}{2}=4.5\) |
| Sum | \(1.75 + 3.75 + 4.5 = 10\) pounds |
Problem 9: 40% of students are 8 or younger; remaining 120 students. How many are 8 or younger?
Solution
| Let total students = \(x\) | \(0.4x\) are ≤8, remaining \(0.6x = 120\) |
| Solve for \(x\) | \(x = \frac{120}{0.6} = 200\) |
| Students ≤8 | \(0.4 \times 200 = 80\) |
Problem 10: 1/5 have no siblings; of remaining, 40% have one sibling. Percent with >1 sibling?
Solution
| Let total = \(x\) | No siblings: \(x/5\); with siblings: \(4x/5\) |
| One sibling: 40% of \(4x/5\) | \(0.4 \cdot \frac{4x}{5} = \frac{1.6x}{5} = 0.32x\) |
| >1 sibling: \( \frac{4x}{5} - 0.32x = 0.8x - 0.32x = 0.48x\) | \(48\%\) |
Problem 11: Car rental: Plan A $20 + $0.05/km, Plan B $15 + $0.07/km. For what km are costs equal?
Solution
| Let \(x\) = km | \(C_A = 20 + 0.05x,\; C_B = 15 + 0.07x\) |
| Set equal | \(20 + 0.05x = 15 + 0.07x\) |
| Solve | \(5 = 0.02x \Rightarrow x = 250\) km |
Problem 12: Joe worked \(x\) hours earned \(y\) dollars. How much for \(z\) hours?
Solution
| Rate per hour | \(\frac{y}{x}\) |
| Earnings for \(z\) hours | \(\frac{y}{x} \cdot z = \frac{yz}{x}\) |
Problem 13: Class of 30 avg 80; 20 girls avg 85. Average of 10 boys?
Solution
| Total marks all | \(30 \times 80 = 2400\) |
| Total marks girls | \(20 \times 85 = 1700\) |
| Total marks boys | \(2400 - 1700 = 700\) |
| Boys average | \(700 / 10 = 70\) |
Problem 14: \( \frac{y}{5} = \frac{10}{25} \), find \(y\).
Solution
| Cross multiply | \(25y = 5 \times 10\) |
| Solve | \(y = 2\) |
Problem 15: \(\sqrt{3^2 + 4^2} =\)
Solution
| Simplify inside | \(9 + 16 = 25\) |
| Square root | \(\sqrt{25} = 5\) |
B - Algebra
Problem 16: If \(a = -2\) and \(b = -2\), find \(\frac{a^3 - 1}{b - 1}\).
Solution
| Substitute values | \(\frac{(-2)^3 - 1}{-2 - 1} = \frac{-8 - 1}{-3} = \frac{-9}{-3} = 3\) |
Problem 17: If \(-2(x + 9) = 20\), then \(-4x = ?\)
Solution
| Solve for \(x\) | \(-2x - 18 = 20 \Rightarrow -2x = 38 \Rightarrow x = -19\) |
| Compute \(-4x\) | \(-4(-19) = 76\) |
Problem 18: If 20% of \(x\) added to its fifth equals seven tenths of \(x\) minus 12, find \(x\).
Solution
| Translate | \(0.2x + 0.2x = 0.7x - 12\) |
| Simplify | \(0.4x = 0.7x - 12 \Rightarrow 12 = 0.3x \Rightarrow x = 40\) |
Problem 19: Trapezoid area \(A = 0.5(b+B)h\). Express \(B\) in terms of \(A, b, h\).
Solution
| Multiply both sides by 2 | \(2A = (b+B)h\) |
| Divide by \(h\) | \(\frac{2A}{h} = b+B\) |
| Solve for \(B\) | \(B = \frac{2A}{h} - b\) |
Problem 20: Tom, Linda, Alex have $120. Alex has 1/3 of Tom, Linda twice Alex. How much does Linda have?
Solution
| Let Tom = \(x\) | Alex = \(x/3\), Linda = \(2x/3\) |
| Sum equation | \(x + \frac{x}{3} + \frac{2x}{3} = 120 \Rightarrow 2x = 120 \Rightarrow x = 60\) |
| Linda's amount | \(\frac{2}{3}\times 60 = 40\) dollars |
Problem 21: Which is equivalent to \(6x^2 - 11x - 2\)?
Solution
| Factor by trial | \((6x+1)(x-2) = 6x^2 -12x + x -2 = 6x^2 -11x -2\) |
Problem 22: Factor of \(x^2 - 7x - 8\)?
Solution
| Factor | \((x+1)(x-8)\), so factor is \(x+1\) or \(x-8\) |
Problem 23: \((2xy^2 - 3x^2y) - (2x^2y^2 - 4x^2y) = \)
Solution
| Distribute minus | \(2xy^2 - 3x^2y - 2x^2y^2 + 4x^2y = 2xy^2 - 2x^2y^2 + x^2y\) |
Problem 24: Stuart drove \(x\) miles for 2 hours, 200 miles for 3 hours. Average speed 70 mph. Find \(x\).
Solution
| Total distance = \(x+200\), total time = 5 h | \(\frac{x+200}{5} = 70 \Rightarrow x+200 = 350 \Rightarrow x=150\) miles |
Problem 25: Which two lines are perpendicular? (I) \(2y+3x=3\), (II) \(-3y-2x=5\), (III) \(-6y+4x=9\), (IV) \(2y+6x=9\).
Solution
| Slopes: (I) \(m=-3/2\), (II) \(m=-2/3\), (III) \(m=2/3\), (IV) \(m=-3\) | Product \((-3/2)\cdot(2/3) = -1\), so (I) and (III) are perpendicular. |
Problem 26: If \(f(x) = (x+1)^2\), then \(f(t+2) = \)
Solution
| Substitute \(t+2\) for \(x\) | \(f(t+2) = (t+2+1)^2 = (t+3)^2 = t^2 + 6t + 9\) |
Problem 27: For \(x>0,y>0\): \((\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y}) - (\sqrt{x}-\sqrt{y})^2 =\)
Solution
| First term = \(x-y\) | Second term = \(x + y - 2\sqrt{xy}\) |
| Subtract | \((x-y) - (x+y-2\sqrt{xy}) = -2y + 2\sqrt{xy}\) |
Problem 28: Slope of line \(\frac{x}{2} - \frac{y}{4} = 7\)
Solution
| Multiply by 4: \(2x - y = 28\) | \(y = 2x - 28\), slope = 2 |
Problem 29: For \(x>3\): \(\left(\frac{x}{x-3} + \frac{1}{2}\right)\left(\frac{2}{x-1}\right) =\)
Solution
| Simplify inside: \(\frac{2x + (x-3)}{2(x-3)} = \frac{3x-3}{2(x-3)}\) | Multiply by \(\frac{2}{x-1}\): \(\frac{3(x-1)}{2(x-3)}\cdot \frac{2}{x-1} = \frac{3}{x-3}\) |
Problem 30: A(2,1), M(3,2) midpoint of AB. Find B coordinates.
Solution
| Midpoint formula | \( (3,2) = \left(\frac{2+x_B}{2}, \frac{1+y_B}{2}\right)\) |
| Solve | \(x_B=4,\; y_B=3\) |
C - College Algebra
Problem 31: Geometric sequence: \(X, 64, \_ , \_ , 8\). Find \(X\).
Solution
| Let ratio \(r\): terms \(X, rX, r^2X, r^3X, r^4X\) | \(rX = 64,\; r^4X = 8\) |
| Divide: \(r^3 = 1/8 \Rightarrow r=1/2\) | \(X = 64 / r = 128\) |
Problem 32: For what \(K\) does the system have no solution? \(2x+5y=9,\; -3x - Ky = 4\)
Solution
| Determinant = 0 for no solution | \(2(-K) - 5(-3) = -2K + 15 = 0 \Rightarrow K = 15/2\) |
Problem 33: \(f(x)=\sqrt{x+1},\; g(x)=|x-1|\). Find \(f(g(9))\).
Solution
| \(g(9)=|9-1|=8\) | \(f(8)=\sqrt{8+1}=3\) |
Problem 34: Range of \(f(x)=x^2-2x+1\)
Solution
| Complete square: \((x-1)^2\) | Minimum value 0, range \([0,\infty)\) |
Problem 35: Linear \(f\) with \(f(-1)=11,\; f(2)=5\). Find \(f(0)\).
Solution
| \(f(x)=mx+b\) | \(-m+b=11,\; 2m+b=5\) |
| Solve: subtract equations → \(3m=-6 \Rightarrow m=-2\) | \(b=9\), so \(f(0)=9\) |
Problem 36: Always increasing functions? (I) \(-x^2+2x+1\), (II) \(2x-1000\), (III) \(|x-4|+255\), (IV) \(-e^{-x}\)
Solution
| (I) quadratic (opens down) no; (II) slope 2 → yes; (III) V-shaped no; (IV) derivative \(e^{-x}>0\) → yes | Always increasing: (II) and (IV) |
Problem 37: \(i^{99} = ?\)
Solution
| \(i^4=1\), \(99=4\cdot24+3\) | \(i^{99}=i^3 = -i\) |
Problem 38: Matrix equation: \(\begin{bmatrix}3&2\\-1&-2\end{bmatrix} \begin{bmatrix}0&a\\b&c\end{bmatrix} = \begin{bmatrix}-4&9\\-4&-7\end{bmatrix}\). Find \(a,b,c\).
Solution
| Multiply: first row, first col: \(3\cdot0+2b=-4 \Rightarrow b=-2\) | First row, second col: \(3a+2c=9\) Second row, second col: \(-a-2c=-7\) |
| Solve system: add equations → \(2a=2 \Rightarrow a=1\), then \(c=3\) | \(a=1,b=-2,c=3\) |
Problem 39: Arithmetic series: 3rd term =114, last term =-27, sum=2325. Find last three terms.
Solution (condensed)
| Let \(a\) first, \(d\) common diff: \(a+2d=114\), \(a+(n-1)d=-27\), sum \(\frac{n}{2}(a-27)=2325\) | Solve to get \(n=50,\; d=-3,\; a=120\). Last three terms: -27, -24, -21 |
D - Geometry
Problem 40: Parallel lines, which set has 4 equal angles?
Solution
Angles j,k vertical; f,g vertical; j and f corresponding. Set {j,k,f,g} all equal → option D.
Problem 41: Cube edge \(x\) and cylinder (r=5,h=5) have equal volume. Find \(x\).
Solution
\(V_{cyl}= \pi (5)^2(5)=125\pi\); \(x^3=125\pi \Rightarrow x=5\sqrt[3]{\pi}\) cm.
Problem 42: Area of equilateral triangle side 20 cm.
Solution
Altitude \(h = \frac{\sqrt{3}}{2}\times 20 = 10\sqrt{3}\); Area = \(\frac{1}{2}\times 20 \times 10\sqrt{3}=100\sqrt{3}\) cm².
Problem 43: Two circles tangent, isosceles triangles. Find \(\angle CBE\).
Solution
Right isosceles triangles: \(\angle CBA=45^\circ\), \(\angle EBD=45^\circ\). Points A,B,D collinear → \(\angle CBE = 180^\circ - 45^\circ - 45^\circ = 90^\circ\).
Problem 44: EB and FC ⟂ AC, AB=x, AC=11, EB=4, FC=9. Find x.
Solution
\(\triangle ABE \sim \triangle ACF\): \(\frac{x}{11}=\frac{4}{9} \Rightarrow x=44/9\).
Problem 45: Right triangle legs \(x,2x\), hypotenuse 5. Find \(x\).
Solution
\(x^2+(2x)^2=5^2 \Rightarrow 5x^2=25 \Rightarrow x=\sqrt{5}\).
Problem 46: Circle radius 10, OA=15, OB=20, AB tangent. Area of \(\triangle ABO\).
Solution
\(AT=\sqrt{15^2-10^2}=5\sqrt{5}\), \(BT=\sqrt{20^2-10^2}=10\sqrt{3}\). Area = \(\frac12 \times 10 \times (5\sqrt{5}+10\sqrt{3}) = 25\sqrt{5}+50\sqrt{3}\) cm².
Problem 47: Trapezoid area 128, bases 10,22, find x (leg).
Solution
Height \(h\): \(\frac12 h(10+22)=128 \Rightarrow h=8\). Projection = \((22-10)/2=6\). Then \(x=\sqrt{8^2+6^2}=10\).
Problem 48: Three tangent circles, large radius \(R\). Diameter C1 twice C3. Shaded area.
Solution
Diameters: \(2R = d_1+d_3\), \(d_1=2d_3 \Rightarrow d_3=2R/3,\; r_3=R/3,\; d_1=4R/3,\; r_1=2R/3\). Shaded area = \(\pi R^2 - \pi (2R/3)^2 - \pi (R/3)^2 = \pi R^2(1 - 4/9 - 1/9)= \frac{4}{9}\pi R^2\).
E - Trigonometry
Problem 49: \(\cos x = 1/3\), \(0^\circ \(\sin x = \sqrt{1-\cos^2 x} = \sqrt{1-1/9} = \frac{2\sqrt{2}}{3}\).Solution
Problem 50: Convert \(135^\circ\) to radians.
Solution
\(135^\circ \times \frac{\pi}{180^\circ} = \frac{3\pi}{4}\).
Problem 51: Right triangle, angle 40°, opposite leg 5 cm. Find other leg.
Solution
\(\tan 40^\circ = \frac{5}{\text{adjacent}} \Rightarrow \text{adjacent} = \frac{5}{\tan 40^\circ}\) cm.
Problem 52: Graph of \(y = \cos(2x)\).
Solution
At \(x=0\), \(y=1\). Period = \(2\pi/2 = \pi\). Graph B matches.
Problem 53: Smallest positive \(x\) for which \(y=\cos(3x)\) has minimum.
Solution
Cosine minimum at argument \(\pi\): \(3x=\pi \Rightarrow x=\pi/3\).
Problem 54: Regular hexagon side 6, made of 6 equilateral triangles. Area?
Solution
Triangle altitude = \(3\sqrt{3}\), area = \(\frac12\times6\times3\sqrt{3}=9\sqrt{3}\). Hexagon area = \(6\times9\sqrt{3}=54\sqrt{3}\) sq units.