5.4 Algebra, Basic Geometry, and Quantitative Word Problems

5.4 Algebra, Basic Geometry, and Quantitative Word Problems

Core Principle: The quantitative word problem domain in the PMA Academic test synthesizes algebra, plane geometry, and physical rates. Rather than testing abstract calculus, the AS&RC computer bank focuses on structured linear relationships, basic 2D and 3D mensuration, algebraic factorization identities, and classical kinematic scenarios (speed, distance, time, and relative motion). Solving these problems under strict countdown pacing requires setting up algebraic models instantly and applying geometric shortcuts.

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Algebraic Equations and Polynomial Factoring

Algebraic questions at AS&RC evaluate speed in manipulating algebraic expressions and solving for unknown variables.

1. Linear Equations with One Variable

A linear equation in one variable takes the standard form $ax + b = c$. Isolation of $x$ is achieved through basic inverse operations: $x = \frac{c - b}{a}$.

Worked Example: Fractional Linear Equation

Solve for $x$:

$$\frac{3x - 5}{4} = \frac{2x + 1}{3}$$

• Step 1 (Cross-multiplication): $3(3x - 5) = 4(2x + 1)$
• Step 2 (Expand terms): $9x - 15 = 8x + 4$
• Step 3 (Group like terms): $9x - 8x = 4 + 15 \implies x = 19$.

2. Simultaneous Linear Equations with Two Variables

Systems of two linear equations ($a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$) are solved on scratch paper via the Elimination Method or the Substitution Method.

Worked Example: Elimination Method

Solve the system:

$$\begin{aligned} 3x + 2y &= 22 \quad \text{--- (Equation 1)} \\ 2x - y &= 3 \quad \text{--- (Equation 2)} \end{aligned}$$

• Step 1: Multiply Equation 2 by $2$ to match the coefficient of $y$:

$$2(2x - y) = 2(3) \implies 4x - 2y = 6 \quad \text{--- (Equation 3)}$$

• Step 2: Add Equation 1 and Equation 3 to eliminate $y$:

$$(3x + 2y) + (4x - 2y) = 22 + 6 \implies 7x = 28 \implies x = 4$$

• Step 3: Substitute $x = 4$ into Equation 2 to find $y$:

$$2(4) - y = 3 \implies 8 - y = 3 \implies y = 5$$

Thus, the solution is $(x, y) = (4, 5)$.

3. Core Algebraic Identities Reference

Memorizing these identities enables rapid simplification without long multiplication:

The Reciprocal Square Archetype

A classic question type in the PMA Academic paper provides $x + \frac{1}{x} = k$ and asks for $x^2 + \frac{1}{x^2}$:

• Square both sides:

$$\left(x + \frac{1}{x}\right)^2 = x^2 + 2(x)\left(\frac{1}{x}\right) + \frac{1}{x^2} = x^2 + 2 + \frac{1}{x^2}$$ $$k^2 = x^2 + \frac{1}{x^2} + 2 \implies \mathbf{x^2 + \frac{1}{x^2} = k^2 - 2}$$

• Numerical Example: If $x + \frac{1}{x} = 6$, then $x^2 + \frac{1}{x^2} = 6^2 - 2 = 36 - 2 = 34$.

4. Quadratic Equations and Middle-Term Factoring

A quadratic equation takes the standard form $ax^2 + bx + c = 0$. In AS&RC testing, quadratic equations are constructed to be factorable by splitting the middle term into two factors whose product equals $a \times c$ and whose sum equals $b$.

• Problem: Solve $x^2 - 8x + 15 = 0$.
• Identify two numbers whose product is $+15$ and sum is $-8$: these are $-3$ and $-5$.
• Factor: $(x - 3)(x - 5) = 0 \implies x = 3 \text{ or } x = 5$.
• The Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• The Discriminant ($D = b^2 - 4ac$):
  • If $D > 0$: Two distinct, real roots.
  • If $D = 0$: Exactly one repeated real root ($x = -b / 2a$).
  • If $D < 0$: No real roots (complex conjugate roots).

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Plane Geometry and Mensuration Formulas

Geometry questions focus on perimeter, interior angles, area, and basic volume formulas.

1. Triangles and Angle Properties

• Sum of Interior Angles: The sum of interior angles in any Euclidean triangle is always $180^\circ$ ($A + B + C = 180^\circ$).
• Exterior Angle Theorem: An exterior angle of a triangle equals the sum of the two opposite interior angles.
• Triangle Inequality Theorem: The sum of the lengths of any two sides of a triangle must strictly exceed the length of the third side ($a + b > c$).
• Area of a Triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
• Equilateral Triangle: All sides equal ($s$), each interior angle is $60^\circ$:

$$\text{Area} = \frac{\sqrt{3}}{4} s^2, \quad \text{Height } h = \frac{\sqrt{3}}{2} s$$

2. Right-Angled Triangles and Pythagoras' Theorem

For any right-angled triangle with perpendicular legs $a$ and $b$ and hypotenuse $c$:

$$a^2 + b^2 = c^2 \implies c = \sqrt{a^2 + b^2}$$

High-Yield Pythagorean Triples

Memorizing these integer triples eliminates the need to calculate square roots on scratch paper:

• $(3, 4, 5)$ and multiples: $(6, 8, 10), (9, 12, 15), (15, 20, 25)$
• $(5, 12, 13)$ and multiples: $(10, 24, 26), (15, 36, 39)$
• $(8, 15, 17)$ and multiples: $(16, 30, 34)$
• $(7, 24, 25)$
• $(9, 40, 41)$

3. Circles and Quadrilaterals

4. Basic 3D Solid Geometry (Mensuration)

• Cube: Volume $V = s^3$; Total Surface Area $TSA = 6s^2$; Longest Internal Diagonal $d = s\sqrt{3}$.
• Cuboid (Rectangular Box): Volume $V = l \times w \times h$; Total Surface Area $TSA = 2(lw + wh + hl)$; Internal Diagonal $d = \sqrt{l^2 + w^2 + h^2}$.
• Cylinder: Volume $V = \pi r^2 h$; Curved Surface Area $CSA = 2\pi rh$; Total Surface Area $TSA = 2\pi r(r + h)$.
• Sphere: Volume $V = \frac{4}{3}\pi r^3$; Total Surface Area $SA = 4\pi r^2$.

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Quantitative Word Problems: Age, Speed, and Train Scenarios

Word problems translate narrative scenarios into algebraic equations. Pacing requires identifying standard problem structures immediately.

1. Age Word Problems

Age problems define relationships across different time frames (past, present, future). Always establish a single variable representing the present age.

Worked Example: Age Relationship

A father is currently three times as old as his son. Eight years ago, the father was five times as old as his son was then. What are their present ages?

• Step 1 (Define present variables): Let the son's present age be $x$. The father's present age is $3x$.
• Step 2 (Formulate past relationship): Eight years ago, the son was $(x - 8)$ and the father was $(3x - 8)$.
• Step 3 (Set up equation):

$$3x - 8 = 5(x - 8)$$ $$3x - 8 = 5x - 40$$ $$40 - 8 = 5x - 3x \implies 32 = 2x \implies x = 16$$

• Conclusion: The son's present age is $16\text{ years}$; the father's present age is $3 \times 16 = 48\text{ years}$.
  (Check: 8 years ago, son was 8 and father was 40; $40 = 5 \times 8$).

2. Speed, Distance, and Time Mechanics

The kinematic relationship is defined by:

$$\text{Distance} = \text{Speed} \times \text{Time}, \quad \text{Speed} = \frac{\text{Distance}}{\text{Time}}, \quad \text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

The Essential Unit Conversion Factor

AS&RC questions routinely give speed in $\text{km/h}$ and distance in meters or time in seconds. Use the $\frac{5}{18}$ ratio for instantaneous conversion:

$$\mathbf{1\text{ km/h} = \frac{1{,}000\text{ m}}{3{,}600\text{ s}} = \frac{5}{18}\text{ m/s}}$$ $$\mathbf{1\text{ m/s} = \frac{18}{5}\text{ km/h} = 3.6\text{ km/h}}$$

• Convert $36\text{ km/h}$ to $\text{m/s}$: $36 \times \frac{5}{18} = 2 \times 5 = 10\text{ m/s}$.
• Convert $54\text{ km/h}$ to $\text{m/s}$: $54 \times \frac{5}{18} = 3 \times 5 = 15\text{ m/s}$.
• Convert $72\text{ km/h}$ to $\text{m/s}$: $72 \times \frac{5}{18} = 4 \times 5 = 20\text{ m/s}$.
• Convert $90\text{ km/h}$ to $\text{m/s}$: $90 \times \frac{5}{18} = 5 \times 5 = 25\text{ m/s}$.

3. Average Speed for Round Trips

When an object traverses a fixed distance $d$ at speed $s_1$ and returns along the same route at speed $s_2$, the average speed is the harmonic mean, never the simple arithmetic average:

$$\mathbf{\text{Average Speed} = \frac{2 \cdot s_1 \cdot s_2}{s_1 + s_2}}$$

Worked Example: A patrol vehicle drives from Base Alpha to Outpost Bravo at $30\text{ km/h}$ and returns along the exact same path at $60\text{ km/h}$. What is the average speed for the entire round trip?

$$\text{Average Speed} = \frac{2 \times 30 \times 60}{30 + 60} = \frac{3{,}600}{90} = 40\text{ km/h}$$

(Note: The common distractor is $\frac{30 + 60}{2} = 45\text{ km/h}$, which is mathematically incorrect because the vehicle spent twice as much time traveling at the slower speed).

4. Relative Speed Rules

• Objects Moving in Opposite Directions (approaching each other or moving away):

$$\mathbf{\text{Relative Speed} = s_1 + s_2}$$

• Objects Moving in the Same Direction (one chasing or overtaking the other):

$$\mathbf{\text{Relative Speed} = |s_1 - s_2|}$$

5. Train Crossing Archetypes

Train problems depend on whether the target being crossed has negligible length or extended length:

1. Crossing a Point Object of Negligible Length (telegraph pole, signal post, stationary sentry):
   • Distance to cover $= L_{\text{train}}$
   • $t = \frac{L_{\text{train}}}{s_{\text{train}}}$
2. Crossing an Extended Stationary Object (platform, railway bridge, tunnel, stationary train):
   • Distance to cover $= L_{\text{train}} + L_{\text{platform}}$
   • $t = \frac{L_{\text{train}} + L_{\text{platform}}}{s_{\text{train}}}$
3. Two Moving Trains Crossing Each Other:
   • Distance to cover $= L_1 + L_2$
   • Time in opposite directions: $t = \frac{L_1 + L_2}{s_1 + s_2}$
   • Time in same direction: $t = \frac{L_1 + L_2}{|s_1 - s_2|}$