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This guide is designed to make you untouchable on GMAT absolute value questions. It covers every possible variation, provides bulletproof strategies, and includes next-level insights that most test-takers miss.
1. Absolute Value Core Concepts (The Foundation)
Definition & Key Properties
Geometric Meaning: Distance from zero on the number line.
Example: |5| = 5, |-3| = 3
Algebraic Definition:
|x| = { x if x ≥ 0
-x if x < 0 }
Critical Properties:
1. Always non-negative: |x| ≥ 0
2. Square root connection: √(x2) = |x| (vital for DS questions)
3. Multiplication & Division rules:
|xy| = |x||y|
|x/y| = |x| / |y| (y ≠ 0)
4. Triangle inequality:
|x + y| ≤ |x| + |y|
Common Misconceptions
|x| ≠ x when x is negative
√(x2) ≠ x if x is negative; it's |x|
|x + y| ≠ |x| + |y| unless x and y have the same sign
2. Solving Absolute Value Equations (All Cases)
Basic Equations
Form: |expression| = k
If k > 0: Solve expression = k and expression = -k
If k = 0: Solve expression = 0
If k < 0: No solution (absolute value can't be negative)
Example:
|2x - 3| = 7
→ 2x - 3 = 7 → x = 5
→ 2x - 3 = -7 → x = -2
Nested Absolute Values
Peel layers like an onion.
Example: ||x - 1| - 3| = 2
Step 1: Solve inner absolute value:
Case 1: |x - 1| - 3 = 2 → |x - 1| = 5 → x = 6 or -4
Case 2: |x - 1| - 3 = -2 → |x - 1| = 1 → x = 2 or 0
Absolute Value on Both Sides
Key Insight: |x| = |y| implies x = y or x = -y
Example:
|2x + 1| = |x - 3|
→ Case 1: 2x + 1 = x - 3 → x = -4
→ Case 2: 2x + 1 = -(x - 3) → x = 2/3
Variable Inside and Outside
Example: |x - 4| = x
Step 1: Recognize x must be ≥ 0
Step 2:
Case 1: x - 4 = x → No solution
Case 2: -(x - 4) = x → -x + 4 = x → x = 2
3. Absolute Value Inequalities (Complete Breakdown)
Basic Forms
1. |x| < k → -k < x < k (conjunction)
2. |x| > k → x < -k or x > k (disjunction)
Memory Tip:
"Less thAND" means AND (conjunction),
"GreatOR" means OR (disjunction)
Variable on One Side
Example: |x + 2| < 2x - 1
Step 1: Constraint → 2x - 1 > 0 → x > 0.5
Step 2: Convert to compound inequality:
-2x + 1 < x + 2 < 2x - 1
Solve:
Right: x + 2 < 2x - 1 → x > 3
Left: -2x + 1 < x + 2 → x > -1/3 (automatically true if x > 3)
Final: x > 3
Special Cases
|x| > -3 → Always true (all real x)
|x| < -2 → Never true (no solution)
4. Advanced Applications (700+ Level)
Optimization Problems
Example:
Find the minimum value of f(x) = |x - 3| + |x + 1| + |x - 5|
Step 1: Identify critical points: x = -1, 3, 5
Step 2: Analyze intervals:
x ≤ -1: f(x) = -3x + 7 → f(-1) = 10
-1 < x ≤ 3: f(x) = -x + 9 → f(3) = 6
3 < x ≤ 5: f(x) = x + 3 → f(3) = 6
x > 5: f(x) = 3x - 7 → f(5) = 8
Answer: Minimum value is 6 at x = 3
Graphical Interpretation
y = |x|: V-shaped graph centered at (0, 0)
y = |x - h| + k: Vertex at (h, k)
Use graphs to quickly estimate or verify solutions.
Piecewise Representation
Convert absolute value expressions into piecewise-defined functions.
Example:
|x - 2| =
x - 2 when x < 2
x - 2 when x ≥ 2
Helps in graphing and integration with inequalities.
5. Data Sufficiency Strategies
Key Approaches
1. Rephrase the Question
Example: "Is |x| = 3?" → "Is x = 3 or -3?"
2. Always Test Both Signs
Example: For |x - 2| < 5, test values on either side of 2
3. Apply Constraints
Example: If |x| = y, then y ≥ 0 must be satisfied
4. Consider Counterexamples
Even if a statement feels sufficient, test multiple values for x
Common Traps
Assuming |x| implies x is positive (don’t forget 0)
Ignoring negative solutions
Overlooking domain restrictions
6. Practice Problems with Explanations
Problem 1 (Medium)
How many solutions does |x + 3| - 2|x - 1| = 0 have?
Step 1: Identify critical points: x = -3 and x = 1
Step 2: Solve in intervals:
x ≤ -3 → both expressions negative
→ -x - 3 - 2(-x + 1) = 0 → x = 5 (not in range)
-3 < x ≤ 1 → mixed signs
→ x + 3 - 2(-x + 1) = 0 → x = -1/3 (valid)
x > 1 → both positive
→ x + 3 - 2(x - 1) = 0 → x = 5 (valid)
Answer: 2 solutions (x = -1/3 and x = 5)
Problem 2 (Hard)
If |2x + 5| = |3x - 2|, what is the sum of all possible x values?
Case 1: 2x + 5 = 3x - 2 → x = 7
Case 2: 2x + 5 = -(3x - 2) → 2x + 5 = -3x + 2 → 5x = -3 → x = -0.6
Sum = 7 + (-0.6) = 6.4
7. Pro Tips for Perfection
Always plug solutions back into the original equation
Draw number lines for inequalities
Memorize the standard forms and response patterns
For |x| = |y|, immediately try x = y and x = -y
8. Challenge Yourself
Try solving without writing equations immediately. Use number sense, estimation, and logical constraints.
Example:
If |x - 3| + |x - 1| = 10, what is the range of possible x values?
Try testing values between 1 and 3 for minimums, then graph.
1. Absolute Value Core Concepts (The Foundation)
Definition & Key Properties
Geometric Meaning: Distance from zero on the number line.
Example: |5| = 5, |-3| = 3
Algebraic Definition:
|x| = { x if x ≥ 0
-x if x < 0 }
Critical Properties:
1. Always non-negative: |x| ≥ 0
2. Square root connection: √(x2) = |x| (vital for DS questions)
3. Multiplication & Division rules:
|xy| = |x||y|
|x/y| = |x| / |y| (y ≠ 0)
4. Triangle inequality:
|x + y| ≤ |x| + |y|
Common Misconceptions
|x| ≠ x when x is negative
√(x2) ≠ x if x is negative; it's |x|
|x + y| ≠ |x| + |y| unless x and y have the same sign
2. Solving Absolute Value Equations (All Cases)
Basic Equations
Form: |expression| = k
If k > 0: Solve expression = k and expression = -k
If k = 0: Solve expression = 0
If k < 0: No solution (absolute value can't be negative)
Example:
|2x - 3| = 7
→ 2x - 3 = 7 → x = 5
→ 2x - 3 = -7 → x = -2
Nested Absolute Values
Peel layers like an onion.
Example: ||x - 1| - 3| = 2
Step 1: Solve inner absolute value:
Case 1: |x - 1| - 3 = 2 → |x - 1| = 5 → x = 6 or -4
Case 2: |x - 1| - 3 = -2 → |x - 1| = 1 → x = 2 or 0
Absolute Value on Both Sides
Key Insight: |x| = |y| implies x = y or x = -y
Example:
|2x + 1| = |x - 3|
→ Case 1: 2x + 1 = x - 3 → x = -4
→ Case 2: 2x + 1 = -(x - 3) → x = 2/3
Variable Inside and Outside
Example: |x - 4| = x
Step 1: Recognize x must be ≥ 0
Step 2:
Case 1: x - 4 = x → No solution
Case 2: -(x - 4) = x → -x + 4 = x → x = 2
3. Absolute Value Inequalities (Complete Breakdown)
Basic Forms
1. |x| < k → -k < x < k (conjunction)
2. |x| > k → x < -k or x > k (disjunction)
Memory Tip:
"Less thAND" means AND (conjunction),
"GreatOR" means OR (disjunction)
Variable on One Side
Example: |x + 2| < 2x - 1
Step 1: Constraint → 2x - 1 > 0 → x > 0.5
Step 2: Convert to compound inequality:
-2x + 1 < x + 2 < 2x - 1
Solve:
Right: x + 2 < 2x - 1 → x > 3
Left: -2x + 1 < x + 2 → x > -1/3 (automatically true if x > 3)
Final: x > 3
Special Cases
|x| > -3 → Always true (all real x)
|x| < -2 → Never true (no solution)
4. Advanced Applications (700+ Level)
Optimization Problems
Example:
Find the minimum value of f(x) = |x - 3| + |x + 1| + |x - 5|
Step 1: Identify critical points: x = -1, 3, 5
Step 2: Analyze intervals:
x ≤ -1: f(x) = -3x + 7 → f(-1) = 10
-1 < x ≤ 3: f(x) = -x + 9 → f(3) = 6
3 < x ≤ 5: f(x) = x + 3 → f(3) = 6
x > 5: f(x) = 3x - 7 → f(5) = 8
Answer: Minimum value is 6 at x = 3
Graphical Interpretation
y = |x|: V-shaped graph centered at (0, 0)
y = |x - h| + k: Vertex at (h, k)
Use graphs to quickly estimate or verify solutions.
Piecewise Representation
Convert absolute value expressions into piecewise-defined functions.
Example:
|x - 2| =
x - 2 when x < 2
x - 2 when x ≥ 2
Helps in graphing and integration with inequalities.
5. Data Sufficiency Strategies
Key Approaches
1. Rephrase the Question
Example: "Is |x| = 3?" → "Is x = 3 or -3?"
2. Always Test Both Signs
Example: For |x - 2| < 5, test values on either side of 2
3. Apply Constraints
Example: If |x| = y, then y ≥ 0 must be satisfied
4. Consider Counterexamples
Even if a statement feels sufficient, test multiple values for x
Common Traps
Assuming |x| implies x is positive (don’t forget 0)
Ignoring negative solutions
Overlooking domain restrictions
6. Practice Problems with Explanations
Problem 1 (Medium)
How many solutions does |x + 3| - 2|x - 1| = 0 have?
Step 1: Identify critical points: x = -3 and x = 1
Step 2: Solve in intervals:
x ≤ -3 → both expressions negative
→ -x - 3 - 2(-x + 1) = 0 → x = 5 (not in range)
-3 < x ≤ 1 → mixed signs
→ x + 3 - 2(-x + 1) = 0 → x = -1/3 (valid)
x > 1 → both positive
→ x + 3 - 2(x - 1) = 0 → x = 5 (valid)
Answer: 2 solutions (x = -1/3 and x = 5)
Problem 2 (Hard)
If |2x + 5| = |3x - 2|, what is the sum of all possible x values?
Case 1: 2x + 5 = 3x - 2 → x = 7
Case 2: 2x + 5 = -(3x - 2) → 2x + 5 = -3x + 2 → 5x = -3 → x = -0.6
Sum = 7 + (-0.6) = 6.4
7. Pro Tips for Perfection
Always plug solutions back into the original equation
Draw number lines for inequalities
Memorize the standard forms and response patterns
For |x| = |y|, immediately try x = y and x = -y
8. Challenge Yourself
Try solving without writing equations immediately. Use number sense, estimation, and logical constraints.
Example:
If |x - 3| + |x - 1| = 10, what is the range of possible x values?
Try testing values between 1 and 3 for minimums, then graph.
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