Inequalities & Absolute Value - GMAT Quantitative Reasoning

If real numbers $a$ and $b$ satisfy $|a-2|=|b+3|$ and $a+b=2,$ what is $|a-b|\ ?$
Answer: 5 When you encounter absolute value equations with multiple variables, the key is systematically considering all possible cases based on the signs of the expressions inside the absolute values. Given $|a-2|=|b+3|$ and $a+b=2$ , you need to analyze when each absolute value expression is positive or negative. The equation $|a-2|=|b+3|$ means either $a-2=b+3$ or $a-2=-(b+3)$ . Case 1: If $a-2=b+3$ , then $a=b+5$ . Substituting into $a+b=2$ gives $(b+5)+b=2$ , so $2b=-3$ and $b=-\frac{3}{2}$ . Therefore $a=\frac{7}{2}$ . Case 2: If $a-2=-(b+3)$ , then $a-2=-b-3$ , so $a=-b-1$ . Substituting into $a+b=2$ gives $(-b-1)+b=2$ , which simplifies to $-1=2$ —impossible. So the only solution is $a=\frac{7}{2}$ and $b=-\frac{3}{2}$ . Therefore $|a-b|=\left|\frac{7}{2}-\left(-\frac{3}{2}\right)\right|=\left|\frac{7}{2}+\frac{3}{2}\right|=|5|=5$ . Answer choice (A) 2 might result from incorrectly calculating $|a+b|$ instead of $|a-b|$ . Choice (B) 3 could come from mishandling the absolute value cases or arithmetic errors. Choice (C) 4 might arise from sign errors when solving the system of equations. Strategy tip: With absolute value systems, always check both cases (positive and negative scenarios) and verify your solutions satisfy the original constraints. Many students rush through case analysis and miss valid solutions or include impossible ones.
Let real number $s$ satisfy $|s+2|+|s-5|<10$ . Which of the following describes all possible values of $s$ ?
Answer: $-3.5<s<6.5$ When you encounter absolute value inequalities on the GMAT, the key is to analyze different cases based on where the expressions inside the absolute values change sign. The critical points here are $s = -2$ and $s = 5$ , which divide the number line into three regions. For $s < -2$ : Both $(s+2)$ and $(s-5)$ are negative, so $|s+2| = -(s+2)$ and $|s-5| = -(s-5)$ . The inequality becomes $-(s+2) - (s-5) < 10$ , which simplifies to $-2s + 3 < 10$ , giving us $s > -3.5$ . Combined with $s < -2$ , we get $-3.5 < s < -2$ . For $-2 \leq s < 5$ : Here $(s+2) \geq 0$ and $(s-5) < 0$ , so the inequality becomes $(s+2) - (s-5) < 10$ , which simplifies to $7 < 10$ . This is always true, so all values in $[-2, 5)$ work. For $s \geq 5$ : Both expressions are non-negative, so we get $(s+2) + (s-5) < 10$ , which simplifies to $2s - 3 < 10$ , giving us $s < 6.5$ . Combined with $s \geq 5$ , we get $5 \leq s < 6.5$ . Combining all valid regions: $-3.5 < s < 6.5$ , which is choice B. Choice A misses the extended range beyond $[-2,5]$ . Choice C incorrectly includes the endpoints $-3.5$ and $6.5$ . Choice D unnecessarily excludes $s = 5$ , which actually satisfies the inequality. Remember: With absolute value inequalities, always check your boundary points by substituting back into the original inequality to verify inclusion or exclusion.
If $a > 0$ and $|x - a| + |x + a| \leq 6a$ , what is the length of the interval of all possible values of $x$ ?
Answer: $6a$ Since a > 0, the critical points are x = a and x = -a. For x ∈ [-a, a], we have |x - a| = a - x and |x + a| = x + a, so |x - a| + |x + a| = (a - x) + (x + a) = 2a. Since 2a ≤ 6a (because a > 0), all x ∈ [-a, a] satisfy the inequality. For x > a: |x - a| + |x + a| = (x - a) + (x + a) = 2x. The inequality 2x ≤ 6a gives x ≤ 3a. For x < -a: |x - a| + |x + a| = (a - x) + (-x - a) = -2x. The inequality -2x ≤ 6a gives x ≥ -3a. Therefore, the solution set is [-3a, 3a], which has length 6a. Choice A results from considering only half the interval. Choice B results from incorrectly calculating the range as [-2a, 2a]. Choice D results from adding the endpoints instead of finding the difference.
The inequality $(x-4)(x+1)(2-x)>0$ is equivalent to which of the following ranges for $x$ ?
Answer: $x<-1 \text{ or } 2<x<4$ When you encounter polynomial inequalities like this one, you need to find where the expression changes sign by identifying the zeros and testing intervals between them. First, find the zeros by setting each factor equal to zero: $x-4=0$ gives $x=4$ , $x+1=0$ gives $x=-1$ , and $2-x=0$ gives $x=2$ . These critical points divide the number line into four intervals: $x<-1$ , $-1<x<2$ , $2<x<4$ , and $x>4$ . Now test a point in each interval to determine the sign of $(x-4)(x+1)(2-x)$ : - For $x=-2$ : $(-6)(-1)(4)=24>0$ ✓ - For $x=0$ : $(-4)(1)(2)=-8<0$ - For $x=3$ : $(-1)(4)(-1)=4>0$ ✓ - For $x=5$ : $(1)(6)(-3)=-18<0$ The expression is positive when $x<-1$ or $2<x<4$ , which matches choice D. Choice A incorrectly includes $x>2$ instead of the bounded interval $2<x<4$ . Choice B misses the critical point at $x=2$ and incorrectly extends beyond $x=4$ . Choice C includes the interval $-1<x<2$ where the expression is actually negative, and incorrectly includes $x>4$ . Strategy tip: Always plot the zeros on a number line and systematically test each interval. Remember that the inequality is strict ( $>0$ ), so the zeros themselves aren't included in the solution. This sign-analysis method works for any polynomial inequality and prevents sign errors.
For how many integer values of $k$ does the system of inequalities $|x - 2| < k$ and $|x - 8| < k$ have at least one solution?
Answer: All integers $k > 3$ The inequality |x - 2| < k is equivalent to 2 - k < x < 2 + k, and |x - 8| < k is equivalent to 8 - k < x < 8 + k. For the system to have a solution, these intervals must overlap: (2 - k, 2 + k) ∩ (8 - k, 8 + k) ≠ ∅. The intervals overlap when max(2 - k, 8 - k) < min(2 + k, 8 + k). Since 2 - k < 8 - k and 2 + k < 8 + k, this becomes 8 - k < 2 + k, which gives 6 < 2k, so k > 3. For integer values, this means k ≥ 4. Choice A is wrong because k = 4 works. Choice B is wrong because k must be positive and greater than 3. Choice C is wrong because k = 3 gives 8 - 3 = 5 and 2 + 3 = 5, so the intervals just touch at a point but don't overlap in their interiors.
Which of the following values of $t$ satisfies both $|t-4|<3$ and $2t+1>9$ ?
Answer: 5 This question tests your ability to solve compound inequalities involving absolute values. When you see absolute value inequalities combined with linear inequalities, you need to solve each constraint separately, then find values that satisfy both conditions simultaneously. Start with the absolute value inequality $|t-4|<3$ . This means the distance between $t$ and 4 is less than 3 units. To solve this, convert it to: $-3 < t-4 < 3$ . Adding 4 to all parts gives you $1 < t < 7$ , so $t$ must be between 1 and 7. Next, solve $2t+1>9$ . Subtract 1 from both sides: $2t > 8$ . Divide by 2: $t > 4$ . So $t$ must be greater than 4. For both conditions to be true, you need $t$ to satisfy both $1 < t < 7$ AND $t > 4$ . The intersection of these conditions is $4 < t < 7$ . Now check each answer choice: Choice A (3) fails because 3 is not greater than 4, so it doesn't satisfy $2t+1>9$ . Choice B (5) works because $4 < 5 < 7$ , satisfying both inequalities. Choice C (7) fails because while $|7-4| = 3$ , we need the absolute value to be strictly less than 3, not equal to it. Choice D (9) fails because 9 is outside the range $1 < t < 7$ . Therefore, B is correct. Strategy tip: When solving compound inequalities, solve each inequality separately first, then find the intersection of the solution sets. Always verify your final answer by substituting back into both original inequalities.
If $a$ and $b$ are real numbers such that $|a - b| = 5$ and $|a + b| = 3$ , what is the value of $|a^2 - b^2|$ ?
Answer: 15 We know that a² - b² = (a + b)(a - b). We have |a - b| = 5 and |a + b| = 3. Therefore, |a² - b²| = |(a + b)(a - b)| = |a + b| · |a - b| = 3 · 5 = 15. Choice A results from adding |a - b| + |a + b| instead of multiplying. Choice C results from squaring |a - b|. Choice D results from computing |a - b|² + |a + b|² = 25 + 9.
If $|3y-7|\le2$ and $y$ is an integer, how many possible values of $2y$ are less than 10?
Answer: 2 When you encounter absolute value inequalities on the GMAT, your goal is to "unwrap" the absolute value by considering what values make the expression inside true. The inequality $|3y-7|\le2$ means the distance between $3y-7$ and zero is at most 2. This translates to: $-2 \le 3y-7 \le 2$ Solving this compound inequality: - Add 7 to all parts: $5 \le 3y \le 9$ - Divide by 3: $\frac{5}{3} \le y \le 3$ Since $\frac{5}{3} = 1.67$ , and $y$ must be an integer, the possible values are $y = 2$ and $y = 3$ . This gives us $2y = 4$ and $2y = 6$ . Both values are less than 10, so there are 2 possible values. Looking at the wrong answers: Choice A suggests only 1 value, likely from missing $y = 2$ or $y = 3$ . Choice C (3 values) might come from incorrectly including $y = 1$ , but $y = 1$ gives $|3(1)-7| = 4 > 2$ , violating our constraint. Choice D (4 values) probably results from solving incorrectly or including values like $y = 1$ and $y = 4$ , both of which don't satisfy the original inequality. Strategy tip: Always verify your integer solutions by substituting back into the original absolute value inequality. This catches calculation errors and ensures you haven't included boundary values that don't actually work.
For all real numbers $x,$ the inequality $|x|<x+2$ is equivalent to which of the following?
Answer: $x>-1$ When you encounter absolute value inequalities, you need to consider that the absolute value expression behaves differently depending on whether the variable inside is positive or negative. This requires analyzing the inequality by cases. For $|x| < x + 2$ , you must examine two scenarios. When $x \geq 0$ , we have $|x| = x$ , so the inequality becomes $x < x + 2$ , which simplifies to $0 < 2$ . This is always true, meaning all non-negative values of $x$ satisfy the original inequality. When $x < 0$ , we have $|x| = -x$ , so the inequality becomes $-x < x + 2$ . Solving this: $-x < x + 2$ leads to $-2x < 2$ , which gives us $x > -1$ . Since we're considering the case where $x < 0$ , we need $-1 < x < 0$ . Combining both cases: all $x \geq 0$ work, plus all $x$ in the interval $(-1, 0)$ . Therefore, the solution is $x > -1$ , which is choice B. Choice A ( $x > -2$ ) is too broad and would include values like $x = -1.5$ that don't satisfy the original inequality. Choice C ( $x < 1$ ) captures some correct values but misses many others, like $x = 2$ . Choice D ( $x < -1$ ) represents exactly the opposite of what we want—these are precisely the values that don't work. Remember: absolute value inequalities almost always require case analysis. Set up your cases based on where the expression inside the absolute value changes sign, then combine your results carefully.
How many integers $n$ satisfy $-20<n<5$ and $|n+3|\ge10$ ?
Answer: 7 This problem combines inequality constraints with absolute value conditions, testing your ability to work systematically through compound restrictions. Start by solving the absolute value inequality $|n+3| \geq 10$ . This means either $n+3 \geq 10$ or $n+3 \leq -10$ . Solving these gives you $n \geq 7$ or $n \leq -13$ . Now you need integers that satisfy both this absolute value condition AND the constraint $-20 < n < 5$ . Let's find the intersection of these conditions: For $n \geq 7$ : Since we also need $n < 5$ , there's no overlap here. For $n \leq -13$ : Combined with $-20 < n < 5$ , we get $-20 < n \leq -13$ . The integers in this range are: $-19, -18, -17, -16, -15, -14, -13$ . That's 7 integers. Let's verify: Each of these satisfies $|n+3| \geq 10$ . For example, $|-19+3| = |-16| = 16 \geq 10$ ✓. The answer is C) 7. Choice A) 5 likely comes from miscounting or missing some boundary values. Choice B) 6 might result from excluding one of the boundary integers like $-13$ or $-19$ . Choice D) 8 could come from incorrectly including $-20$ (which violates $n > -20$ ) or making an error in the absolute value setup. Strategy tip: With compound inequalities involving absolute values, always solve the absolute value condition first, then find the intersection with other constraints. Double-check your boundary values carefully—they're common sources of counting errors.
If $|u-3|+|v+4|=0,$ what is $|u+v|\ ?$
Answer: 1 When you encounter an equation where the sum of absolute values equals zero, remember that absolute values are always non-negative. The only way for a sum of non-negative terms to equal zero is if each term individually equals zero. Given $|u-3|+|v+4|=0$ , both $|u-3|=0$ and $|v+4|=0$ must be true simultaneously. From $|u-3|=0$ , we get $u-3=0$ , so $u=3$ . From $|v+4|=0$ , we get $v+4=0$ , so $v=-4$ . Therefore, $|u+v|=|3+(-4)|=|-1|=1$ . Looking at the wrong answers: Choice A) 0 might tempt you if you incorrectly thought that since the original equation equals zero, the answer should also be zero. Choice C) 3 could trap you if you only solved for $u$ and forgot about $v$ , or if you confused $|u+v|$ with just $|u|$ . Choice D) 7 might result from incorrectly calculating $|u|+|v|=|3|+|-4|=3+4=7$ instead of $|u+v|$ . The key insight is recognizing that when absolute values sum to zero, each component must be zero. This is a fundamental property that appears frequently on the GMAT. Always remember: if $|A|+|B|+...=0$ , then each absolute value expression equals zero individually.
For all real $x$ such that $x^{2}-6x+5\le0,$ which statement must be true?
Answer: $|x-3|\le2$ When you encounter inequality problems involving quadratic expressions, your first step should be to find where the inequality holds by factoring and analyzing the solution set. Start by solving $x^2 - 6x + 5 \leq 0$ . Factor this quadratic: $x^2 - 6x + 5 = (x-1)(x-5)$ . So you need $(x-1)(x-5) \leq 0$ . This inequality holds when one factor is positive and the other negative, which occurs when $1 \leq x \leq 5$ . This is your constraint set. Now you must determine which statement is always true for every $x$ in the interval $[1,5]$ . For choice D, $|x-3| \leq 2$ , notice that when $1 \leq x \leq 5$ , the distance from $x$ to 3 is at most 2. The farthest points from 3 in this interval are $x = 1$ and $x = 5$ , both giving $|x-3| = 2$ . Since the maximum value of $|x-3|$ on $[1,5]$ is 2, we have $|x-3| \leq 2$ for all $x$ in our constraint set. Choice A fails because $x^2 - 10x + 21 = (x-3)(x-7) < 0$ when $3 < x < 7$ , so it's negative (not $\geq 0$ ) for $x \in (3,5]$ . Choice B contradicts choice D since $|x-3| \leq 2$ means $|x-3| \geq 2$ is false for most values in our interval. Choice C fails because $x^2 - 4x + 3 = (x-1)(x-3) \leq 0$ when $1 \leq x \leq 3$ , making it non-positive in part of our interval. Strategy tip: When checking which statement must be true over an interval, test the endpoints and any critical points within that interval to verify your answer.
If real numbers $x$ and $y$ satisfy $|x-3|+|y+1|\le4,$ what is the maximum possible value of $x+y$ ?
Answer: 6 When you encounter absolute value inequalities on the GMAT, visualize them geometrically. The expression $|x-3|+|y+1|\le4$ represents the set of points $(x,y)$ whose combined distances from $x=3$ and $y=-1$ don't exceed 4. This creates a diamond-shaped region centered at $(3,-1)$ . To maximize $x+y$ , you need the point in this feasible region where the sum $x+y$ is largest. Lines of constant $x+y$ have slope $-1$ , so you want the point where the line $x+y=k$ (for maximum $k$ ) just touches the boundary of your diamond region. The diamond's vertices occur where the absolute value expressions change sign. The critical point is $(7,-1)$ , found by setting $x-3=4$ and $y+1=0$ . Let's verify: $|7-3|+|-1+1|=4+0=4\le4$ ✓ At this point, $x+y=7+(-1)=6$ . Choice (A) 4 represents a point well inside the feasible region, not the maximum. Choice (B) 5 could be achieved at points like $(6,-1)$ , but this isn't optimal. Choice (D) 7 would require points like $(8,-1)$ , but $|8-3|+|-1+1|=5>4$ , violating the constraint. Strategy tip: For optimization problems with absolute value constraints, always check the "corner" points where the absolute value expressions equal zero or reach their maximum allowed values. These boundary points typically yield optimal solutions.
If $\dfrac1{|z-2|}\le0.2$ and $z\neq2,$ which of the following must be true?
Answer: $z\le-3 \text{ or } z\ge7$ When you encounter an inequality involving absolute values in denominators, your first step is to recognize that since we're dealing with $\frac{1}{|z-2|}$ , the absolute value $|z-2|$ must be positive (it's never zero since $z \neq 2$ ). To solve $\frac{1}{|z-2|} \leq 0.2$ , multiply both sides by $|z-2|$ . Since $|z-2| > 0$ , the inequality direction stays the same: $1 \leq 0.2|z-2|$ . Dividing by 0.2 gives us $5 \leq |z-2|$ , or equivalently $|z-2| \geq 5$ . The absolute value inequality $|z-2| \geq 5$ means the distance between $z$ and 2 is at least 5 units. This occurs when $z-2 \geq 5$ or $z-2 \leq -5$ , giving us $z \geq 7$ or $z \leq -3$ . Therefore, answer D is correct. Looking at the wrong answers: A) $-5 \leq z \leq 5$ represents values close to 2, which would make $\frac{1}{|z-2|}$ large, violating our inequality. B) $z \leq -5$ or $z \geq 5$ uses the wrong boundary values—this comes from incorrectly solving $|z| \geq 5$ instead of $|z-2| \geq 5$ . C) $-3 \leq z \leq 7$ includes values near 2, which again would make the fraction too large. Strategy tip: When solving inequalities with absolute values, always check whether you need the "close to" or "far from" interpretation. Inequalities like $\frac{1}{|expression|} \leq \text{small number}$ typically require the variable to be "far from" the center value.

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