DAT Quantitative Reasoning Quiz: Equations And Inequalities

What this quiz covers

This quiz focuses on Equations And Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for DAT Quantitative Reasoning.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\dfrac{28}{9}$ (correct answer)
2. $\dfrac{26}{9}$
3. $\dfrac{14}{9}$
4. $\dfrac{7}{3}$

Explanation: This problem tests your ability to solve linear equations involving fractions, a fundamental algebra skill that appears regularly on quantitative reasoning exams. To solve $\frac{3}{4}(x-2)=\frac{5}{6}$, you need to isolate $x$ by undoing the operations in reverse order. First, eliminate the fraction coefficient by multiplying both sides by $\frac{4}{3}$ (the reciprocal of $\frac{3}{4}$): $\frac{4}{3} \cdot \frac{3}{4}(x-2) = \frac{4}{3} \cdot \frac{5}{6}$ The left side simplifies to $(x-2)$, and the right side becomes $\frac{4 \cdot 5}{3 \cdot 6} = \frac{20}{18} = \frac{10}{9}$: $x-2 = \frac{10}{9}$ Now add 2 to both sides: $x = \frac{10}{9} + 2 = \frac{10}{9} + \frac{18}{9} = \frac{28}{9}$ Answer A ($\frac{28}{9}$) is correct. Answer B ($\frac{26}{9}$) results from incorrectly subtracting 2 instead of adding it, giving $\frac{10}{9} - 2 = -\frac{8}{9}$ then making sign errors. Answer C ($\frac{14}{9}$) comes from incorrectly multiplying by $\frac{3}{4}$ instead of its reciprocal, then adding 2. Answer D ($\frac{7}{3}$) represents $\frac{21}{9}$, which results from adding 1 instead of 2 to $\frac{10}{9}$. When solving equations with fractions, always multiply by the reciprocal to eliminate fraction coefficients, and double-check your fraction arithmetic by finding common denominators carefully.

Question 2

1. $\left\{-\dfrac12,\;4\right\}$ (correct answer)
2. $\{-2,\;4\}$
3. $\left\{\dfrac12,\;-4\right\}$
4. $\{2,\;-4\}$

Explanation: When you encounter a quadratic equation like $2x^2-7x-4=0$, you need to find the values of $x$ that make the equation true. The most reliable approach is using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. For this equation, $a=2$, $b=-7$, and $c=-4$. Substituting these values: $x = \frac{-(-7) \pm \sqrt{(-7)^2-4(2)(-4)}}{2(2)} = \frac{7 \pm \sqrt{49+32}}{4} = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4}$ This gives us two solutions: $x = \frac{7+9}{4} = \frac{16}{4} = 4$ and $x = \frac{7-9}{4} = \frac{-2}{4} = -\frac{1}{2}$ Therefore, the solution set is $\left\{-\frac{1}{2}, 4\right\}$, which matches choice A. Choice B gives $\{-2, 4\}$ - this keeps the correct positive solution but incorrectly gives $-2$ instead of $-\frac{1}{2}$. Choice C gives $\left\{\frac{1}{2}, -4\right\}$ - this has sign errors on both solutions. Choice D gives $\{2, -4\}$ - this also has incorrect values for both solutions, possibly from calculation errors in applying the quadratic formula. Always double-check your solutions by substituting them back into the original equation. For quadratic equations, the quadratic formula is foolproof when factoring seems difficult, so master this formula and be careful with your arithmetic.

Question 3

1. $x>5$ (correct answer)
2. $x<5$
3. $x>-5$
4. $x<-5$

Explanation: When you encounter an inequality with parentheses, you need to distribute first, then isolate the variable while carefully handling the inequality direction. Starting with $-3(2-x)>9$, distribute the $-3$ to both terms inside the parentheses: $-3(2) + (-3)(-x) = -6 + 3x$. So the inequality becomes $-6 + 3x > 9$. Next, add 6 to both sides: $3x > 15$. Finally, divide both sides by 3: $x > 5$. Since we divided by a positive number, the inequality direction stays the same. Looking at the wrong answers: Answer B ($x < 5$) represents the most common error - flipping the inequality sign when dividing by a positive number. Remember, you only flip the inequality when multiplying or dividing by a negative number. Answer C ($x > -5$) likely comes from incorrectly distributing the negative sign, perhaps getting $-6 - 3x > 9$, which would lead to $x < -5$, then incorrectly flipping it. Answer D ($x < -5$) results from the same distribution error but without the extra incorrect flip. You can verify the answer by testing a value: if $x = 6$, then $-3(2-6) = -3(-4) = 12$, and indeed $12 > 9$. The key strategy here is to work systematically through algebraic steps and remember that inequality direction only changes when multiplying or dividing by negative numbers - not positive ones.

Question 4

1. $\{2,\;5\}$ (correct answer)
2. $\{1,\;3\}$
3. $\{4,\;7\}$
4. $\{5\}$

Explanation: When you encounter absolute value equations, remember that the absolute value represents distance from zero on the number line. The equation $|2x-7|=3$ means the expression $2x-7$ is exactly 3 units away from zero, which gives us two possibilities. To solve $|2x-7|=3$, you need to consider both cases where the expression inside equals positive and negative 3: Case 1: $2x-7=3$ Adding 7 to both sides: $2x=10$ Dividing by 2: $x=5$ Case 2: $2x-7=-3$ Adding 7 to both sides: $2x=4$ Dividing by 2: $x=2$ You can verify: when $x=5$, we get $|2(5)-7|=|3|=3$ ✓, and when $x=2$, we get $|2(2)-7|=|-3|=3$ ✓. Looking at the answer choices: Choice A gives $\{2,5\}$, which matches our solutions perfectly. Choice B $\{1,3\}$ would give us $|2(1)-7|=5$ and $|2(3)-7|=1$, neither of which equals 3. Choice C $\{4,7\}$ yields $|2(4)-7|=1$ and $|2(7)-7|=7$, again missing our target of 3. Choice D $\{5\}$ only includes one of the two correct solutions. Remember that absolute value equations typically have two solutions unless the expression inside the absolute value bars equals zero at the boundary. Always set up both the positive and negative cases, and don't forget to check your work by substituting back into the original equation.

Question 5

1. $-5\le x\le3$ (correct answer)
2. $x\le-5\text{ or }x\ge3$
3. $-3\le x\le5$
4. $x<-5\text{ or }x>3$

Explanation: When you encounter absolute value inequalities, remember that the absolute value represents distance on a number line. The inequality $|x+1|\le4$ asks: "For which values of x is the distance from x to -1 equal to or less than 4?" To solve this systematically, use the definition that $|A| \le B$ is equivalent to $-B \le A \le B$ when B is positive. Here, we have $|x+1| \le 4$, so this becomes: $-4 \le x+1 \le 4$ Subtracting 1 from all parts: $-4-1 \le x \le 4-1$ $-5 \le x \le 3$ You can verify this by testing boundary values: when $x = -5$, we get $|-5+1| = |-4| = 4 \le 4$ ✓, and when $x = 3$, we get $|3+1| = |4| = 4 \le 4$ ✓. Choice A gives $-5 \le x \le 3$, which matches our solution perfectly. Choice B represents the solution to $|x+1| \ge 4$, the opposite inequality. Choice C shifts the interval incorrectly—this would be the solution if the original expression were $|x-1| \le 4$. Choice D uses strict inequalities and represents values where $|x+1| > 4$. Remember this key pattern: $|expression| \le number$ always gives you a single interval (AND condition), while $|expression| \ge number$ gives you two separate intervals (OR condition). Don't confuse the direction of the inequality when converting from absolute value form.

Question 6

1. $\left\{\dfrac{1-\sqrt{17}}{2},\;\dfrac{1+\sqrt{17}}{2}\right\}$ (correct answer)
2. $\left\{\dfrac{-1-\sqrt{17}}{2},\;\dfrac{-1+\sqrt{17}}{2}\right\}$
3. $\left\{\dfrac{1-\sqrt{13}}{2},\;\dfrac{1+\sqrt{13}}{2}\right\}$
4. $\left\{\dfrac{-1-\sqrt{13}}{2},\;\dfrac{-1+\sqrt{13}}{2}\right\}$

Explanation: When you encounter a system of equations where one equation is linear and the other is quadratic, you're looking for intersection points. The solution requires setting the equations equal to each other since both expressions equal $y$. Setting $x + 1 = x^2 - 3$, you need to rearrange this into standard quadratic form. Moving all terms to one side gives you $x^2 - x - 4 = 0$. Now apply the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = -1$, and $c = -4$. Substituting these values: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)} = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}$ This gives you the two solutions: $\frac{1 - \sqrt{17}}{2}$ and $\frac{1 + \sqrt{17}}{2}$, which matches choice A. Choice B uses the correct discriminant ($\sqrt{17}$) but incorrectly shows $-1$ in the numerator instead of $+1$, which would result from mishandling the sign when applying the quadratic formula. Choices C and D both incorrectly calculate the discriminant as $13$ instead of $17$, likely from computational errors when evaluating $1 + 16$. Choice D compounds this error with the same sign mistake as choice B. Remember: when solving systems with linear and quadratic equations, always double-check your algebra when rearranging to standard form, and be extra careful with signs when applying the quadratic formula—these are the most common sources of error.

Question 7

1. $10$ (correct answer)
2. $9$
3. $11$
4. $8$

Explanation: This question tests your understanding of exponential growth and requires finding when a power of 2 exceeds a given threshold. When you encounter problems asking for the "least integer n such that..." you're looking for the smallest value that satisfies the inequality. To solve $2^n > 1000$, you need to systematically check powers of 2 around the target value. Since $2^{10} = 1024$ is a common benchmark you should memorize, start there. We have $2^{10} = 1024 > 1000$, so $n = 10$ satisfies the inequality. But is this the smallest such integer? Check $n = 9$: $2^9 = 512 < 1000$. Since 512 is less than 1000, $n = 9$ doesn't work. Therefore, $n = 10$ is indeed the least integer where $2^n > 1000$, making choice (A) correct. Looking at the wrong answers: (B) $n = 9$ fails because $2^9 = 512 < 1000$. (C) $n = 11$ gives $2^{11} = 2048 > 1000$, which satisfies the inequality but isn't the least such integer. (D) $n = 8$ fails because $2^8 = 256 < 1000$. Study tip: Memorize key powers of 2 up to $2^{10} = 1024$. These appear frequently on quantitative reasoning exams. When finding "least" or "greatest" values satisfying inequalities, always check the boundary cases to ensure you haven't found a value that works but isn't optimal.

Question 8

1. $x\ge\dfrac25$ (correct answer)
2. $x\le\dfrac25$
3. $x\ge-\dfrac25$
4. $x\le-\dfrac25$

Explanation: When you encounter linear inequalities involving fractions, your goal is to isolate the variable while preserving the inequality's direction. The key is treating inequalities like equations, except when multiplying or dividing by negative numbers. Starting with $\frac{5-2x}{3} \le x+1$, first eliminate the fraction by multiplying both sides by 3: $5-2x \le 3(x+1)$. This gives you $5-2x \le 3x+3$. Next, collect all terms with $x$ on one side and constants on the other. Subtract $3x$ from both sides: $5-2x-3x \le 3$, which simplifies to $5-5x \le 3$. Then subtract 5 from both sides: $-5x \le -2$. Here's the crucial step: when dividing by $-5$ (a negative number), you must flip the inequality sign. Dividing both sides by $-5$ gives $x \ge \frac{2}{5}$. Looking at the wrong answers: Choice B ($x \le \frac{2}{5}$) results from forgetting to flip the inequality sign when dividing by a negative. Choice C ($x \ge -\frac{2}{5}$) comes from an arithmetic error, likely incorrectly handling the signs during algebraic manipulation. Choice D ($x \le -\frac{2}{5}$) combines both errors—wrong arithmetic and failing to flip the inequality. The correct answer is A: $x \ge \frac{2}{5}$. Strategy tip: Always remember that dividing or multiplying an inequality by a negative number flips the inequality sign. This is the most common mistake students make with linear inequalities, so double-check this step every time.

Question 9

1. $\dfrac32$ (correct answer)
2. $2$
3. $\dfrac43$
4. $1$

Explanation: When you encounter exponential equations like this one, the key strategy is to express both sides using the same base, which allows you to equate the exponents directly. Notice that both 9 and 27 can be written as powers of 3: $9 = 3^2$ and $27 = 3^3$. Substituting these into the original equation gives us $(3^2)^x = 3^3$. Using the power rule for exponents, $(3^2)^x = 3^{2x}$, so our equation becomes $3^{2x} = 3^3$. Since the bases are now equal, the exponents must be equal: $2x = 3$. Solving for $x$ gives us $x = \frac{3}{2}$, which is answer choice A. Let's examine why the other options don't work. Choice B ($x = 2$) would give us $9^2 = 81$, which is far too large. Choice C ($x = \frac{4}{3}$) yields $9^{4/3} = (3^2)^{4/3} = 3^{8/3} = 3^{2.67...}$, which equals approximately 15.6, still too small. Choice D ($x = 1$) simply gives us $9^1 = 9$, which is much smaller than our target of 27. You can verify: $9^{3/2} = (3^2)^{3/2} = 3^3 = 27$ ✓ Study tip: When solving exponential equations, always look for a common base first. Most DAT problems will use bases like 2, 3, 5, or 10, so practice recognizing powers of these numbers quickly.

Question 10

1. $\{2,4\}$ (correct answer)
2. $\{-2,-4\}$
3. $\{1,8\}$
4. $\{-1,4\}$

Explanation: When you encounter a quadratic equation, you need to find the values of x that make the equation true. For $\frac{1}{2}x^2 - 3x + 4 = 0$, you can solve this using the quadratic formula or by testing the given options through substitution. Let's verify by substituting the values from option A into the original equation. For x = 2: $\frac{1}{2}(2)^2 - 3(2) + 4 = \frac{1}{2}(4) - 6 + 4 = 2 - 6 + 4 = 0$ ✓ For x = 4: $\frac{1}{2}(4)^2 - 3(4) + 4 = \frac{1}{2}(16) - 12 + 4 = 8 - 12 + 4 = 0$ ✓ Both values satisfy the equation, confirming that A is correct. Option B gives negative values that, when substituted, don't satisfy the equation. For x = -2: $\frac{1}{2}(-2)^2 - 3(-2) + 4 = 2 + 6 + 4 = 12 \neq 0$. Option C fails because x = 1 gives $\frac{1}{2}(1) - 3(1) + 4 = 2.5 \neq 0$. Option D is incorrect since x = -1 yields $\frac{1}{2}(1) + 3 + 4 = 7.5 \neq 0$. For quadratic equations on standardized tests, substitution is often faster than using the quadratic formula, especially when answer choices are provided. Always check both solutions when given a pair, as partial credit traps are common where only one value from a pair might work.

Question 11

1. $14\le x\le 26$
2. $16\le x\le 26$ (correct answer)
3. $16\le x\le 28$
4. $20\le x\le 26$
5. $26\le x\le 16$

Explanation: This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves a clinic budgeting for masks, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, 16≤x≤26, is obtained by subtracting 120 from all parts and dividing by 30. A common mistake is 14≤x≤26, which results from miscalculating the lower bound. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.

Question 12

1. $4$ (correct answer)
2. $-4$
3. $\dfrac12$
4. $2$

Explanation: When you encounter a rational equation like this, your goal is to eliminate the fraction by isolating the variable. The restriction $x \ne 1$ tells you that $x = 1$ would make the denominator zero, creating an undefined expression. To solve $\dfrac{x+2}{x-1}=2$, multiply both sides by $(x-1)$ to clear the fraction: $x + 2 = 2(x - 1)$ Distribute the 2 on the right side: $x + 2 = 2x - 2$ Subtract $x$ from both sides: $2 = x - 2$ Add 2 to both sides: $x = 4$ You can verify: $\dfrac{4+2}{4-1} = \dfrac{6}{3} = 2$ ✓ Now let's examine why the other answers fail. Choice B) $-4$ would give $\dfrac{-4+2}{-4-1} = \dfrac{-2}{-5} = \dfrac{2}{5}$, not 2. Choice C) $\dfrac{1}{2}$ would yield $\dfrac{\frac{1}{2}+2}{\frac{1}{2}-1} = \dfrac{\frac{5}{2}}{-\frac{1}{2}} = -5$, which is far from 2. Choice D) $2$ gives $\dfrac{2+2}{2-1} = \dfrac{4}{1} = 4$, which is double what we need. The key strategy for rational equations is to multiply both sides by the denominator early in your solution process. This transforms the problem into a linear equation that's much easier to solve. Always check your final answer by substituting back into the original equation to catch any arithmetic errors.

Question 13

1. $N(3)=990.0$
2. $N(3)=1{,}089.0$
3. $N(3)=1{,}197.9$ (correct answer)
4. $N(3)=1{,}209.0$
5. $N(3)=1{,}331.0$

Explanation: This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves predicting growth of a disinfectant-resistant bacteria strain, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, N(3)=1,197.9, is obtained by substituting t=3 into 900*(1.1)^3 and calculating 900*1.331. A common mistake is N(3)=1,089.0, which results from using (1.1)^2 instead of ^3. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.

Question 14

1. $-3<x<3$ (correct answer)
2. $x<-3\text{ or }x>3$
3. $x<3$
4. $x>-3$

Explanation: When you encounter compound inequalities like this, you're looking for values of $x$ that satisfy both conditions simultaneously. This requires solving each inequality separately, then finding their intersection. Start with the first inequality: $x + 5 > 2$. Subtracting 5 from both sides gives you $x > -3$. For the second inequality: $3x - 1 < 8$. Adding 1 to both sides yields $3x < 9$, and dividing by 3 gives you $x < 3$. Since you need both conditions to be true, you're looking for values where $x > -3$ AND $x < 3$. This means $x$ must be greater than -3 but less than 3, written as $-3 < x < 3$. This is answer choice A. Let's examine why the other options fail. Answer B ($x < -3$ or $x > 3$) represents the complement of the correct answer - it includes all values outside the solution interval. Answer C ($x < 3$) only satisfies the second inequality but ignores the constraint from the first inequality that $x > -3$. Answer D ($x > -3$) only satisfies the first inequality but ignores the upper bound from the second inequality. Remember that "and" in compound inequalities means intersection - you need the overlap of both solution sets. Always solve each inequality independently first, then find where both conditions are satisfied simultaneously. A number line sketch can help visualize the intersection when you're unsure.

Question 15

1. $2$ (correct answer)
2. $3$
3. $\dfrac32$
4. $\dfrac52$

Explanation: When you encounter exponential equations like this one, the key strategy is to express both sides using the same base, then equate the exponents. Start by recognizing that $27$ can be written as a power of $3$: $27 = 3^3$. This transforms your equation into $3^{2x-1} = 3^3$. Since the bases are now equal, the exponents must also be equal: $2x - 1 = 3$. Solving for $x$: $2x = 4$, so $x = 2$. Let's verify: $3^{2(2)-1} = 3^3 = 27$ ✓ Now let's examine why the other answers don't work: B) $x = 3$: If you substitute this value, you get $3^{2(3)-1} = 3^5 = 243$, which is far too large. This might result from incorrectly setting $2x - 1 = 5$ instead of $3$. C) $x = \frac{3}{2}$: Substituting gives $3^{2(\frac{3}{2})-1} = 3^{3-1} = 3^2 = 9$, which is too small. This error could arise from confusing the relationship between the base and exponent. D) $x = \frac{5}{2}$: This yields $3^{2(\frac{5}{2})-1} = 3^{5-1} = 3^4 = 81$, again too large. This might come from misidentifying $27$ as $3^4$ or making an arithmetic error. Strategy tip: Always express both sides of exponential equations using the same base when possible. Common powers to memorize: $3^2 = 9$, $3^3 = 27$, $3^4 = 81$. After solving, substitute your answer back into the original equation to verify it works.

Question 16

1. $1<x<4$ (correct answer)
2. $x<1\text{ or }x>4$
3. $x\le1\text{ or }x\ge4$
4. $x\le1\text{ and }x\ge4$

Explanation: When you encounter quadratic inequalities, your goal is to find where the quadratic expression is positive or negative by analyzing its roots and the parabola's shape. Start by factoring the quadratic $x^2 - 5x + 4$. You need two numbers that multiply to 4 and add to -5: those are -1 and -4. So $x^2 - 5x + 4 = (x-1)(x-4)$. Now solve $(x-1)(x-4) < 0$. The roots are $x = 1$ and $x = 4$, which divide the number line into three intervals: $x < 1$, $1 < x < 4$, and $x > 4$. Since the coefficient of $x^2$ is positive, this parabola opens upward, meaning it's negative between the roots and positive outside them. Test a point in each interval: when $x = 0$: $(0-1)(0-4) = 4 > 0$; when $x = 2$: $(2-1)(2-4) = -2 < 0$; when $x = 5$: $(5-1)(5-4) = 4 > 0$. The inequality is satisfied only when $1 < x < 4$, which is choice A. Choice B ($x < 1$ or $x > 4$) gives you where the expression is positive, not negative. Choice C includes the endpoints where the expression equals zero, but we need strictly less than zero. Choice D uses "and" instead of "or," which would mean $x$ must simultaneously be less than 1 and greater than 4—an impossible condition. Remember: for quadratic inequalities, always factor first, find the roots, then test intervals to determine the sign pattern.

Question 17

1. $x=13$
2. $x=14$
3. $x=15$ (correct answer)
4. $x=16$
5. $x=17$

Explanation: This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves calculating the number of crowns based on a setup fee and per-crown cost, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, x=15, is obtained by subtracting 120 from both sides to get 15x=225, then dividing by 15. A common mistake is x=14, which results from miscalculating 15*14 +120 as 345 instead of 330. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.

Question 18

1. $N(3)=450$
2. $N(3)=600$
3. $N(3)=675$ (correct answer)
4. $N(3)=900$
5. $N(3)=1{,}350$

Explanation: This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves modeling bacteria growth in a dental waterline, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, N(3)=675, is obtained by substituting t=3 into 200*(1.5)^3 and computing 200*3.375. A common mistake is N(3)=600, which results from miscalculating (1.5)^3 as 3 instead of 3.375. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.

Question 19

1. $\dfrac54$ (correct answer)
2. $1$
3. $\dfrac32$
4. $2$

Explanation: When you encounter exponential equations with different bases, your goal is to rewrite both sides using the same base so you can set the exponents equal to each other. Notice that both 5 and 125 are powers of 5: $125 = 5^3$. Let's rewrite the equation using base 5 throughout: $5^{x+1} = 125^{2-x}$ $5^{x+1} = (5^3)^{2-x}$ Using the power rule $(a^m)^n = a^{mn}$, the right side becomes: $5^{x+1} = 5^{3(2-x)}$ $5^{x+1} = 5^{6-3x}$ Since the bases are equal, the exponents must be equal: $x + 1 = 6 - 3x$ Solving for $x$: $x + 3x = 6 - 1$ $4x = 5$ $x = \frac{5}{4}$ This confirms that A) $\frac{5}{4}$ is correct. Let's check why the other options fail. If you substitute B) $x = 1$: the left side gives $5^2 = 25$ while the right side gives $125^1 = 125$ - not equal. For C) $x = \frac{3}{2}$: you get $5^{2.5} \approx 55.9$ versus $125^{0.5} \approx 11.2$ - also unequal. For D) $x = 2$: the left side is $5^3 = 125$ but the right side is $125^0 = 1$ - clearly not equal. Strategy tip: Always look for common bases when solving exponential equations. Convert everything to the smallest possible base (here, 5 instead of 125) to simplify your work and avoid calculation errors.

Question 20

1. $4\le x\le 6$
2. $4\le x\le 5.5$ (correct answer)
3. $3\le x\le 5.5$
4. $3\le x\le 6$
5. $5.5\le x\le 6$

Explanation: This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves budgeting for dental office chairs within a range, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, 4≤x≤5.5, is obtained by subtracting 800 from all parts and dividing by 800. A common mistake is 4≤x≤6, which results from incorrect division or boundary miscalculation. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.