Word Problems and Applications

Using distance = speed × time, we get x(x+20)=1125, or x²+20x−1125=0. Factoring gives (x−25)(x+45)=0, so x=25 or x=−45. Since time cannot be negative, x=25 hours is the valid answer. Exam tip: In speed–distance–time problems, first write distance = speed × time and reject any negative time root.

Total work is (x(x+18)=1440), giving (x=30). For total work, multiply days and work per day.

The total number of questions gives \(x(x+22)=1400\). Thus, \(x^2+22x-1400=0\), which factors as \((x-28)(x+50)=0\). Therefore, \(x=28\) or \(x=-50\); since the number of days cannot be negative, the valid answer is 28. Exam tip: In word problems, reject algebraic roots that are impossible in the given context.

The total number of books gives the equation \(x(x+27)=1980\), or \(x^2+27x-1980=0\). Factoring gives \((x-33)(x+60)=0\), so \(x=33\) or \(x=-60\). Since the number of shelves cannot be negative, \(x=33\) is valid. Exam tip: In word problems, reject a negative root when the variable represents a count or measurement.

The total number of chairs gives the equation
x(x+30)=2275
, or
x^2+30x-2275=0
. Factoring,
(x-35)(x+65)=0
, so
x=35
or
x=-65
. Since the number of rows cannot be negative, the valid answer is 35. Exam tip: In arrangement word problems, reject any negative root and retain the positive value that fits the context.

The total-number condition gives \(x(x-16)=1536\), so \(x^2-16x-1536=0\). Factoring yields \((x-48)(x+32)=0\), giving \(x=48\) or \(x=-32\). Since the number of rows cannot be negative, \(x=48\) is correct. Exam tip: In word problems, reject roots that are not meaningful for the given quantity.

The total number of question papers gives \(x(x-18)=1600\). Thus, \(x^2-18x-1600=0\), which factors as \((x-50)(x+32)=0\). Therefore, \(x=50\) or \(x=-32\). Since the number of students cannot be negative, the valid answer is 50. Exam tip: reject any root that is not meaningful in the given context.

Let the original number be \(x\). Then \((x+8)(x+15)=2340\), which gives \(x^2+23x-2220=0\), or \((x-37)(x+60)=0\). Thus, \(x=37\) or \(x=-60\); since the number is specified as positive, the correct answer is 37. Exam tip: Always check both roots and select the one satisfying the condition stated in the question.

Let the original number be \(x\). Then \((x-11)(x+20)=2112\), which gives \(x^2+9x-2332=0\). Factoring, \((x-44)(x+53)=0\), so \(x=44\) or \(x=-53\). Since the question asks for the positive original number, the answer is 44. Exam tip: use a minus sign for ‘less than’ and a plus sign for ‘more than’, then apply the stated positivity condition.

Let the positive number be x. Then \((x-14)(x+25)=2700\). On expanding and simplifying, \(x^2+11x-3050=0\), which factors as \((x-50)(x+61)=0\). Thus, \(x=50\) or \(x=-61\); since the number is specified as positive, the correct answer is 50. Exam tip: Always check the roots of a quadratic against the conditions stated in the word problem, such as positivity.

Let the number be \(x\). Then \(x^2=30x+2800\), so \(x^2-30x-2800=0\). Factoring gives \((x-70)(x+40)=0\), hence \(x=70\) or \(x=-40\). Since the question asks for the positive number, the answer is 70. In an exam, find both roots and then apply the condition stated in the question, such as positivity.

Let the number be \(x\). Then \(x^2-40x=3825\), so \(x^2-40x-3825=0\). Factoring gives \((x-85)(x+45)=0\), hence \(x=85\) or \(x=-45\). Since the number is positive, the valid answer is \(85\). Exam tip: After solving a quadratic equation, always apply the condition given in the question, such as positivity, to select the valid root.

Let the number be x. Then x² + x = 2352, so x² + x − 2352 = 0. Factoring gives (x − 48)(x + 49) = 0, hence x = 48 or x = −49. Since the question asks for the positive number, the answer is 48. As an exam check, 48² + 48 = 2304 + 48 = 2352, whereas 49 gives 2450.

Let the number be x. Then x² + 399 = 3208, so x² = 3208 − 399 = 2809. Since 2809 = 53², x = ±53; therefore, the required positive number is 53. Exam tip: Taking a square root gives both signs, but select the positive value when the question asks for a positive number.

Let the number be \(x\). Then \(x^2-975=2625\), so \(x^2=2625+975=3600\). Hence \(x=\pm60\), but the positive number is \(60\). Exam tip: Translate “subtracting from the square” as \(x^2-975\), and select the positive root when asked for a positive number.

((x+18)^2=4096) and (x+18=64), so (x=46). If the positive case is clear, take the positive square root.

((x-22)^2=2500) and (x-22=50), so (x=72). After taking square root, choose the sign according to the condition.

Let the son's age be \(x\) years. Then the father's age is \(x+42\) years, so \(x(x+42)=2160\). This gives \(x^2+42x-2160=0\), which factors as \((x-30)(x+72)=0\). Since age must be positive, \(x=30\), and the father's age is \(30+42=72\) years. Option 60 does not satisfy both the age difference and the product conditions. Exam tip: In age problems, reject the negative root and retain only the positive age.

If the daughter's age is (x), then (x(x+38)=2240), giving (x=32). In age problems, ignore the negative root.

According to the statement, \((x+12)(x-12)=2160\). Using the difference-of-squares identity, \(x^2-144=2160\), so \(x^2=2304=48^2\). Thus \(x=48\) or \(-48\), but age cannot be negative; hence the present age is 48 years. Check: \((48+12)(48-12)=60\times36=2160\). Exam tip: reject the negative root in age-based problems.