Word Problems and Applications

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.

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.

The direct answer is option C: 32 years. Let the daughter’s age be x years. The mother is 38 years older, so her age is x+38. Their product is 2240, hence x(x+38)=2240. Expanding gives x^2+38x-2240=0. Factoring, (x-32)(x+70)=0, so x=32 or x=-70. An age cannot be negative, therefore the daughter is 32 years old. Option A, 28, does not work because 28×66=1848, not 2240. Option B, 30, gives 30×68=2040, also incorrect. Option C gives 32×70=2240 and exactly satisfies both conditions. Option D, 70, would make the mother 108 and the product 7560, so it is wrong; 70 is the mother’s age in the correct pair, not the daughter’s age. Always reject a negative mathematical root in an age problem because age is a positive quantity. A useful exam cue is to check the answer in both conditions: difference 70−32=38 and product 70×32=2240.

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.

((x+10)(x-10)=1500) gives (x^2-100=1500), so (x=40). The difference of squares idea helps in such questions.

The area increase is (x + 12)² − x² = 2448. Expanding gives 24x + 144 = 2448, so 24x = 2304 and x = 96 m. Therefore, the original side length is 96 m.

The reduced side is \((x-9)\) cm, so \((x-9)^2=1681=41^2\). Since a side length is positive and the reduced side must also be positive, \(x-9=41\), not \(-41\). Hence, \(x=50\) cm. Exam tip: In quadratic equations involving lengths, retain only the positive root that fits the situation.

The direct answer is B, 60 cm. Let the original breadth be x cm; then the length is x+22 cm. After both dimensions increase by 10, the new dimensions are x+10 and x+32. The area increase is (x+10)(x+32)−x(x+22)=1520. Expanding gives x²+42x+320−x²−22x=1520, so 20x+320=1520 and x=60. Option A, 54, gives an increase of 20(54)+320=1400. Option B works exactly. Option C, 66, gives 1640. Option D, 82, gives 1960. The key idea is that the common x² terms cancel; this is a linear equation, not a needlessly complicated quadratic. Memory cue: increasing both sides by 10 adds 10(length+breadth)+100.

Let the original breadth be x m; then the original length is (x + 20) m. After the increases, the dimensions are (x + 28) m and (x + 6) m. Thus, (x + 28)(x + 6) = 3360, giving x² + 34x − 3192 = 0 = (x − 42)(x + 76). The possible breadths are 42 m and −76 m; a breadth must be positive, so the original breadth was 42 m.

If breadth is (x), then (x(x+30)=2800), giving (x=40) and length (70). The wire length is (2(40+70)=220) cm.

Since the path surrounds the garden on all four sides, 5 + 5 = 10 m must be added to both dimensions. Thus, the outer length is 60 + 10 = 70 m and the outer width is 42 + 10 = 52 m. The total area is \(70 \times 52 = 3640\) square metres, so option C is correct. Exam tip: for a path around all sides, add twice the path width to each original dimension.

The outer dimensions, including the path, are \(64+2x\) metres and \(44+2x\) metres. Thus, \((64+2x)(44+2x)=3996\). On simplification, \(x^2+54x-295=0\), which factors as \((x-5)(x+59)=0\). The roots are \(5\) and \(-59\), but a width cannot be negative, so \(x=5\) metres. Exam tip: for a uniform path around all four sides, add \(2x\) to each original dimension.

The border adds x metres on each side of the inner square, so the outer side is 60 + 2x. Hence, 60 + 2x = 92, giving 2x = 32 and x = 16 m. The border is 16 m wide.

Let the breadth be (x\text{ m}), so (x(x+5)=300), giving (x=15). In exams, reject the negative dimension.

Let the faster pipe take (x\text{ h}), then (\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}). This gives (x^2-7x-30=0), so (x=10).