Word Problems and Applications

The total number of plants gives the equation \(x(x-6)=667\), or \(x^2-6x-667=0\). Factoring, \((x-29)(x+23)=0\), so \(x=29\) or \(x=-23\). Since the number of rows cannot be negative, \(x=29\) is the valid answer. Exam tip: In word problems, retain only the root that is meaningful in the given context.

The total number of questions gives \(x(x-5)=414\). Thus, \(x^2-5x-414=0\), which factors as \((x-23)(x+18)=0\). Therefore, \(x=23\) or \(x=-18\). Since the number of students cannot be negative, the valid answer is \(23\). Exam tip: in word problems, reject algebraic roots that are impossible in the given context.

Let the original number be \(x\). Then \((x+3)(x+7)=600\), which gives \(x^2+10x-579=0\). Using the quadratic formula, \(x=-5\pm2\sqrt{151}\). Since the number is positive, the valid value is \(x=-5+2\sqrt{151}\approx14.6\). Option 15 is a close distractor, but it gives \(18\times22=396\), not 600. Exam tip: Form the equation first and then use the stated condition to select the valid root.

Let the original number be \(x\). Then \((x-5)(x+8)=464\), which simplifies to \(x^2+3x-504=0\), or \((x-21)(x+24)=0\). Thus, \(x=21\) or \(x=-24\); the positive original number is 21. In such problems, carefully translate ‘5 less’ as \(x-5\) and ‘8 more’ as \(x+8\).

Let the original number be x. Then \((x-2)(x+9)=620\), so \(x^2+7x-18=620\), or \(x^2+7x-638=0\). Factoring gives \((x-22)(x+29)=0\). Thus x=22 or x=-29; since the number is specified as positive, the correct answer is 22. Exam tip: verify the result in the original condition: \((22-2)(22+9)=20\times31=620\).

Let the number be \(x\). Then \(x^2=14x+240\), so \(x^2-14x-240=0\). Factoring gives \((x-24)(x+10)=0\), hence \(x=24\) or \(x=-10\). Since the question asks for the positive number, the answer is 24. Check: \(24^2=576\) and \(14\times24+240=576\). Exam tip: Form the quadratic equation directly from the statement, find both roots, and then apply the condition given in the question.

Let the number be \(x\). Then \(x^2-16x=425\), so \(x^2-16x-425=0\). Adding 64 to both sides gives \((x-8)^2=489\), hence \(x=8\pm\sqrt{489}\). Since \(8-\sqrt{489}<0\), the positive number is \(8+\sqrt{489}\). Exam tip: substitute the selected value back into the original equation; a quadratic word problem need not have an integer answer.

Let the number be \(x\). Then \(x^2+x=930\), so \(x^2+x-930=0\). Factoring gives \((x-30)(x+31)=0\), hence \(x=30\) or \(x=-31\). The positive number is therefore 30. As an exam tip, verify the answer by checking \(30^2+30=900+30=930\); choosing 31 gives 992, not 930.

(x^2+156=1057) gives (x^2=901), not a listed perfect square, so this is a data-error type question. In exams, check consistency before solving.

(x^2-175=550) gives (x^2=725), so no listed option should be correct. In such questions, check data consistency.

Direct answer: C, 26. Let the number be x. “Four more than the number” is x+4, and its square is 900: (x+4)^2=900. Taking square roots gives x+4=±30. Thus x=26 or x=-34. The question asks for the positive case, so x=26. Option A, 22, makes x+4=26 and its square 676. Option B, 24, gives 28^2=784. Option C, 26, gives 30^2=900 and is correct. Option D, 30, gives 34^2=1156. The negative solution -34 is not listed and is excluded by the positive condition. Memory cue: after a square equation, consider both roots, then apply the wording of the problem.

Let the original number be x. The phrase “6 less than a number” means x − 6, so the condition is (x − 6)² = 784. Since 784 = 28², taking both square-root possibilities gives x − 6 = 28 or x − 6 = −28. Hence x = 34 or x = −22. The question specifically asks for the positive case, so the required number is 34. Substitution verifies the result: (34 − 6)² = 28² = 784. Option A is only the positive square-root value and is not the original number. Options B and D give expressions whose squares are not 784. Thus option C is the unique correct answer.

Let the son's age be x years. Then the father's age is x + 28 years. Thus, x(x + 28) = 1485, or x² + 28x − 1485 = 0. Factoring gives (x − 27)(x + 55) = 0. Since age cannot be negative, x = 27, so the father's age is 27 + 28 = 55 years. Exam tip: In age problems, reject the negative root and verify the positive root in the original conditions.

Direct answer: A, 16 years. Let the daughter’s age be x years. The mother is x+24, and their product is 640: x(x+24)=640. Hence x^2+24x-640=0. Factoring gives (x-16)(x+40)=0, so x=16 or x=-40. An age cannot be negative, so the daughter is 16 and the mother is 40; 16×40=640 and their difference is 24. Option A satisfies both facts. Option B, 20, would pair with 44 and product 880. Option C, 24, would pair with 48 and product 1152. Option D, 40, is the mother’s age, not the daughter’s, and pairs with 64 in the stated model. Memory cue: reject negative age and identify exactly which person the question asks about.

From the condition, \((x+5)(x-5)=875\). Using the difference-of-squares identity gives \(x^2-25=875\), so \(x^2=900\) and \(x=\pm 30\). Since age cannot be negative, the present age is 30 years. In age problems, always reject the negative root.

((x+4)(x-4)=609) gives (x^2-16=609) and (x=25). The difference of squares idea helps in such questions.

The original area is \(x^2\), while the new area is \((x+5)^2\). Thus, \((x+5)^2-x^2=425\), which simplifies to \(10x+25=425\). Hence, \(10x=400\) and \(x=40\) metres. Therefore, option C is correct. Exam tip: For such problems, equate the difference of the two areas before solving the resulting linear equation.

The reduced side is \((x-3)\) cm, so \((x-3)^2=225\). This gives \(x-3=\pm15\). Since a side length must be positive and the reduced side is 15 cm, \(x-3=15\), hence \(x=18\) cm. The negative possibility would give \(x=-12\) cm, which is not physically valid. Exam tip: reject any root that gives a negative length in a geometric word problem.

The correct answer is B, 21 cm. Let the original breadth be x cm. Then the original length is x + 10 cm. After increasing both dimensions by 4 cm, the new breadth is x + 4 and the new length is x + 14. The increase in area is therefore (x + 4)(x + 14) − x(x + 10) = 224. Expanding gives x² + 18x + 56 − x² − 10x = 224, so 8x + 56 = 224, 8x = 168, and x = 21. Option A, 18, gives an increase of 200 cm², not 224. Option B gives the required increase: 25 × 35 − 21 × 31 = 875 − 651 = 224. Option C gives 232 cm², and option D gives 248 cm². Memory cue: when both sides increase, compare new area minus old area; do not compare only one dimension.

Let the original breadth be \(x\) m. Then the original length is \(x+5\) m. After the increase, the length becomes \(x+8\) m and the breadth becomes \(x+2\) m. Hence, \((x+8)(x+2)=462\), which gives \(x^2+10x-446=0\). Using the quadratic formula, \(x=-5\pm\sqrt{471}\). Since a breadth must be positive, \(x=-5+\sqrt{471}\approx16.70\) m. The distractor 18 m is incorrect because it would give a new area of \(26\times20=520\) square metres. Exam tip: discard the negative root and verify the remaining root in the original area equation.

Use the governing square formulas: area = side² and perimeter = 4 × side. Let the side of the square be s centimetres. The given area gives s² = 625, so the positive side length is s = 25 cm. Because the wire is bent around the whole square, its length is the complete perimeter, not just one side. Thus wire length = 4s = 4 × 25 = 100 cm. Option C is correct. The value 25 cm would be the length of one side, so it is not an option for the total wire. Values 80 cm and 90 cm do not equal the required perimeter, and 125 cm is also inconsistent with a square having area 625 square centimetres.

If breadth is (x), then (x(x+9)=700), giving (x=25) and length (34). The wire length is (2(25+34)=118), so the correct option is (118).

Since the path surrounds all four sides, its width is added twice to both dimensions. Outer length = 20 + 2 + 2 = 24 m and outer width = 14 + 2 + 2 = 18 m. Therefore, total outer area = 24 × 18 = 432 square metres. Values such as 384 or 408 result from failing to add the path width correctly in one or both directions. Exam tip: For a path around all four sides, add twice the path width to each original dimension.

The consecutive integers are \(x\) and \(x+1\). Thus, \(x(x+1)=56\), which gives \(x^2+x-56=0\). Option B incorrectly represents the second integer as \(x-1\). Exam tip: translate each word condition into algebra first.

The border reduces the outer side by \(x\) on both opposite sides, so the inner side is \(28-2x\). Thus, \(28-2x=20\), giving \(2x=8\) and \(x=4\) m. Therefore, option B is correct. Exam tip: for a uniform border around a square, subtract twice the border width from the outer side.