Word Problems and Applications

Let the original number be \(x\). Then \((x-7)(x+12)=756\), which gives \(x^2+5x-840=0\). Applying the quadratic formula yields \(x=\frac{-5\pm\sqrt{3385}}{2}\). The positive value is \(\frac{-5+\sqrt{3385}}{2}\); the other root is negative. Exam tip: translate ‘7 less’ as \(x-7\) and ‘12 more’ as \(x+12\) before forming the equation.

Let the positive number be \(x\). Then \((x-5)(x+14)=1242\). Substituting \(x=32\), we get \((32-5)(32+14)=27\times46=1242\), so the original number is 32. Exam tip: In such word problems, testing the options in the equation can be a quick alternative to solving the quadratic equation completely.

Let the number be \(x\). Then \(x^2=18x+432\), so \(x^2-18x-432=0\). Using the quadratic formula, \(x=\frac{18\pm\sqrt{18^2+4\cdot432}}{2}=9\pm3\sqrt{57}\). Since \(9-3\sqrt{57}\) is negative, the positive number is \(9+3\sqrt{57}\). Exam tip: form the quadratic equation first, calculate both roots, and then select the root that satisfies the condition in the question.

Let the number be \(x\). Then \(x^2-20x=621\), so \(x^2-20x-621=0\). Using the quadratic formula, \(x=\frac{20\pm\sqrt{400+2484}}{2}=10\pm\sqrt{721}\). Since the number is positive and \(10-\sqrt{721}<0\), the required value is \(10+\sqrt{721}\). Exam tip: translate the word statement into a quadratic equation first, then use the given condition, such as positivity, to select the valid root.

Let the number be \(x\). Then \(x^2+x=1260\), so \(x^2+x-1260=0\). Factoring gives \((x-35)(x+36)=0\), hence \(x=35\) or \(x=-36\). The positive number is \(35\). Exam tip: quickly recognise that \(1260=35\times36\), the product of two consecutive integers; 36 is incorrect because \(36^2+36=1332\).

(x^2+175=1076) gives (x^2=901), not a perfect square. If the data are inconsistent, do not choose options blindly.

Let the number be \(x\). According to the statement, \(x^2-256=1040\), so \(x^2=1296=36^2\). Thus, \(x=36\) or \(x=-36\), but the positive number is \(36\). For instance, choosing 34 gives \(34^2-256=900\), so it is not correct. Exam tip: Interpret “subtracted from the square” as \(x^2-\text{given value}\).

Let the number be \(n\). “\(9\) more than a number” is \((n+9)\), so \((n+9)^2=1681\). Taking square root in the positive case: \(n+9=\sqrt{1681}=41\). Hence \(n=41-9=32\).

Let the son's age be \(x\) years. Then the father's age is \(x+32\) years, so \(x(x+32)=1280\). This gives \(x^2+32x-1280=0\), or \((x-32)(x+64)=0\). Since age cannot be negative, \(x=32\), and the father's age is \(32+32=64\) years. Exam tip: In age problems, reject any negative root before selecting the answer.

The direct answer is A, 24 years. Let the daughter’s age be x years. The mother is x+30, so x(x+30)=1476. Thus x²+30x−1476=0, or (x−24)(x+54)=0. The possible algebraic roots are 24 and −54; age cannot be negative, so x=24. Check: 24×54=1476. Option A works because it gives the required product and age difference. Option B gives ages 26 and 56, whose product is 1456. Option C gives 30 and 60, whose product is 1800. Option D is the mother’s possible age when the daughter is 24, not the daughter’s age. Memory cue: in age-product questions, write younger age as x and older age as x+the difference, then reject the negative root.

The equation is \((x+7)(x-7)=1551\). Using the difference of squares, \(x^2-49=1551\), so \(x^2=1600\) and \(x=\pm 40\). Since age cannot be negative, the present age is 40 years. Exam tip: apply \((x+a)(x-a)=x^2-a^2\) directly in such age problems.

((x+6)(x-6)=1333) gives (x^2-36=1333) and (x=37). The difference of squares idea helps in such questions.

The increase in area is (x + 8)² − x² = 1216. Simplifying gives 16x + 64 = 1216, so 16x = 1152 and x = 72 metres. Therefore, option B is correct.

The reduced side is \((x-5)\) cm, so \((x-5)^2=784=28^2\). Since a side length must be positive, take \(x-5=28\), giving \(x=33\) cm. The other algebraic possibility, \(x-5=-28\), gives \(x=-23\) cm, which is not physically meaningful. Exam tip: After finding the reduced side from the area, add back the length that was removed.

Let the original breadth be x cm. Then the original length is x + 12 cm. The original area is x(x + 12), and after increasing both dimensions by 6 cm, the new area is (x + 6)(x + 18). Their difference is 480: (x + 6)(x + 18) − x(x + 12) = 480. Expanding gives x² + 24x + 108 − x² − 12x = 480, so 12x + 108 = 480 and x = 31. Thus the breadth is 31 cm, option B. Substitution confirms the increase: 37×24 − 31×43 = 888 − 1333? The dimensions must be ordered correctly: new length 49 and new breadth 37, giving 1813 − 1333 = 480.

Let the original breadth be \(x\) m. Then the original length is \(x+9\) m. After the increases, the new length and breadth are \(x+13\) and \(x+3\) m, respectively. Thus, \((x+13)(x+3)=1190\), which gives \(x^2+16x-1151=0\). Using the quadratic formula, \(x=\frac{-16\pm\sqrt{4860}}{2}=-8\pm9\sqrt{15}\). Since a breadth must be positive, the valid value is \(x=-8+9\sqrt{15}\) m; the other root is negative. Exam tip: write the changed dimensions carefully before forming the area equation.

If breadth is (x), then (x(x+16)=1344), giving (x=32) and length (48). The wire length is (2(32+48)=160) cm.

Since the path surrounds the garden, its width is added twice to both the length and the breadth. The outer length is \(34+2(3)=40\) m and the outer breadth is \(22+2(3)=28\) m. Therefore, the total area is \(40\times28=1120\) square metres, so option C is correct. Option B results from an incorrect calculation of the outer dimensions or their product. Exam tip: for a path around all four sides, add twice the path width to each dimension.

The outer dimensions, including the path, are \((42+2x)\) m and \((28+2x)\) m. Thus, \((42+2x)(28+2x)=1848\), which gives \(4x^2+140x-672=0\), or \(x^2+35x-168=0\). Hence, \(x=\frac{-35\pm\sqrt{1897}}{2}\). Since a width must be positive, the valid value is \(x=\frac{-35+\sqrt{1897}}{2}\approx4.28\) m. The negative root is not physically meaningful. Exam tip: when a path surrounds a rectangle externally, add \(2x\) to both its length and breadth.

The strip is removed from both opposite sides, so the inner side is \(52-2x\). Therefore, \(52-2x=36\), giving \(2x=16\) and \(x=8\) m. Using areas gives the equivalent quadratic equation \((52-2x)^2=36^2\); the geometrically valid root is \(x=8\). Exam tip: divide the difference between the outer and inner side lengths by 2. For example, 10 m would reduce the inner side to 32 m, not 36 m.

If the smaller integer is (x), then (x^2+(x+1)^2=1201), giving (x=24). Write consecutive integers as (x) and (x+1).

Let the smaller integer be \(x\). The next consecutive odd integer is therefore \(x+2\). Thus, \(x^2+(x+2)^2=2050\), which gives \(2x^2+4x-2046=0\) or \(x^2+2x-1023=0\). Factoring, \((x-31)(x+33)=0\), so \(x=31\) or \(x=-33\). Since the integers are positive, \(x=31\). Therefore, the correct answer is 31. Exam tip: consecutive odd integers differ by 2, not by 1.

The direct answer is C, 36. Consecutive positive even numbers differ by 2, so let them be x and x+2. Their square sum gives x²+(x+2)²=2452, hence 2x²+4x−2448=0, or x²+2x−1224=0. Factoring gives (x−34)(x+36)=0. The positive value is x=34, so the larger number is 36. Option A, 32, would pair with 34 and gives 1024+1156=2180. Option B, 34, is the smaller number, and 34²+36²=2452. Option C is correct because 36 is the larger member of that pair. Option D, 38, would pair with 36 and give 2740. Exam cue: consecutive even numbers are x and x+2, not x and x+1.

If the smaller number is (x) and the larger is (x+17), then (x(x+17)=1260). This gives smaller number (28) and larger number (45).

Let the numbers be \(x\) and \(80-x\). Then \(x^2+(80-x)^2=3232\), which gives \(2x^2-160x+3168=0\), or \(x^2-80x+1584=0\). Factoring, \((x-36)(x-44)=0\), so the numbers are 36 and 44. Therefore, the smaller number is 36; 44 is the larger number, not the required answer. Exam tip: When the sum and the sum of squares are given, use \(a^2+b^2=(a+b)^2-2ab\) to find the product quickly.