Word Problems and Applications

Let the son's present age be \(x\) years. The father's present age is then \(x+28\) years. After 4 years, their ages will be \(x+4\) and \(x+32\), respectively. Thus, \((x+4)(x+32)=960\), which gives \(x^2+36x-832=0\). Its positive solution is \(x=16\); the negative solution cannot represent an age. Check: after 4 years, the ages will be 20 and 48, and \(20\times48=960\). Exam tip: In age problems, reject any negative or otherwise unrealistic root.

Let one friend’s age be x years. Then the other friend’s age is (44 − x) years. Thus, x(44 − x) = 480, giving x² − 44x + 480 = 0. Factoring, (x − 24)(x − 20) = 0, so the two ages are 24 years and 20 years. Therefore, the older friend is 24 years old. Exam tip: For such problems, form the quadratic equation from the given sum and product before solving it.

If the middle integer is (x), the first and third are (x-1) and (x+1). From ((x-1)(x+1)=195), (x=14).

The question is made unambiguous by specifying positive even integers. Let the middle integer be x. Then the consecutive even integers are x-2, x, and x+2. The product condition gives (x-2)(x+2)=480, so x^2-4=480 and x^2=484. Thus x=22 or x=-22 algebraically, but positivity excludes -22. Therefore the middle integer is 22, making option A correct. A quick check gives the sequence 20, 22, 24, and 20×24=480. Option B is the first integer rather than the middle one, option C is the third integer, and option D does not produce the required product with its neighboring even integers. The quadratic setup also demonstrates the symmetric structure around the middle term.

Let the breadth be \(x\) units. Then the length is \(x+17\) units. By the Pythagorean theorem, \(x^2+(x+17)^2=25^2\), which simplifies to \(x^2+17x-168=0\). Factoring gives \((x-7)(x+24)=0\), so \(x=7\) or \(x=-24\). Since a breadth cannot be negative, the valid answer is 7. Exam tip: For rectangle problems involving the length, breadth, and diagonal, apply the Pythagorean theorem first.

Let the number be \(x\). Then \(x^2=15x+54\), so \(x^2-15x-54=0\). Factoring gives \((x-18)(x+3)=0\), hence \(x=18\) or \(x=-3\). The positive number is therefore 18. Exam tip: Find both roots of a quadratic equation and then apply the condition given in the question, such as ‘positive’.

Direct answer: A, 14. Let the numbers be x and 26-x because their sum is 26. Their squared sum gives x^2+(26-x)^2=340. Expanding: 2x^2-52x+676=340, so 2x^2-52x+336=0; dividing by 2 gives x^2-26x+168=0=(x-12)(x-14). Thus the numbers are 12 and 14, and the larger is 14. Option A is correct. Option B, 12, is the smaller number. Option C, 13, would make the squared sum 169+169=338. Option D, 16, would require the other number to be 10 and gives 356. Memory cue: use sum and squared sum, then check both the sum and the larger-number wording.

Let the smaller number be \(x\). Then the larger number is \(x+3\). Thus, \(x(x+3)=378\), which gives \(x^2+3x-378=0\). Factoring, \((x-18)(x+21)=0\), so \(x=18\) or \(x=-21\). Since both numbers are positive, \(-21\) is rejected and the smaller number is 18. Exam tip: In word problems, always check the given condition, such as positivity, before accepting a quadratic root.

Let the breadth be \(x\) units. Then the length is \(x+6\) units, so the area gives \(x(x+6)=432\), or \(x^2+6x-432=0\). Factoring, \((x+24)(x-18)=0\), so \(x=18\) or \(x=-24\). Since a length cannot be negative, the breadth is 18 units. Exam tip: Reject any negative root when solving for a physical dimension.

The breadth is 9 units. According to the statement, the length is \(4(9-3)=4\times6=24\) units. Therefore, the area is \(\text{length}\times\text{breadth}=24\times9=216\) square units. The value 108 does not result from multiplying the correct length and breadth. In such questions, first evaluate the quantity inside the brackets and then apply the stated multiplier.

Let the positive number be \(x\). According to the question, \(x(x+8)=209\), so \(x^2+8x-209=0\). Factoring gives \((x-11)(x+19)=0\), hence \(x=11\) or \(x=-19\). Since the number must be positive, the correct answer is 11. Although 19 is a factor of 209, it does not satisfy the condition because \(19(19+8)=513\). Exam tip: after solving a quadratic, check the given condition, such as positivity, before selecting the root.

Let the number be \(x\). According to the question, \(x(x-6)=135\), so \(x^2-6x-135=0\). Factoring gives \((x-15)(x+9)=0\), hence \(x=15\) or \(x=-9\). Since the question asks for the positive number, the answer is \(15\). In an exam, find both roots and then apply the stated condition, such as ‘positive’.

Let the breadth be \(x\) units. Then the length is \(x+6\) units, so \(x(x+6)=216\), or \(x^2+6x-216=0\). Factoring gives \((x-12)(x+18)=0\), so \(x=12\) or \(x=-18\). Since a length cannot be negative, the breadth is \(12\) units. Option 18 is the length, not the breadth. Exam tip: In rectangle-area problems, let one side be \(x\) and express the other side in terms of \(x\) before forming the quadratic equation.

Let the border width be \(w\). Then inner dimensions are \(24-2w\) and \(18-2w\). So
\[ (24-2w)(18-2w)=280.\]
Expand:
\[ 432-48w-36w+4w^2=280\]
\[ 4w^2-84w+152=0.\]
Divide by \(4\): \(w^2-21w+38=0\). Solving:
\[ w=\frac{21\pm\sqrt{21^2-4\cdot 38}}{2}=\frac{21\pm\sqrt{441-152}}{2}=\frac{21\pm\sqrt{289}}{2}=\frac{21\pm 17}{2}.\]
So \(w=19\) or \(w=2\). If \(w=19\), then \(24-2w<0\), not possible for an actual frame. Hence \(w=2\,\text{m}\).

Let the path width be \(x\) metres. The remaining rectangle has length \(36-2x\) and width \(24-2x\). Thus, \((36-2x)(24-2x)=540\), which gives \(x^2-30x+81=0\). Its roots are \(x=3\) and \(x=27\), but the path width must be less than half the garden's breadth, so \(x=27\) is not feasible. Therefore, the correct width is 3 metres. Exam tip: for a path inside all four sides, subtract \(2x\) from each original dimension.

Let the width of the path be \(x\) metres. Then the outer rectangle has length \(30+2x\) and width \(18+2x\). Thus, \((30+2x)(18+2x)=748\). On simplifying, \(x^2+24x-52=0\), or \((x-2)(x+26)=0\). Since a width must be positive, \(x=2\) metres is the valid answer; \(x=-26\) is not physically meaningful. Exam tip: because the path is added on both opposite sides, add \(2x\) to each original dimension.

Total revenue equals price per item multiplied by the number of items, so \(x(80-x)=1500\). Rearranging gives \(x^2-80x+1500=0\), or \((x-30)(x-50)=0\). Thus, \(x=30\) or \(x=50\); the smaller value is 30. In such problems, first form the revenue equation and then compare both roots.

The given equation is \(x(30-x)=216\). On simplifying, we get \(x^2-30x+216=0\), which factors as \((x-12)(x-18)=0\). Therefore, \(x=12\) or \(x=18\), and the larger value is 18. Exam tip: after finding both roots of a quadratic equation, select the larger or smaller one exactly as asked.

If the number of people is \(n\), each handshake corresponds to one pair of people. Therefore, the total number of handshakes is \(\frac{n(n-1)}{2}\). Thus, \(\frac{n(n-1)}{2}=66\), giving \(n(n-1)=132\). The positive solution is \(n=12\), since \(\frac{12\times11}{2}=66\). The closest distractor, 11, is incorrect because 11 people would make \(\frac{11\times10}{2}=55\) handshakes. Exam tip: use the formula \(\frac{n(n-1)}{2}\) for one-time handshake problems.

If the number of teams is \(n\), each match corresponds to one pair of distinct teams. Therefore, the number of matches is \(\frac{n(n-1)}{2}\). Thus, \(\frac{n(n-1)}{2}=28\), giving \(n(n-1)=56\) and \(n^2-n-56=0\). Factoring, \((n-8)(n+7)=0\), so \(n=8\) or \(n=-7\). Since the number of teams cannot be negative, \(n=8\) is the correct answer. Exam tip: For a single round-robin tournament, use \(\frac{n(n-1)}{2}\) and reject any negative root.

The sum of the two sides of a rectangle is half its perimeter, so the sum is \(60/2=30\) metres. If one side is \(x\) metres, the other is \(30-x\) metres. Thus, \(x(30-x)=216\), giving \(x^2-30x+216=0\). Factoring gives \((x-18)(x-12)=0\), so the sides are 18 metres and 12 metres. Therefore, the longer side is 18 metres. In exams, first use half the perimeter to find the sum of the two sides.

If (B) alone takes (x) days, then (A) takes (x-5) days. From (\frac{1}{x}+\frac{1}{x-5}=\frac{1}{6}), (x=15).

Let the smaller positive number be x. The larger number is then x + 13. Using the product condition, x(x + 13) = 558, so x² + 13x − 558 = 0. This factors as (x + 31)(x − 18) = 0, giving x = 18 or x = −31. Since both numbers are positive, x = 18 is the valid smaller number. Therefore the larger number is x + 13 = 31, and option C is correct. Option A is the smaller number rather than the requested larger one. Option B has difference 6 from 31 if paired appropriately and does not satisfy the product condition, while option D does not produce the given difference and product together.

Let the smaller positive number be x. Then the other number is 65−x. The condition on the squares gives x²+(65−x)²=2125. Expanding, x²+4225−130x+x²=2125, so 2x²−130x+2100=0. Dividing by 2 gives x²−65x+1050=0. This factors as (x−30)(x−35)=0, so the two numbers are 30 and 35. Since x was defined as the smaller number, x=30 and option B is correct. A check confirms that 30+35=65 and 30²+35²=900+1225=2125. Option C is the larger number, while options A and D do not form a pair satisfying both given conditions.

Total work equals number of days multiplied by work completed per day. Therefore x(x + 9) = 736, giving x² + 9x − 736 = 0. The expression factors as (x − 23)(x + 32) = 0, so x = 23 or x = −32. A number of days cannot be negative, hence x = 23 is the valid solution. The daily output is then x + 9 = 32 units, and 23 × 32 = 736 confirms the result. Thus option B is correct. Option D is the daily work amount, not the number of days. Options A and C fail the product condition and do not give the stated total work.