Word Problems and Applications

Let the smaller number be \(x\); then the larger number is \(x+6\). Thus, \(x(x+6)=216\), or \(x^2+6x-216=0\). Factoring gives \((x-12)(x+18)=0\), so \(x=12\) or \(x=-18\). Since the numbers are positive, the smaller number is 12. Check: \(12\times18=216\) and \(18-12=6\). Exam tip: In such problems, let the smaller number be \(x\) and express the other number using the given difference.

Let the consecutive numbers be \(x\) and \(x+1\). Then \(x^2+(x+1)^2=365\), which gives \(2x^2+2x-364=0\) or \(x^2+x-182=0\). Factoring, \((x-13)(x+14)=0\), so \(x=13\) or \(x=-14\). Since the numbers are natural, \(x=13\), giving 13 and 14. Check: \(13^2+14^2=169+196=365\). Exam tip: Represent consecutive numbers as \(x\) and \(x+1\) before forming the quadratic equation.

Let the first number be x. The consecutive even numbers are then x and x+2. Thus, x(x+2)=224, or x²+2x−224=0. Factoring gives (x−14)(x+16)=0, so x=14 or x=−16. Since the numbers must be positive, x=14 and the next number is 16. Therefore, the correct answer is 14 and 16. Exam tip: consecutive even numbers differ by 2, not by 1.

Let the breadth of the rectangle be \(x\) cm. Its length is then \(x+5\) cm. By the Pythagorean theorem, \(x^2+(x+5)^2=25^2\), which simplifies to \(x^2+5x-300=0\). Factoring gives \((x+20)(x-15)=0\), so \(x=15\) or \(x=-20\). Since a length cannot be negative, the breadth is 15 cm. Exam tip: identify the length, breadth and diagonal as the sides of a right triangle before applying the Pythagorean theorem.

If the number of rows is \(x\), then each row has \(x+3\) students. Thus, \(x(x+3)=154\), giving \(x^2+3x-154=0\). Factoring, \((x+14)(x-11)=0\), so \(x=11\) or \(x=-14\). Since the number of rows cannot be negative, the valid answer is 11. In such word problems, always reject a root that is not meaningful in the given context.

Let the uniform frame width be \(x\) cm. Then the inner length is \((12-2x)\) cm and the inner breadth is \((8-2x)\) cm. Thus, \((12-2x)(8-2x)=60\), which gives \(x^2-10x+9=0\), so \(x=1\) or \(x=9\). The value \(x=9\) cm is impossible because it exceeds the outer breadth of 8 cm. Therefore, the frame width is 1 cm. Exam tip: When a border is present on both sides, subtract \(2x\) from each outer dimension.

Let the number be \(x\). Then \(x^2+5x=84\), so \(x^2+5x-84=0\). Factoring gives \((x+12)(x-7)=0\), hence \(x=-12\) or \(x=7\). Since the number is positive, \(-12\) is rejected, making \(7\) the correct answer. In an exam, always apply the condition stated in the question when selecting the valid root.

The perimeter of the rectangle is 120 m, so \(2(l+b)=120\), giving \(l+b=60\). Its area gives \(lb=875\). If one dimension is \(x\), the other is \(60-x\), so \(x(60-x)=875\). Thus \(x^2-60x+875=0\), whose roots are 25 and 35. Therefore, the dimensions are 25 m and 35 m. Option A gives an area of 800 square metres, while option C gives 900 square metres. Exam tip: In fencing problems, form the perimeter equation first and the area equation second.

Let one number be x; then the other is 19 − x. Using the product condition, x(19 − x) = 84, which gives x² − 19x + 84 = 0. Factoring, (x − 7)(x − 12) = 0, so the numbers are 7 and 12. Option A has the correct sum but its product is 78, not 84. Exam tip: always verify both the sum and the product of the selected pair.

Let the smaller positive number be \(x\). The other number is \(x+4\), so \(x(x+4)=221\), giving \(x^2+4x-221=0\). Factoring, \((x-13)(x+17)=0\), so the solutions are \(13\) and \(-17\). Since the number is specified as positive, \(x=13\) is valid. Therefore, the smaller number is 13. Exam tip: In such word problems, represent the smaller number by \(x\) and express the other number directly from the given difference.

Let the smaller number be \(x\); then the larger number is \(x+9\). Thus, \(x^2+(x+9)^2=585\), which simplifies to \(2x^2+18x-504=0\) and then \(x^2+9x-252=0\). Factoring gives \((x+21)(x-12)=0\), so \(x=12\) or \(x=-21\). Since both numbers are positive, \(x=12\) is valid. In the exam, reject any root that violates the stated positivity condition.

The father's present age is \(x+28\) years. After 4 years, their ages will be \(x+4\) and \(x+32\) years, respectively. Thus, \((x+4)(x+32)=960\), which gives \(x^2+36x-832=0\). Its solutions are \(x=16\) and \(x=-52\). Since age cannot be negative, the son's present age is 16 years, so option C is correct. In an exam, substitute the answer back into the original condition to verify the product.

The combined work rate is \(\frac{1}{2}\) task per day, so \(\frac{1}{x}+\frac{1}{x+3}=\frac{1}{2}\). On simplifying, \(2(2x+3)=x(x+3)\), which gives \(x^2-x-6=0\). Thus, \((x-3)(x+2)=0\), so \(x=3\) or \(x=-2\). Since time cannot be negative, \(x=3\) days is valid. Exam tip: In work-rate problems, use the reciprocal of time for the rate and reject any negative time value.

Adding the filling rates gives \(\frac{1}{x}+\frac{1}{x+4}=\frac{5}{24}\). On simplifying, \(24(2x+4)=5x(x+4)\), so \(5x^2-28x-96=0\). Factoring gives \((x-8)(5x+12)=0\), yielding \(x=8\) or \(x=-\frac{12}{5}\). Since time cannot be negative, \(x=8\) hours. The other pipe takes 12 hours, so the 8-hour pipe is the faster one. Exam tip: In filling-rate problems, add the reciprocals of the individual times and reject any negative time value.

Let the original speed be \(x\) km/h. Since the time decreases by 1 hour, \(\frac{150}{x}-\frac{150}{x+5}=1\). On simplifying, \(x^2+5x-750=0\), so \((x-25)(x+30)=0\). Speed cannot be negative; hence \(x=25\) km/h. For comparison, at 20 km/h the time difference would be \(\frac{150}{20}-\frac{150}{25}=1.5\) hours, not 1 hour. Exam tip: use time = distance ÷ speed and reject any negative root.

Let the breadth be \(x\) metres. Then the length is \(2x+2\) metres. Using area, \(x(2x+2)=180\), which gives \(x^2+x-90=0\). Factoring, \((x+10)(x-9)=0\), so \(x=9\) or \(x=-10\). Since a length cannot be negative, the breadth is \(9\) metres. Exam tip: In dimension-based problems, reject any negative root as physically meaningless.

Let the original price of each notebook be x rupees. Then the number of notebooks bought is \(\frac{720}{x}\), while at a price reduced by 6 rupees, the number would be \(\frac{720}{x-6}\). Hence, \(\frac{720}{x-6}-\frac{720}{x}=4\). On simplifying, \(x^2-6x-1080=0\), or \((x-36)(x+30)=0\). Since a price must be positive, \(x=36\) rupees. Thus, option C is correct. Exam tip: In price-based word problems, express the number of items as total cost divided by the price per item, and reject any negative root.

Let the shorter leg be \(x\) cm. Then the longer leg is \((x+7)\) cm. By the Pythagorean theorem, \(x^2+(x+7)^2=17^2\), which simplifies to \(x^2+7x-120=0\). Factoring gives \((x-8)(x+15)=0\). Since a length cannot be negative, \(x=8\) cm. For instance, choosing \(9\) cm would give legs of \(9\) cm and \(16\) cm, which do not produce a hypotenuse of \(17\) cm. Exam tip: In right-triangle word problems, represent the unknown side by \(x\), apply the Pythagorean theorem, and reject any negative length.

Let the smaller multiple be \(x\); then the next consecutive multiple is \(x+3\). Thus, \(x(x+3)=270\), giving \(x^2+3x-270=0\). Factoring gives \((x-15)(x+18)=0\). Since the multiples are positive, \(x=15\), so the two multiples are 15 and 18. Exam tip: You can also quickly verify the options by multiplying each pair and checking which product equals 270.

By the Pythagorean theorem, \(x^2+(x+7)^2=(x+8)^2\). On simplifying, \(x^2-2x-15=0\), so \((x-5)(x+3)=0\). Thus, \(x=5\) or \(x=-3\); a length cannot be negative, so \(x=5\) cm. Since \(x\) is smaller than \(x+7\), the smallest side is 5 cm. Exam tip: Apply the Pythagorean theorem first and reject any negative root because side lengths are positive.

Let the number be \(x\). Then \(x-\frac{1}{x}=\frac{3}{2}\). Multiplying by \(2x\) gives \(2x^2-3x-2=0\), which factors as \((2x+1)(x-2)=0\). Thus, \(x=-\frac{1}{2}\) or \(x=2\). Since the number is positive, \(x=2\) is the valid answer. Check: \(2-\frac{1}{2}=\frac{3}{2}\). Exam tip: After clearing the reciprocal, always apply the condition given in the question, such as positivity.

Let one person’s age be x years; then the other person’s age is 30 − x years. Using the product, x(30 − x) = 216, which gives x² − 30x + 216 = 0. Factoring, (x − 12)(x − 18) = 0, so the ages are 12 years and 18 years. Option B has the correct sum, but 11 × 19 = 209, not 216. Exam tip: in such questions, verify both the sum and the product for the options.

Let the original speed be \(x\) km/h. The reduced speed is \(x-15\) km/h, so \(\frac{180}{x-15}-\frac{180}{x}=1\). On simplifying, \(x^2-15x-2700=0\), giving \(x=60\) or \(x=-45\). Speed cannot be negative, and it must be greater than 15 km/h, so the original speed is 60 km/h. Verification: \(\frac{180}{45}-\frac{180}{60}=4-3=1\) hour. Exam tip: use \(\text{time}=\frac{\text{distance}}{\text{speed}}\) and discard any physically impossible root.

The wire forms the complete perimeter of the rectangle, so 2(length + breadth) = 40, giving length + breadth = 20. Also, length × breadth = 96. The two dimensions are 8 m and 12 m because their sum is 20 and their product is 96. Therefore, the difference is 12 − 8 = 4 m. Exam tip: When a wire forms a rectangle, treat the wire length as the perimeter.

Let the width of the path be \(x\) metres. The outer rectangle then has length \(18+2x\) and breadth \(12+2x\). Hence, \((18+2x)(12+2x)=352\), which gives \(x^2+15x-34=0\). Factoring, \((x-2)(x+17)=0\), so \(x=2\) or \(x=-17\). A width cannot be negative, so \(x=2\) metres. Exam tip: Since the path is added on both sides of each dimension, use \(18+2x\) and \(12+2x\), not \(18+x\) and \(12+x\).