Word Problems and Applications

If the smaller number is (x), the larger is (x+8). From (x(x+8)=273), (x=13), so the larger number is (21).

If the smaller number is (x), the larger is (x+4). From (x^2+(x+4)^2=250), (x=9), so the larger number is (13).

Let the positive integer be \(x\). Its product with the next integer is \(x(x+1)=552\), so \(x^2+x-552=0\). Factoring gives \((x-23)(x+24)=0\), yielding \(x=23\) or \(x=-24\). Since the number is specified as positive, \(x=23\) is correct. Check: \(23\times24=552\). Exam tip: Represent the number and its next integer as \(x\) and \(x+1\) before forming the quadratic equation.

Let the number be \(x\). Then \(x^2-17=127\), so \(x^2=144\) and \(x=\pm 12\). Since the question asks for the positive number, the answer is 12. The values 10 and 14 are incorrect because their squares are 100 and 196, respectively. Exam tip: after taking a square root, consider both signs and then select the value required by the question.

Let the height reached on the wall be \(x\) units. Then the distance of the foot of the ladder from the wall is \(x-17\) units. By the Pythagorean theorem, \(x^2+(x-17)^2=25^2\). This gives \(x^2-17x-168=0\), or \((x-24)(x+7)=0\). Thus \(x=24\) units, since a physical length cannot be negative. Exam tip: reject any negative root when solving a word problem involving length or distance.

Let the shorter leg be \(x\). Then the other leg is \(x+31\). By the Pythagorean theorem, \(x^2+(x+31)^2=41^2\), which simplifies to \(x^2+31x-360=0\). Factoring gives \((x-9)(x+40)=0\), so the roots are 9 and −40. Since a length cannot be negative, the shorter leg is 9. Exam tip: always reject a negative root when solving for a physical length.

Let the breadth be \(x\) units. Then the length is \(x+4\) units, so the area gives \(x(x+4)=360\). Thus, \(x^2+4x-360=0\), which factors as \((x-18)(x+20)=0\). Therefore, \(x=18\) or \(x=-20\). Since a length cannot be negative, the breadth is 18 units. Exam tip: Always reject a negative root when solving a geometrical measurement problem.

Let the breadth be \(x\) units. Then the length is \(2x\) units. Using area, \(x\times 2x=242\), so \(2x^2=242\), giving \(x^2=121\) and \(x=\pm 11\). Since length and breadth must be positive, the breadth is \(11\) units. The value \(22\) represents the length, not the breadth. Exam tip: reject the negative root when solving measurement problems.

We have \(x(70-x)=1200\). Rearranging gives \(x^2-70x+1200=0\), which factors as \((x-30)(x-40)=0\). Thus, \(x=30\) or \(x=40\), and the larger value is 40. Exam tip: In such problems, form the quadratic equation first, find both roots, and then select the one requested.

Let the breadth of the field be \(x\) metres. Then its length is \(x+8\) metres. Using the area, \(x(x+8)=384\), so \(x^2+8x-384=0\). Factoring gives \((x-16)(x+24)=0\), hence \(x=16\) or \(x=-24\). Since a length cannot be negative, the breadth is 16 m. Option B (24 m) is the corresponding length, not the breadth. Exam tip: Always reject negative roots when solving geometrical measurement problems.

Let the original speed be \(x\) km/h. Then \(\frac{360}{x}-\frac{360}{x+15}=2\). On simplifying, \(x(x+15)=2700\), so \(x^2+15x-2700=0\). Factoring gives \((x-45)(x+60)=0\), hence \(x=45\) or \(x=-60\). Since speed cannot be negative, the original speed is \(45\) km/h. Check: at 45 km/h the time is 8 hours, and at 60 km/h it is 6 hours, a decrease of exactly 2 hours. Exam tip: reject any negative root when the variable represents speed, distance, or time.

Let the original speed be \(x\) km/h. The original travel time is \(\frac{300}{x}\) hours, while the time at the reduced speed is \(\frac{300}{x-10}\) hours. Therefore, \(\frac{300}{x-10}-\frac{300}{x}=1\). On simplifying, \(x^2-10x-3000=0\), giving \(x=60\) or \(x=-50\). Since speed cannot be negative, the original speed is \(60\) km/h. Check: at 60 km/h the time is 5 hours, and at 50 km/h it is 6 hours, an increase of exactly 1 hour. Exam tip: reject physically impossible negative roots and verify the remaining value in the original condition.

The equation is (\frac{48}{20+x}+\frac{32}{20-x}=4). Substituting (x=4) gives time (2+2=4) hours.

If the number of students is (x), then (\frac{1800}{x}-\frac{1800}{x+10}=6). Solving gives (x=50).

If there are (x) students, then (\frac{1680}{x-4}-\frac{1680}{x}=10). Solving gives (x=28).

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).

If the middle one is (x), the first and third are (x-2) and (x+2). From ((x-2)(x+2)=480), (x=22).

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.