Word Problems and Applications

If breadth is (x), then (x(x+5)=150), so (x=10). In word problems choose the variable carefully first.

Let the smaller integer be \(x\). Then the other integer is \(x+1\). Thus, \(x(x+1)=240\), or \(x^2+x-240=0\). Factoring gives \((x-15)(x+16)=0\). Since the integers are positive, \(x=15\), so the pair is 15 and 16; indeed, \(15\times16=240\). Option A is close but \(14\times15=210\), not 240. Exam tip: For consecutive integers, represent them as \(x\) and \(x+1\) before forming the quadratic equation.

The mother's present age is \(x+20\) years. Five years ago, their ages were \(x-5\) and \(x+15\) years, respectively. Thus, \((x-5)(x+15)=189\), which gives \(x^2+10x-264=0\). Factoring, \((x-12)(x+22)=0\). Since age cannot be negative, \(x=12\) years, so option C is correct. In age problems, remember to subtract the stated number of years from both present ages.

Let the car’s original speed be \(x\) km/h. The original time is \(\frac{240}{x}\) hours, and after increasing the speed to \(x+20\) km/h, the new time is \(\frac{240}{x+20}\) hours. Therefore, \(\frac{240}{x}-\frac{240}{x+20}=1\). On simplifying, \(x^2+20x-4800=0\), which factors as \((x-60)(x+80)=0\). Since speed must be positive, \(x=60\) km/h is the valid answer; \(-80\) km/h is rejected. Exam tip: In speed–time–distance problems, write the times for both situations first and then form their difference equation.

Let the width of the path be \(x\) m. The dimensions of the remaining grassy rectangle are \((30-2x)\) m and \((20-2x)\) m. Thus, \((30-2x)(20-2x)=416\), which gives \(x^2-25x+46=0\), or \((x-2)(x-23)=0\). The root \(x=23\) m is not possible because the path width must be less than half the breadth, so the valid value is \(x=2\) m. Exam tip: for a path inside all four sides, subtract \(2x\) from both original dimensions.

Let the first odd number be \(x\). The next consecutive odd number is \(x+2\), since consecutive odd numbers differ by 2. Thus, \(x(x+2)=255\), giving \(x^2+2x-255=0\). Factoring, \((x-15)(x+17)=0\), so the positive value is \(x=15\). Therefore, the numbers are 15 and 17, and \(15\times17=255\). Exam tip: for consecutive odd numbers, use a difference of 2, not 1.

Let the other side be \(x\) cm. Then the hypotenuse is \((x+5)\) cm. By the Pythagorean theorem, \(x^2+10^2=(x+5)^2\), so \(x^2+100=x^2+10x+25\). Hence, \(10x=75\) and \(x=7.5\) cm. Therefore, option D is correct. Exam tip: in a right triangle, the hypotenuse is the longest side, so it must be represented by \(x+5\), not by \(x\).

Let the tens digit be \(x\). Then the units digit is \(x+1\). Therefore, \(x^2+(x+1)^2=61\), which gives \(2x^2+2x-60=0\) or \(x^2+x-30=0\). Thus, \((x-5)(x+6)=0\), so \(x=5\) or \(x=-6\). Since a digit cannot be negative, the tens digit is 5 and the units digit is 6; hence the number is 56. Exam tip: In digit problems, reject roots that are negative or greater than 9. The nearby option 67 has consecutive digits, but their squared sum is 36+49=85, not 61.

The combined work rate is \(\frac{1}{6}\) of the task per day, so \(\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}\). Clearing the denominators gives \(6(2x+5)=x(x+5)\), or \(x^2-7x-30=0\). Factoring, \((x-10)(x+3)=0\), so \(x=10\) or \(x=-3\). Since a number of days cannot be negative, \(x=10\), making option B correct. Exam tip: In work-rate problems, add the reciprocals of the individual completion times.

Let the breadth of the rectangle be \(x\) metres. Then its length is \(x+4\) metres. Using the area, \(x(x+4)=96\), so \(x^2+4x-96=0\). Factoring gives \((x-8)(x+12)=0\), hence \(x=8\) or \(x=-12\). Since a length cannot be negative, the breadth is 8 metres. Exam tip: In rectangle word problems, assign a variable to one dimension and express the other using the given relationship before applying the area formula.

The equation (x+\frac{1}{x}=\frac{5}{2}) gives (2x^2-5x+2=0). The solutions are (x=2) or (x=\frac{1}{2}).

Let the number be \(x\). Then \(x^2=3x+28\), so \(x^2-3x-28=0\). Factoring gives \((x-7)(x+4)=0\), hence \(x=7\) or \(x=-4\). Since the number is positive, the correct answer is 7. Exam tip: when a word problem specifies a positive number, discard any negative root.

Let the height of the triangle be \(x\) cm. Then its base is \((x+4)\) cm. Using the area formula, \(\frac{1}{2}x(x+4)=60\), which gives \(x^2+4x-120=0\). Factoring, \((x-10)(x+12)=0\), so \(x=10\) or \(x=-12\). Since a length cannot be negative, the height is \(10\) cm. Exam tip: Always reject a negative root when the variable represents a physical length.

The wall, ground and ladder form a right-angled triangle. By the Pythagorean theorem, \(x^2+(x+7)^2=13^2\). Simplifying gives \(2x^2+14x-120=0\), or \((x-5)(x+12)=0\). Since a distance cannot be negative, \(x=5\) metres is valid; \(x=-12\) is rejected. In such word problems, always choose the root that is physically meaningful.

Let the length and breadth be x and y. From the perimeter, 2(x+y)=50, so x+y=25; from the area, xy=150. Substituting y=25-x gives \(x(25-x)=150\), or \(x^2-25x+150=0\). Factoring, \((x-10)(x-15)=0\), so the dimensions are 10 m and 15 m. Option B has the correct sum of 25 but its product is 144, not 150, so it does not satisfy the area condition. Exam tip: In rectangle problems, use the perimeter to find the sum of the dimensions and the area to find their product.

Let the original price per item be \(x\) rupees. The original number of items is \(\frac{1200}{x}\), while after the price reduction it is \(\frac{1200}{x-10}\). Therefore, \(\frac{1200}{x-10}-\frac{1200}{x}=4\). Simplifying gives \(x(x-10)=3000\), or \(x^2-10x-3000=0\). Thus, \(x=60\) or \(x=-50\); a price cannot be negative, so the original price was 60 rupees. Exam tip: In price-quantity problems, express the number of items as total money divided by price and reject any impossible root.

Let the speed of the boat in still water be \(x\) km/h. Its upstream and downstream speeds are then \(x-4\) and \(x+4\) km/h, respectively. Therefore, \(\frac{48}{x-4}-\frac{48}{x+4}=3\). On simplifying, we get \(x^2=144\), so \(x=12\) or \(x=-12\). Since speed is positive and must exceed the stream speed, the valid answer is 12 km/h. Exam tip: In boat-and-stream problems, use \(x-v\) for upstream speed and \(x+v\) for downstream speed.

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.