Word Problems and Applications

Correct answer: C. 12

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.

Correct answer: C. 13 and 14

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.

Correct answer: C. 14 and 16

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.

Correct answer: C. 15 cm

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.

Correct answer: C. 11

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.

Correct answer: A. 1 cm

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.

Correct answer: B. 7

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.

Correct answer: B. 25 m and 35 m

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.

Correct answer: B. 7 and 12

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.

Correct answer: C. 13

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.

Correct answer: C. 12

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.

Correct answer: C. 16 years

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.

Correct answer: A. 3 days

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.

Correct answer: B. 8 hours

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.

Correct answer: B. 25 km/h

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.

Correct answer: C. \(9\ \text{m}\)

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.

Correct answer: C. 36 rupees

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.

Correct answer: C. \(8\text{ cm}\)

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.

Correct answer: C. 15 and 18

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.

Correct answer: B. 5 cm

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.

Correct answer: C. \(2\)

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.

Correct answer: C. 12 years and 18 years

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.

Correct answer: D. 60 किमी/घंटा

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.

Correct answer: B. 4 m

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.

Correct answer: B. 2 m

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 \(12+x\).