Word Problems and Applications

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 \(12+x\).

Let the breadth be \(x\) cm. Then the length is \((x+5)\) cm. Using the area, \(x(x+5)=300\), so \(x^2+5x-300=0\). Factoring gives \((x-15)(x+20)=0\), hence \(x=15\) or \(x=-20\). Since a length cannot be negative, the breadth is \(15\) cm. In such word problems, always reject the negative root as physically meaningless.

Let the smaller number be \(x\). The next consecutive odd number is \(x+2\), so \(x^2+(x+2)^2=394\). Simplifying gives \(2x^2+4x-390=0\), or \(x^2+2x-195=0=(x+15)(x-13)\). Thus, \(x=13\) or \(-15\); since the number must be positive, the smaller number is \(13\). Exam tip: consecutive odd numbers differ by 2, not by 1.

Let the original speed be \(x\) km/h. The decrease in travel time gives \(\frac{240}{x}-\frac{240}{x+10}=2\). Simplifying, \(x^2+10x-1200=0\), or \((x-30)(x+40)=0\). Thus \(x=30\) or \(-40\); the negative value cannot represent a speed. Therefore, the original speed was 30 km/h. Exam tip: For a fixed distance, use \(\text{time}=\frac{\text{distance}}{\text{speed}}\) and reject any negative root.

Let the tens digit be \(x\). Then the units digit is \(x+1\). Therefore, \(x^2+(x+1)^2=41\), which gives \(2x^2+2x-40=0\) or \((x-4)(x+5)=0\). Since a digit cannot be negative, \(x=4\); hence the units digit is 5 and the number is 45. Although 56 also has digits differing by 1, its digit-square sum is \(25+36=61\), not 41. Exam tip: reject any negative root when the variable represents a digit.

Let the breadth of the hall be \(x\) metres. Its length will then be \(x+2\) metres. Using the area, \(x(x+2)=168\), so \(x^2+2x-168=0\). Factoring gives \((x+14)(x-12)=0\), yielding \(x=12\) or \(x=-14\). Since a length cannot be negative, the breadth is 12 metres. For example, 10 metres would give an area of \(10\times12=120\) square metres, so it is not correct. Exam tip: reject any negative root when solving for a physical dimension.

Let the number of students be \(x\). Each student donated \(x+2\) rupees, so \(x(x+2)=168\). Thus, \(x^2+2x-168=0\), which factors as \((x+14)(x-12)=0\). Therefore, \(x=12\) or \(x=-14\); the number of students cannot be negative, so the correct answer is 12, option B. Exam tip: In word problems, reject roots that are impossible in the given context, such as a negative number of students.

If the still-water speed is (x), then (\frac{30}{x-2}+\frac{30}{x+2}=8). Solving gives (x=8).

Let the smaller integer be (x); then the next integer is (x+1). Thus, x(x+1)=552, giving x^2+x-552=0. Factoring gives (x-23)(x+24)=0, so x=23 or x=-24. Since the integers are positive, x=23 is valid, and the numbers are 23 and 24. Exam tip: represent consecutive integers as x and x+1 before forming the quadratic equation.

If one number is (x), the other is (31-x). From (x(31-x)=234), the numbers are (13) and (18).

Here (l+b=25) and (lb=150). This gives (t^2-25t+150=0), so the measures are (10) cm and (15) cm.

Let the breadth of the rectangle be \(x\) cm. Its length is then \(x+3\) cm. Using the area, \(x(x+3)=180\), so \(x^2+3x-180=0\). Factoring gives \((x+15)(x-12)=0\), hence \(x=12\) or \(x=-15\). Since a length cannot be negative, \(-15\) is rejected and the breadth is 12 cm. Exam tip: In geometrical word problems, discard any negative root because dimensions must be positive.

Let the breadth be \(x\) m. Then the length is \(x+5\) m, so \(x(x+5)=336\), giving \(x^2+5x-336=0\). Factoring, \((x+21)(x-16)=0\), so \(x=16\) or \(x=-21\). Since a breadth cannot be negative, the correct answer is 16 m. Exam tip: reject any negative root when the variable represents a physical length.

Let the original side be \(x\) m. The increase in area gives \((x+4)^2-x^2=96\). Expanding, \(x^2+8x+16-x^2=96\), so \(8x=80\) and \(x=10\) m. Therefore, option B is correct. Exam tip: if the side of a square increases by \(a\), the area increase is \(2ax+a^2\); here, \(2\times4\times x+4^2=96\) gives the answer directly.

Let the smaller integer be \(x\). Since the integers are consecutive and even, the other integer is \(x+2\). Thus, \(x(x+2)=360\), giving \(x^2+2x-360=0\). Factoring, \((x-18)(x+20)=0\), so \(x=18\) or \(x=-20\). Because the integers are positive, the smaller integer is 18. Exam tip: consecutive even integers always differ by 2.

Let the smaller integer be \(x\). Then the larger integer is \(x+2\). Thus, \(x(x+2)=483\), or \(x^2+2x-483=0\). Factoring gives \((x-21)(x+23)=0\), so \(x=21\) or \(x=-23\). Since the integers are positive, \(x=21\), and the larger integer is \(21+2=23\). Exam tip: consecutive odd integers always differ by 2, not 1.

Let the positive number be \(x\). Then \(x(x+7)=330\), so \(x^2+7x-330=0\). Factoring gives \((x-15)(x+22)=0\), yielding \(x=15\) or \(x=-22\). Since the number must be positive, \(x=15\) is correct. Exam tip: verify the answer using \(15\times(15+7)=15\times22=330\).

Let the number be \(x\). Then \(x^2+9x=400\), or \(x^2+9x-400=0\). Factoring gives \((x-16)(x+25)=0\), so \(x=16\) or \(x=-25\). Since the question asks for the positive number, the answer is 16. Exam tip: Find both roots of the quadratic and then apply the condition specified in the question.

The equation is (x^2-14x=51). From (x^2-14x-51=0), (x=17) or (x=-3), so the positive number is (17).