Word Problems and Applications

If the original side is (x), then (x^2=196), so (x=14). The new side is (18), and the new area is (18^2=324).

Let the number be \(x\). Then \(x^2=15x+54\), so \(x^2-15x-54=0\). Factoring gives \((x-18)(x+3)=0\), hence \(x=18\) or \(x=-3\). The positive number is therefore 18. Exam tip: Find both roots of a quadratic equation and then apply the condition given in the question, such as ‘positive’.

If one number is (x), the other is (26-x). From (x^2+(26-x)^2=340), the numbers are (14) and (12).

Let the smaller number be \(x\). Then the larger number is \(x+3\). Thus, \(x(x+3)=378\), which gives \(x^2+3x-378=0\). Factoring, \((x-18)(x+21)=0\), so \(x=18\) or \(x=-21\). Since both numbers are positive, \(-21\) is rejected and the smaller number is 18. Exam tip: In word problems, always check the given condition, such as positivity, before accepting a quadratic root.

Let the breadth be \(x\) units. Then the length is \(x+6\) units, so the area gives \(x(x+6)=432\), or \(x^2+6x-432=0\). Factoring, \((x+24)(x-18)=0\), so \(x=18\) or \(x=-24\). Since a length cannot be negative, the breadth is 18 units. Exam tip: Reject any negative root when solving for a physical dimension.

The breadth is 9 units. According to the statement, the length is \(4(9-3)=4\times6=24\) units. Therefore, the area is \(\text{length}\times\text{breadth}=24\times9=216\) square units. The value 108 does not result from multiplying the correct length and breadth. In such questions, first evaluate the quantity inside the brackets and then apply the stated multiplier.

Let the positive number be \(x\). According to the question, \(x(x+8)=209\), so \(x^2+8x-209=0\). Factoring gives \((x-11)(x+19)=0\), hence \(x=11\) or \(x=-19\). Since the number must be positive, the correct answer is 11. Although 19 is a factor of 209, it does not satisfy the condition because \(19(19+8)=513\). Exam tip: after solving a quadratic, check the given condition, such as positivity, before selecting the root.

Let the number be \(x\). According to the question, \(x(x-6)=135\), so \(x^2-6x-135=0\). Factoring gives \((x-15)(x+9)=0\), hence \(x=15\) or \(x=-9\). Since the question asks for the positive number, the answer is \(15\). In an exam, find both roots and then apply the stated condition, such as ‘positive’.

Let the breadth be \(x\) units. Then the length is \(x+6\) units, so \(x(x+6)=216\), or \(x^2+6x-216=0\). Factoring gives \((x-12)(x+18)=0\), so \(x=12\) or \(x=-18\). Since a length cannot be negative, the breadth is \(12\) units. Option 18 is the length, not the breadth. Exam tip: In rectangle-area problems, let one side be \(x\) and express the other side in terms of \(x\) before forming the quadratic equation.

Let the border width be (x), then the inner area is ((24-2x)(18-2x)=280). (x=2) is the correct value.

Let the path width be \(x\) metres. The remaining rectangle has length \(36-2x\) and width \(24-2x\). Thus, \((36-2x)(24-2x)=540\), which gives \(x^2-30x+81=0\). Its roots are \(x=3\) and \(x=27\), but the path width must be less than half the garden's breadth, so \(x=27\) is not feasible. Therefore, the correct width is 3 metres. Exam tip: for a path inside all four sides, subtract \(2x\) from each original dimension.

Let the width of the path be \(x\) metres. Then the outer rectangle has length \(30+2x\) and width \(18+2x\). Thus, \((30+2x)(18+2x)=748\). On simplifying, \(x^2+24x-52=0\), or \((x-2)(x+26)=0\). Since a width must be positive, \(x=2\) metres is the valid answer; \(x=-26\) is not physically meaningful. Exam tip: because the path is added on both opposite sides, add \(2x\) to each original dimension.

Total revenue equals price per item multiplied by the number of items, so \(x(80-x)=1500\). Rearranging gives \(x^2-80x+1500=0\), or \((x-30)(x-50)=0\). Thus, \(x=30\) or \(x=50\); the smaller value is 30. In such problems, first form the revenue equation and then compare both roots.

The given equation is \(x(30-x)=216\). On simplifying, we get \(x^2-30x+216=0\), which factors as \((x-12)(x-18)=0\). Therefore, \(x=12\) or \(x=18\), and the larger value is 18. Exam tip: after finding both roots of a quadratic equation, select the larger or smaller one exactly as asked.

If the number of people is \(n\), each handshake corresponds to one pair of people. Therefore, the total number of handshakes is \(\frac{n(n-1)}{2}\). Thus, \(\frac{n(n-1)}{2}=66\), giving \(n(n-1)=132\). The positive solution is \(n=12\), since \(\frac{12\times11}{2}=66\). The closest distractor, 11, is incorrect because 11 people would make \(\frac{11\times10}{2}=55\) handshakes. Exam tip: use the formula \(\frac{n(n-1)}{2}\) for one-time handshake problems.

If the number of teams is \(n\), each match corresponds to one pair of distinct teams. Therefore, the number of matches is \(\frac{n(n-1)}{2}\). Thus, \(\frac{n(n-1)}{2}=28\), giving \(n(n-1)=56\) and \(n^2-n-56=0\). Factoring, \((n-8)(n+7)=0\), so \(n=8\) or \(n=-7\). Since the number of teams cannot be negative, \(n=8\) is the correct answer. Exam tip: For a single round-robin tournament, use \(\frac{n(n-1)}{2}\) and reject any negative root.

The sum of the two sides of a rectangle is half its perimeter, so the sum is \(60/2=30\) metres. If one side is \(x\) metres, the other is \(30-x\) metres. Thus, \(x(30-x)=216\), giving \(x^2-30x+216=0\). Factoring gives \((x-18)(x-12)=0\), so the sides are 18 metres and 12 metres. Therefore, the longer side is 18 metres. In exams, first use half the perimeter to find the sum of the two sides.

If the units digit is (x), then the tens digit is (x+3). From (x(x+3)=18), (x=3), so the number is (63).

If (B) alone takes (x) days, then (A) takes (x-5) days. From (\frac{1}{x}+\frac{1}{x-5}=\frac{1}{6}), (x=15).

If the smaller integer is (x), then (x^2+(x+1)^2=841), giving (x=20). Write consecutive numbers as (x) and (x+1).