Word Problems and Applications

If the middle integer is (x), the first and third are (x-1) and (x+1). From ((x-1)(x+1)=143), (x=12).

Let the middle even integer be n. Since the integers are consecutive and even, the first and third are n − 2 and n + 2. Their product is (n − 2)(n + 2) = 320. Using the difference-of-squares identity gives n² − 4 = 320, so n² = 324. Thus n = 18 or n = −18. The options and the usual positive-integer interpretation select 18. Verification is immediate: the three integers are 16, 18, and 20, and the product of the first and third is 16 × 20 = 320. Option B is the first integer, option C is the third integer, and option D does not produce the stated product with its neighboring even integers.

Let the breadth be \(x\) units; then the length is \(x+7\) units. By the Pythagorean theorem, \(x^2+(x+7)^2=13^2\). On simplifying, \(x^2+7x-60=0\), or \((x+12)(x-5)=0\). Thus \(x=5\) or \(x=-12\). Since a breadth cannot be negative, the breadth is 5 units. Option 12 is a distractor related to the negative root \(-12\). Exam tip: Reject negative values when a variable represents a physical length or distance.

For a square, area equals side squared. Let the original side be s. Since s² = 144 and a side length is positive, s = √144 = 12. Increasing the side by 3 gives a new side of 15, so the new area is 15² = 225. The same result follows from the identity (12 + 3)² = 12² + 2(12)(3) + 3² = 144 + 72 + 9 = 225. Therefore option A is correct. Option B, 169, would correspond to a side of 13, meaning an increase of only 1. Option C corresponds to side 14, and option D is not the square of the new side. The positive length condition also rules out taking −12 as the original side.

Let the number be \(x\). Then \(x^2=10x+24\), so \(x^2-10x-24=0\). Factoring gives \((x-12)(x+2)=0\), hence \(x=12\) or \(x=-2\). Since the question asks for the positive number, the correct answer is 12. Exam tip: Find both roots of a quadratic equation and then apply any condition given in the question, such as positivity.

Let the number be \(x\). Then \(x+x^2=72\), so \(x^2+x-72=0\). Factoring gives \((x-8)(x+9)=0\), hence \(x=8\) or \(x=-9\). The positive number is therefore 8. In such questions, find both roots and then apply the stated positive or negative condition.

The correct answer is A. Let the numbers be a and b. We know a+b=22 and a²+b²=274. Using (a+b)²=a²+b²+2ab, we get 22²=274+2ab, so 484−274=2ab and ab=105. Therefore a and b are roots of t²−22t+105=0. Factoring gives (t−15)(t−7)=0, so the two numbers are 15 and 7. The larger number is 15. Option B is the smaller number, C is the average of the two numbers, and D satisfies neither the required pair nor the square-sum condition with 9. Verification: 15+7=22 and 15²+7²=225+49=274. The identity prevents guessing and is useful in many word problems.

Let the smaller number be \(x\). Then the larger number is \(x+6\). Thus, \(x(x+6)=216\), giving \(x^2+6x-216=0\). Factoring, \((x-12)(x+18)=0\), so \(x=12\) or \(x=-18\). Since the numbers are positive, \(-18\) is rejected. Therefore, the smaller number is 12. Exam tip: In word problems involving positive numbers, discard any negative root.

Let the breadth of the room be \(x\) units. Then its length is \(x+10\) units. Using the area, \(x(x+10)=300\), so \(x^2+10x-300=0\). Factoring gives \((x-15)(x+20)=0\), hence \(x=15\) or \(x=-20\). Since a length cannot be negative, the breadth is 15 units. In word problems, always reject roots that are not physically meaningful.

The breadth is 8, so the length is \(3(8-2)=3\times6=18\). Therefore, the area is length \(\times\) breadth = \(18\times8=144\) square units. Exam tip: “three times the number obtained by subtracting 2” means \(3(8-2)\), not \(3\times8-2\).

Let the positive number be \(x\). The statement gives \(x(x+6)=135\), so \(x^2+6x-135=0\). Factoring, \((x-9)(x+15)=0\), gives \(x=9\) or \(x=-15\). Since the number must be positive, the valid answer is \(9\). Check: \(9\times(9+6)=9\times15=135\). Exam tip: Translate the wording into an equation such as \(x(x+6)\) first, then apply the condition that the required number is positive.

Let the number be \(x\). According to the question, \(x(x-4)=77\), so \(x^2-4x-77=0\). Factoring gives \((x-11)(x+7)=0\), hence \(x=11\) or \(x=-7\). The positive number is 11; the close distractor 7 is incorrect because \(7(7-4)=21\), not 77. Exam tip: Translate the wording into a quadratic equation first, then select the solution that satisfies the stated condition.

Let the breadth of the poster be \(x\) units. Then its length is \(x+3\) units. Using the area, \(x(x+3)=154\), so \(x^2+3x-154=0\). Factoring gives \((x+14)(x-11)=0\), yielding \(x=11\) or \(x=-14\). Since a length cannot be negative, the breadth is 11 units. Exam tip: In dimension-based problems, reject any negative root.

Let the uniform border width be x units. Since the border is removed from both opposite sides in each direction, the inner length becomes 20-2x and the inner breadth becomes 16-2x. The given inner area therefore gives \\(20-2x)(16-2x)=192\\. The width must also be positive and less than 8.

Testing option A, x=2 gives inner dimensions 16 and 12, whose area is 16 times 12=192. Thus option A is correct. Algebraically, expansion gives 4x^2-72x+128=0, or x^2-18x+32=0, with roots 2 and 16. The root 16 is impossible because it would make the inner dimensions invalid, so the physical border width is 2 units.

Let the path width be (x), then ((30-2x)(20-2x)=384). Substitution must be checked carefully before choosing the option.

The correct answer is B. Let the inside path width be w metres. Because the path is on both sides, the remaining rectangle has length 30−2w and width 20−2w. Its area is 416, so (30−2w)(20−2w)=416. Expanding gives 600−100w+4w²=416, or 4w²−100w+184=0. Dividing by 4 gives w²−25w+46=0, which factors as (w−2)(w−23)=0. The root w=23 is impossible because it makes the inner dimensions negative. Thus w=2 m. Check: the inner dimensions are 26 m and 16 m, and 26×16=416. Option A, C and D give different remaining areas. The factor 2 is essential because the path reduces each dimension at both ends.

If the path width is (x), then ((28-2x)(18-2x)=352). Substituting (x=2) gives (24\times14=336), so match the data carefully.

Let the path width be (x), then ((32-2x)(22-2x)=504). Substituting (x=2) gives (28\times18=504), so the width is (2).

Let the path width be (x), then ((25+2x)(15+2x)=589). Substituting (x=3) gives (31\times21=651), so this option is not valid.

Let the uniform outside-path width be x metres. The path adds x metres on each side, so the outside rectangle has length 25 + 2x and width 15 + 2x. Its area therefore satisfies (25 + 2x)(15 + 2x) = 651. Expanding gives 4x² + 80x + 375 = 651, or 4x² + 80x − 276 = 0; dividing by 4 gives x² + 20x − 69 = 0. This factors as (x − 3)(x + 23) = 0. The width must be positive, so x = 3. Checking gives outside dimensions 31 m and 21 m, with area 31 × 21 = 651. Thus option A is correct; the other options fail the area condition.

Total revenue equals price per item multiplied by the number of items, so \(x(60-x)=875\). Rearranging gives \(x^2-60x+875=0\), or \((x-25)(x-35)=0\). Thus, \(x=25\) or \(x=35\), and the smaller value is 25. Exam tip: In such problems, first form the revenue equation and then select the required root from the two solutions.

From the given relation, \(x(20-x)=96\). Rearranging gives \(20x-x^2=96\), or \(x^2-20x+96=0\). Factoring, \((x-8)(x-12)=0\), so \(x=8\) or \(x=12\). Therefore, the larger value of \(x\) is 12. Exam tip: after finding both roots of a quadratic equation, select the larger or smaller one according to what the question asks.

If the number of persons is (n), then the number of handshakes is (\frac{n(n-1)}{2}=45). This gives (n=10).

Let the original speed be \(x\) kilometres per hour. The original travel time is \(\frac{180}{x}\) hours, while the time at the increased speed \(x+15\) is \(\frac{180}{x+15}\) hours. Hence, \(\frac{180}{x}-\frac{180}{x+15}=1\). On simplifying, \(x^2+15x-2700=0\), or \((x-45)(x+60)=0\). Thus, \(x=45\) or \(x=-60\); since speed cannot be negative, the original speed was 45 kilometres per hour. Exam tip: In distance–speed–time problems, use \(\text{time}=\frac{\text{distance}}{\text{speed}}\) before forming the quadratic equation.

Let the shorter leg be \(x\) units. Then the longer leg is \(x+14\) units. By the Pythagorean theorem, \(x^2+(x+14)^2=34^2\), which simplifies to \(x^2+14x-480=0\). Factoring gives \((x-16)(x+30)=0\), so \(x=16\) or \(x=-30\). Since a length cannot be negative, the shorter leg is 16 units. Remember that 30 units is the longer leg, not the shorter one.