Consecutive Numbers And Path Width A Mathematical Exploration

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In the realm of mathematics, problem-solving often involves deciphering relationships between numbers and geometric figures. This article delves into two intriguing mathematical puzzles. The first focuses on identifying consecutive positive numbers based on the sum of their reciprocals, while the second explores the dimensions of a path surrounding a rectangular lawn. Let's embark on this mathematical journey and unravel the solutions to these captivating problems.

To solve this problem, we need to translate the word problem into an algebraic equation. Let the two consecutive positive numbers be n and n + 1. The reciprocals of these numbers are 1/n and 1/(n + 1), respectively. According to the problem, the sum of these reciprocals is 17/72. Thus, we have the equation:

1n+1n+1=1772\frac{1}{n} + \frac{1}{n+1} = \frac{17}{72}

To solve this equation, we first find a common denominator for the fractions on the left side, which is n(n + 1). Multiplying both sides of the equation by this common denominator, we get:

(n+1)+n=1772n(n+1)(n+1) + n = \frac{17}{72}n(n+1)

Simplifying the left side, we have:

2n+1=1772n(n+1)2n + 1 = \frac{17}{72}n(n+1)

Now, multiply both sides by 72 to eliminate the fraction:

72(2n+1)=17n(n+1)72(2n + 1) = 17n(n+1)

Expanding both sides, we get:

144n+72=17n2+17n144n + 72 = 17n^2 + 17n

Rearranging the terms to form a quadratic equation, we have:

17n2+17n−144n−72=017n^2 + 17n - 144n - 72 = 0

17n2−127n−72=017n^2 - 127n - 72 = 0

Now, we need to solve this quadratic equation for n. We can use the quadratic formula, factoring, or other methods to find the solutions. In this case, factoring seems like a viable option. We look for two numbers that multiply to 17 * -72 = -1224 and add up to -127. These numbers are -136 and 9. So, we can rewrite the middle term as -136n + 9n:

17n2−136n+9n−72=017n^2 - 136n + 9n - 72 = 0

Now, we factor by grouping:

17n(n−8)+9(n−8)=017n(n - 8) + 9(n - 8) = 0

(17n+9)(n−8)=0(17n + 9)(n - 8) = 0

This gives us two possible solutions for n:

17n+9=0⇒n=−91717n + 9 = 0 \Rightarrow n = -\frac{9}{17}

n−8=0⇒n=8n - 8 = 0 \Rightarrow n = 8

Since we are looking for positive numbers, we discard the negative solution. Therefore, n = 8. The next consecutive positive number is n + 1 = 8 + 1 = 9. Thus, the two consecutive positive numbers are 8 and 9. We can verify this by substituting these numbers back into the original equation:

18+19=9+872=1772\frac{1}{8} + \frac{1}{9} = \frac{9 + 8}{72} = \frac{17}{72}

This confirms that our solution is correct.

The second problem involves a rectangular lawn with a path of uniform width surrounding it. The dimensions of the lawn are 46 m by 34 m, and the area of the path is 425 square meters. Our task is to find the width of the path. This problem combines geometric concepts with algebraic problem-solving, requiring a clear understanding of area calculations and equation manipulation. Let's solve this problem step by step.

Let the width of the path be x meters. The lawn has dimensions 46 m by 34 m. When a path of uniform width x is added around the lawn, the new dimensions of the outer rectangle (including the lawn and the path) will be (46 + 2x) meters and (34 + 2x) meters. The area of the outer rectangle is the product of these new dimensions, which is (46 + 2x)(34 + 2x). The area of the lawn is simply the product of its dimensions, which is 46 * 34.

The area of the path is the difference between the area of the outer rectangle and the area of the lawn. We are given that the area of the path is 425 square meters. Therefore, we can set up the following equation:

(46+2x)(34+2x)−(46imes34)=425(46 + 2x)(34 + 2x) - (46 imes 34) = 425

Expanding the product (46 + 2x)(34 + 2x), we get:

1564+92x+68x+4x2−1564=4251564 + 92x + 68x + 4x^2 - 1564 = 425

Simplifying the equation by combining like terms and canceling out 1564, we have:

4x2+160x=4254x^2 + 160x = 425

Now, we rearrange the equation to form a quadratic equation in the standard form:

4x2+160x−425=04x^2 + 160x - 425 = 0

To solve this quadratic equation, we can use the quadratic formula, factoring, or completing the square. The quadratic formula is given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where a = 4, b = 160, and c = -425. Substituting these values into the formula, we get:

x=−160±1602−4(4)(−425)2(4)x = \frac{-160 \pm \sqrt{160^2 - 4(4)(-425)}}{2(4)}

x=−160±25600+68008x = \frac{-160 \pm \sqrt{25600 + 6800}}{8}

x=−160±324008x = \frac{-160 \pm \sqrt{32400}}{8}

x=−160±1808x = \frac{-160 \pm 180}{8}

This gives us two possible solutions for x:

x1=−160+1808=208=2.5x_1 = \frac{-160 + 180}{8} = \frac{20}{8} = 2.5

x2=−160−1808=−3408=−42.5x_2 = \frac{-160 - 180}{8} = \frac{-340}{8} = -42.5

Since the width of the path cannot be negative, we discard the negative solution. Therefore, the width of the path is 2.5 meters. To verify this solution, we can substitute x = 2.5 back into the original equation:

(46+2(2.5))(34+2(2.5))−(46imes34)=(46+5)(34+5)−1564(46 + 2(2.5))(34 + 2(2.5)) - (46 imes 34) = (46 + 5)(34 + 5) - 1564

=(51)(39)−1564= (51)(39) - 1564

=1989−1564= 1989 - 1564

=425= 425

This confirms that our solution is correct.

In conclusion, we have successfully solved two distinct mathematical problems. The first involved finding two consecutive positive numbers whose reciprocals sum to 17/72, and we determined these numbers to be 8 and 9. The second problem required us to find the width of a path surrounding a rectangular lawn, given the area of the path, and we found the width to be 2.5 meters. These problems highlight the importance of translating word problems into algebraic equations, applying relevant formulas and techniques, and verifying the solutions. The process of mathematical problem-solving is not only about finding the right answers but also about developing logical thinking, analytical skills, and a deeper appreciation for the elegance and power of mathematics. By tackling such challenges, we enhance our mathematical prowess and our ability to approach and solve real-world problems with confidence.

In summary, this article has explored two engaging mathematical problems, each requiring a unique approach and set of problem-solving skills. The first problem focused on number theory, specifically the relationship between consecutive positive integers and the sum of their reciprocals. By setting up and solving a quadratic equation, we successfully identified the two consecutive numbers as 8 and 9. This exercise underscored the importance of algebraic manipulation and equation-solving techniques in number theory problems. The second problem shifted our focus to geometry, challenging us to determine the width of a path surrounding a rectangular lawn. This problem involved understanding the relationship between areas and dimensions, as well as the ability to form and solve a quadratic equation based on the given information. Through the application of the quadratic formula, we found the width of the path to be 2.5 meters. This problem highlighted the practical applications of geometry and algebra in real-world scenarios. Both problems exemplify the beauty and versatility of mathematics as a problem-solving tool. By engaging with these challenges, readers can enhance their mathematical intuition, strengthen their problem-solving abilities, and develop a deeper appreciation for the interconnectedness of mathematical concepts. Ultimately, the ability to approach and solve mathematical problems effectively is a valuable skill that can be applied across various disciplines and aspects of life.