We are given that the sum of the reciprocals of the two parts equals (3)/(20). We set up the equation by equating the sum of (1)/(x) and (1)/(27 - x) to (3)/(20).

Combining the fractions on the left-hand side gives (27)/(27x - x²). Dividing both numerators by 3 gives (9)/(27x - x²) = (1)/(20). Cross-multiplying yields 27x - x² = 180, which rearranges to x² - 27x + 180 = 0.

Combining the fractions on the left-hand side gives (27)/(27x - x²). Dividing both numerators by 3 gives (9)/(27x - x²) = (1)/(20). Cross-multiplying yields 27x - x² = 180, which rearranges to x² - 27x + 180 = 0.

16. The sum of the squares of two consecutive multiples of $7$ is $1225$ . Find the multiples. [CBSE 2014] 17. The sum of a natural number and its reciprocal is $\frac{65}{8}$ . Find the number. 18. Divide $57$ into two parts whose product is $680$ . 19. Divide $27$ into two parts such that the sum of their reciprocals is $\frac{3}{20}$ . 20. Divide $16$ into two parts such that twice the square of the larger part exceeds the square of the smaller part by $164$ . 21. Find two natural numbers, the sum of whose squares is $25$ times their sum and also equal to $50$ times their difference. 22. The difference of the squares of two natural numbers is $45$ . The square of the smaller number is four times the larger number. Find the numbers. [CBSE 2007]

Answer: The two parts are $15$ and $12$ .

Step-by-step solution

Step 1: Set up the algebraic representation

Since the total sum of the two parts is $27$ , if we let the first part be $x$ , the remaining part must be $27 - x$ . Here, $x$ must lie strictly between $0$ and $27$ .

Step 2: Formulate the equation from the given condition

We are given that the sum of the reciprocals of the two parts equals $\frac{3}{20}$ . We set up the equation by equating the sum of $\frac{1}{x}$ and $\frac{1}{27 - x}$ to $\frac{3}{20}$ .

Step 3: Simplify and form the quadratic equation

Combining the fractions on the left-hand side gives $\frac{27}{27x - x^2}$ . Dividing both numerators by $3$ gives $\frac{9}{27x - x^2} = \frac{1}{20}$ . Cross-multiplying yields $27x - x^2 = 180$ , which rearranges to $x^2 - 27x + 180 = 0$ .

Step 4: Solve the quadratic equation by factorisation

We find two numbers whose product is $180$ and whose sum is $-27$ . These numbers are $-15$ and $-12$ . Factoring the quadratic expression gives $(x - 15)(x - 12) = 0$ , so $x = 15$ or $x = 12$ .

Step 5: Determine the two parts

When one part is $15$ , the other part is $27 - 15 = 12$ . Similarly, when one part is $12$ , the other part is $15$ . Thus, the two required parts are $15$ and $12$ .

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