Question

divide 27 into two parts such that the sum of their reciprocals is 3 20 .

Answer 1

Question:

Divide 27 into two parts such that the sum of their reciprocals is 3/20 .

Answer with explanation:

27 is divided into two parts, so the sum of the two numbers is 27, isn't it?

Let the two numbers be a and b.

$\mathsf{a+b=27\ \ \ \ \ \longrightarrow\ \ \ (1)}$

Given that the sum of their reciprocals is 3/20.

$\displaystyle \mathsf{\frac{1}{a}+\frac{1}{b}=\frac{3}{20}}\\ \\ \\ \mathsf{\frac{a+b}{ab}=\frac{3}{20}} \\ \\ \\ \mathsf{\frac{27}{ab}=\frac{3}{20}} \ \ \ \ \ \ \ \ \ \ \textsf{[From (1)]} \\ \\ \\ \mathsf{\frac{27}{ab}=\frac{3 \times 9}{20 \times 9}} \\ \\ \\ \mathsf{\frac{27}{ab}=\frac{27}{180}} \\ \\ \\ \\ \mathsf{\therefore\ ab=180\ \ \ \ \ \longrightarrow\ \ \ (2)}$

$\mathsf{(1)^2} \\ \\ \\ \mathsf{(a+b)^2=27^2} \\ \\ \\ \mathsf{a^2+2ab+b^2=729\ \ \ \ \ \longrightarrow\ \ \ (3)}$

$\mathsf{4 \times (2)} \\ \\ \\ \mathsf{4(ab)=4\times 180} \\ \\ \\ \mathsf{4ab=720\ \ \ \ \ \longrightarrow\ \ \ (4)}$

$\mathsf{(3)-(4)} \\ \\ \\ \mathsf{a^2+2ab+b^2-4ab=729-720} \\ \\ \\ \mathsf{a^2-2ab+b^2=9} \\ \\ \\ \mathsf{(a-b)^2=9} \\ \\ \\ \mathsf{a-b=3\ \ \ \ \ \longrightarrow\ \ \ (5)}$

$\mathsf{(3)+(5)} \\ \\ \\ \begin{aligned}\mathsf{a+b}&=\mathsf{27}\\ +\ \ \ \mathsf{a-b}&=\mathsf{3}\\ \cline{1-3}\mathsf{2a}&=\mathsf{30}\\ \mathsf{a}&=\bold{15}\end{aligned}$

$\mathsf{(3)-(5)} \\ \\ \\ \begin{aligned}\mathsf{a+b}&=\mathsf{27}\\ -\ \ \ \mathsf{a-b}&=\mathsf{3}\\ \cline{1-3}\mathsf{2b}&=\mathsf{24}\\ \mathsf{a}&=\bold{12}\end{aligned}$

Thus 27 is divided into 15 and 12.

Answer 2

Hope it helps u ! ...............