Question

Question :- 8

Divide 27 into two Parts such that the sum of their reciprocal is 3/20​

Answer 1

• Let the number be M.

» We have to divide the number such that their sum came 27.

One number is M.

So, another number is 27 - M

Also, said in question the sum of their reciprocal is 3/20

Reciprocals are : $\dfrac{1}{M}$ and $\dfrac{1}{27\:-\:M}$

A.T.Q.

=> $\dfrac{1}{M}$ + $\dfrac{1}{27\:-\:M}$ = $\dfrac{3}{20}$

=> $\dfrac{27\:-\:M\:+\:M}{(M)(27\:-\:M)}$ = $\dfrac{3}{20}$

=> $\dfrac{27}{(M)(27\:-\:M)}$ = $\dfrac{3}{20}$

Cross-multiply them

=> 20(27) = 3M(27 - M)

=> 540 = 81M - 3M²

=> 3M² - 81M + 540 = 0

=> M² - 27M + 180 = 0

=> M² - 15M - 12M + 180 = 0

=> M(M - 15) -12(M - 15) = 0

=> (M - 12) (M - 15) = 0

=> M = +12, +15

» If M = 12

Then;

=> 27 - M = 27 - 12

=> 15

Similarly

» If M = 15

Then,

=> 27 - M = 27 - 15

=> 12

____________________________

12 and 15 are the two numbers that divide 27 into two parts.

________ [ ANSWER ]

____________________________

Answer 2

Let number = x

Another number = 27 - x

According to question;

1/x + 1/(27 - x) = 3/20

(27 - x + x)/(27x - x²) = 3/20

Cross-multiply,

20(27) = 3(27x - x²)

540 = 81M - 3x²

x² - 27x + 180 = 0

x² - 15x - 12x + 180 = 0

x(x - 15) -12(x - 15) = 0

(x - 12) (x - 15) = 0

x = 12, 15