Question

divide 27 into the parts such that the sum of their reciprocals is 3/20​

Answer 1

Let the two parts be a and b.

1/a + 1/b = 3/20

(a+b)/ab = 3/20     ------ (1)

a+b = 27    ------ (2)

Substitute (2) in (1).

(27)/ab = 3/20

540 = 3ab

ab = 180

Now,

(a-b)^2 = (a+b)^2 -4ab

(a-b)^2 = (27)^2 - 4(180)

(a - b)^2 = 729 - 720

a - b = 3    ----- (3)

Solving (2) & (3), we get

a + b = 27

a - b = 3

----------------

2a = 30

a = 15

Then:

a - b = 3

15 - b = 3

b = 12.

Hence, the required parts are 12 and 15