Let two parts be x and y So, x+y = 27 (1)A2q 1/x + 1/ y = 3/20 y+x/xy= 3/20 27/xy = 3/20 {from (1) } 3xy = 540 y = 540/3x y= 180/x (2)Substituting y= 180/x in eqn (1) x+180/x = 27 x^ +180/x =27 x^ + 180 =27x x^ -27x+180= 0 x^ – (12 + 15) x + 180 =0 x^ – 12x – 15x + 180 = 0 x (x- 12) – 15 (x -12) =0 ( x-15) (x-12) =0 Either, Or x=15 x =12 Therefore two parts are 15 and 12 HOPE IT HELPS YOU

Let the one part be x. The other\xa0part will be (27 – x).According to question,{tex}\\frac { 1 } { x } + \\frac { 1 } { 27 – x } = \\frac { 3 } { 20 }{/tex}{tex}\\Rightarrow \\frac { 27 – x + x } { x ( 27 – x ) } = \\frac { 3 } { 20 }{/tex}{tex}\\Rightarrow \\frac { 27 } { 27 x – x ^ { 2 } } = \\frac { 3 } { 20 }{/tex}After cross multiplication,{tex}\\Rightarrow{/tex}\xa027\xa0{tex}\\times{/tex}\xa020 = 3(27x) – 3×2{tex}\\Rightarrow{/tex}\xa0540 = 81x – 3×2{tex}\\Rightarrow{/tex}\xa03x2\xa0- 81x + 540 = 0{tex}\\Rightarrow{/tex}\xa0x2\xa0- 27x + 180 = 0Factorise the equation,{tex}\\Rightarrow{/tex}\xa0x2\xa0- 15x – 12x + 180 = 0{tex}\\Rightarrow{/tex}\xa0x(x – 15) – 12(x – 15) = 0{tex}\\Rightarrow{/tex}\xa0x – 15 = 0 or x – 12 = 0{tex}\\Rightarrow{/tex}\xa0x = 15 or x = 12Therefore, the required two parts are 15 and 12.