Question

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

Answer 1

$\huge{\bold{\underline{\underline{AnSwEr:-}}}}$

$\large\tt\star\ 12 \: and \: 15$

$\large{\bold{\underline{\underline{STeP \: bY \:sTep \: eXpLanAtiOn:}}}}$

•Let the first part be 'x'

•Let the second part be '27-x'

Now ; According To given question ....

__________

Let us solve ;

$\large\tt\to\frac{1}{x} + \frac{1}{27-x}= \frac{3}{20} \\ \\ \large\tt\to\frac{27-x+x}{x(27-x)}= \frac{3}{20} \\ \\ \large\tt\to\frac{27}{27x-x^{2} }= \frac{3}{20} \\ \\ \large\sf\star\ Cross \: multiplication \star \\ \\ \large\tt\to\ {71x - 3x^{2} }=540 \\ \\ \large\tt\to\ {-3x^{2} +71x -540 =0} \\ \\ \large\sf\star\ TaKe \:negative \: sign \:common \star \\ \\ \large\tt\to\ 3x^{2}-71x+540=0 \\ \\ \large\tt\to\ 3(x^{2}- 27x +180)= 0 \\ \\ \large\tt\to\ x^{2} -27x + 180 = 0 \\ \\ \large\tt\to\ x^{2}-15x -12x +180 \\ \\ \large\tt\to\ x(x-15)-12(x-15)\\ \\ \large\tt\to\ (x-15)(x-12)=0\\ \\ \large\tt\to\ x=12,15$

$\large\red{\boxed{\tt\blue{x}\purple{=} \green {12} \orange,\pink{15} }}$

Answer 2

$see \: solution \: in \: image$