$24$

Lakshman Patel RJIT
asked
in Quantitative Aptitude
Apr 3, 2020
edited
Nov 1, 2020
by soujanyareddy13

334 views
0 votes

You can verify this question with the options also.

let tan’s digits =$x$

then unit digit=$\frac{8}{x}$

so the number is $10x+\frac{8}{x}$

now according to the question;

$\implies 10x+\frac{8}{x}+18=10*\frac{8}{x}+x$

$\implies \frac{10x^2+8+18x}{x}=\frac{80+x^2}{x}$

$\implies 9x^2+18x-72=0$

$\implies x^2+2x-8=0$

$\implies x(x+4)-2(x+4)=0$

$\implies(x+4)(x_2)=0$

$\implies x=-4,2$

$\therefore \text{tan’s digit=2, unit digit=$\frac{8}{2}$=4}$

So the number is $24$

Option $B$ is correct here.

let tan’s digits =$x$

then unit digit=$\frac{8}{x}$

so the number is $10x+\frac{8}{x}$

now according to the question;

$\implies 10x+\frac{8}{x}+18=10*\frac{8}{x}+x$

$\implies \frac{10x^2+8+18x}{x}=\frac{80+x^2}{x}$

$\implies 9x^2+18x-72=0$

$\implies x^2+2x-8=0$

$\implies x(x+4)-2(x+4)=0$

$\implies(x+4)(x_2)=0$

$\implies x=-4,2$

$\therefore \text{tan’s digit=2, unit digit=$\frac{8}{2}$=4}$

So the number is $24$

Option $B$ is correct here.