Question

The sum of all the solutions of the equation $|505x-1010|+|1515x+505|=2020$, is

• A. $-\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. $-\frac{1}{4}$

Answer 1

The correct option is D $-\frac{1}{4}$

$|505x-1010|+|1515x+505|=2020$
$\Rightarrow |505(x-2)|+|505(3x+1)|=505\times 4$
$\Rightarrow |x-2|+|3x+1|=4$

Case-$1:$ If $x<-\frac{1}{3}$
$|x-2|+|3x+1|=4$
$\Rightarrow 2-x-3x-1=4$
$\Rightarrow -4x=3$
$\Rightarrow x=-\frac{3}{4},$ which is valid for $x<-\frac{1}{3}$
$\therefore x=-\frac{3}{4}...\left(1\right)$

Case-$2:$ If $-\frac{1}{3}\le x<2$
$|x-2|+|3x+1|=4$
$\Rightarrow 2-x+3x+1=4$
$\Rightarrow 2x=1$
$\Rightarrow x=\frac{1}{2},$ which is valid for $-\frac{1}{3}\le x<2$
$\therefore x=\frac{1}{2}...\left(2\right)$

Case-$3:$ If $x\ge 2$
$x-2+3x+1=4$
$\Rightarrow 4x=5$
$\Rightarrow x=\frac{5}{4},$ which is not valid for $x\ge 2$
$\therefore x\ne \frac{5}{4}...\left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$,
solutions of the given equation are $x=-\frac{3}{4}$ and $x=\frac{1}{2}$
Sum of solutions $=-\frac{3}{4}+\frac{1}{2}=-\frac{1}{4}$