Question

$P=\{x:10<3x-2<34,x\in R\}Q=\{x:28\le 5X+3\le 68,x\in R\}$
The solution for $P\cap Q$ = .

• A. $4\le x<13$
• B. $4\le x\le 13$
• C. $5\le x\le 12$
• D. $5\le x<12$

Answer 1

The correct option is D $5\le x<12$

Given, $P=\{x:10<3x-2<34,x\in R\}$
$\Rightarrow$ 10 < 3x -2
$\Rightarrow$ x > 4
and 3x -2 < 34
$\Rightarrow$ x < 12
Hence, the solution for P is 4 < x < 12.

Given,
$Q=\{x:28\le 5x+3\le 68,x\in R\}$
$\Rightarrow$ $28\le 5x+3$
$\Rightarrow$ $5\le x$
and $5x+3\le 68$
$\Rightarrow$ $x\le 13$
Hence, the solution for Q is $5\le x\le 13$.

Hence, the solution for $P\cap Q$ = $5\le x<12$.