Answer
$108$
Work Step by Step
The intensity level is given as
$\beta=10log(\frac{I}{I_{\circ}}) $...eq(1)
The intensity level for 120 machines is given as
$\beta_{120}=10log (\frac{120I}{I_{\circ}})$.....eq(2)
The intensity level for $nI$ machines is
$\beta_{nI}=10log(\frac{nI}{I_{\circ}})$....eq(3)
$\implies 92dB=10log(\frac{120I}{I_{\circ}})$
$\implies 92dB=10log120+10log\frac{I}{I_{\circ}}$.....eq(4)
Now we substitute $82db$for $\beta_{nI}$ in eq(3) to obtain:
$82dB=10logn+10log(\frac{I}{I_{\circ}})$...eq(5)
Subtracting eq(4) from eq(5), we obtain:
$10dB=10log120-10logn$
$\implies 10dB=10log\frac{120}{n}$
This simplifies to:
(number of machines that will produce 82 db)
$n=12$
Thus,
turned off machines = total machines - number of machines that produce 82 db
$=120-12=108$