At a certain time, the mass of a radioactive substance was recorded as 150 g. This mass decreased with time due to decay and after $t$ hours, the recorded mass was $M$ g. It is known that $M$ can be modelled by the formula $M = 150e^{-kt}$, where $k$ is a positive constant. After 50 hours, its mass has decreased to 120 g.

1. Substitute the given conditions at $t = 50$ to form an equation for $k$.

When $t = 50$, $M = 120$.
$120 = 150e^{-50k}$

2. Solve for the constant $k$.

$e^{-50k} = \frac{120}{150} = 0.8$
$-50k = \ln(0.8)$
$k = -\frac{\ln(0.8)}{50} \approx 0.00446287$

3. Substitute $t = 120$ and the value of $k$ into the model to find $M$.

When $t = 120$,
$M = 150e^{-120(0.00446287)}$

4. Calculate the final mass.

$M = 150e^{-0.53554} \approx 87.815$
$M \approx 87.8 \text{ g (to 3 s.f.)}$

1. Determine the value of $k$ from the given information in the stem

When $t = 50$, $M = 120$.
$120 = 150e^{-50k}$
$e^{-50k} = \frac{120}{150} = 0.8$
$-50k = \ln 0.8$
$k = -\frac{\ln 0.8}{50} \approx 0.00446287$

2. Determine the target mass when one third of the substance has decayed

Initial mass = $150$ g.
Mass decayed = $\frac{1}{3} \times 150 = 50$ g.
Remaining mass $M = 150 - 50 = 100$ g.

3. Solve for $t$ when $M = 100$

$100 = 150e^{-kt}$
$e^{-kt} = \frac{100}{150} = \frac{2}{3}$
$-kt = \ln\left(\frac{2}{3}\right)$
$t = \frac{\ln(2/3)}{-k} = \frac{\ln(2/3)}{\frac{\ln 0.8}{50}} = 50 \frac{\ln(2/3)}{\ln 0.8}$
$t \approx 90.619 \approx 90.6$ hours (to 3 s.f.)