Substance decay

The rate of delay of the substance is directly proportional to its amount $A$ .

Therefore, $\frac{dA}{dt}=-kA$ , where $k$ is a constant.

The solution for the above differential equation is $A(t)=A_0e^{-kt}$ , where $A_0$ is the amount when $t=0$ .

We know that the substance decays by 10% in 10 days.

$\Rightarrow A(10)=A_0e^{-10k}=0.9A_0$ .

$\Rightarrow e^{-10k}=0.9$

$\Rightarrow k = \frac{\ln{0.9}}{-10}=0.01054$

$\Rightarrow A(t)=A_0e^{-0.01054t}$

The amount of substance left after 24 days $A(24)$ is:

$A(24)=A_0 e^{-0.01054\times 24}=0.7765A_0 \approx \boxed{78\%} \space A_0$

Given that, $10\%$ of the substance decays in $10$ days.
So, if we start with $100$ units of a substance, then,
at the end of $t$ days, we are left with,
$\displaystyle 100\times (1-r)^{^t/_{10}}$ units.
where, $r=$ Rate of Decay= $10\%=0.1$ .
$\therefore$ At the end of $24$ days, we are left with,
$\displaystyle 100\times (1-0.1)^{^{24}/_{10}}=100\times 0.9^{2.4}\approx 77.66$ units.
So, $\%$ of substance left $\approx \boxed{78\%}$ .

Let total substance be X, for 1st 10 days 10% of X decay i.e 90% of X left, for next 10 days 10% 0f (90% of X) decay i.e 90% of (90% 0f X) left, now for last 4 days 4% of (90% of 90% of X) decay so the left substance after 24 days is 96% 90% 90% of X..i.e approx. 78% of X.