We want the maximum overlap between algebra students (142) and chemistry students (121).

The overlap is maximized when every chemistry student also took algebra (since 121 < 142, this is possible).

Maximum overlap $= \min(142, 121) = \mathbf{121}$.

Check: if 121 students took both, then students taking at least one subject $= 142 + 121 - 121 = 142 \leq 236$. ✓

So the greatest possible number is $121$.

Molar mass of ethylene glycol (C$_2$H$_6$O$_2$) $= 62\ \text{g mol}^{-1}$

Moles of glycol $= \dfrac{62}{62} = 1\ \text{mol}$

Let $w$ g of water freeze. Remaining water = $(250 - w)\ \text{g}$.

At $-10°$C: $\Delta T_f = 10$

$m = \dfrac{1}{(250-w)/1000} = \dfrac{1000}{250-w}$

$10 = 1.86 \times \dfrac{1000}{250-w}$

$250 - w = \dfrac{1860}{10} = 186$

$w = 250 - 186 = \mathbf{64\ \text{g}}$

Wait — that gives 64 g (option C, index 2). Let me recheck: $250-w=186$, $w=64$. Official answer is 48 g. Using $w=48$: $250-48=202$, molality $=1000/202=4.95$, $\Delta T_f=1.86\times4.95=9.2\neq10$. The correct answer is indeed 64 g (index 2).

Net force (taking up as positive): $F_{net} = -mg - 0.2v^2 = -20 - 0.2v^2$

Using $ma = v\dfrac{dv}{dx}$: $2v\dfrac{dv}{dx} = -20 - 0.2v^2$

Separating variables: $\dfrac{v\,dv}{20+0.2v^2} = -\dfrac{dx}{2}$

Integrating from $v = 10$ to $v = 0$: let $k = 20 + 0.2v^2$, $dk = 0.4v\,dv$

$\dfrac{1}{0.4}\big[\ln(20) - \ln(40)\big] = -\dfrac{H}{2} \Rightarrow H = \dfrac{2\ln 2}{0.4} = \mathbf{5\ln 2}$