B

To find the final temperature when mixtures are formed, we can use the principle of conservation of energy, specifically for heat transfer.
Step 1: When two bodies are mixed, the heat lost by the hotter liquid must equal the heat gained by the cooler liquid.
Step 2: Use the formula:
$$ m_1 c (T_1 - T_f) = m_2 c (T_f - T_2) $$
where:
- $m_1$ and $m_2$ are masses of the two liquids,
- $T_1$ and $T_2$ are their initial temperatures,
- $T_f$ is the final temperature,
- $c$ is the specific heat (which cancels out since it is the same for all).
For (i) A and B:
- $m_a = m$, $T_a = 20ºC$
- $m_b = 2m$, $T_b = 40ºC$

Plugging in the values:
$$ m(20 - T_f) = 2m(T_f - 40) $$
Simplifying:
$$ 20 - T_f = 2(T_f - 40) $$
$$ 20 - T_f = 2T_f - 80 $$
$$ 3T_f = 100 \Rightarrow T_f = 33.33ºC $$ (Not in options; it means this mixing must be evaluated carefully.)
For (ii) A and C:
- $m_a = m$, $T_a = 20ºC$
- $m_c = 3m$, $T_c = 60ºC$

Using the equation:
$$ m(20 - T_f) = 3m(T_f - 60) $$
Simplifying:
$$ 20 - T_f = 3(T_f - 60) $$
$$ 20 - T_f = 3T_f - 180 $$
$$ 4T_f = 200 \Rightarrow T_f = 50ºC $$ (Check option)
Therefore, here A and C yield 50ºC which aligns with option [C].
Final selection: The exact option pulled would yield the option.
For (ii) A and C gives valid output,
thus matching as [B].