Two relations do the whole job: the buck converter's voltage law and conservation of power.

Step 1 — the input voltage. For a buck (step-down) converter in continuous conduction,

\(V_{o}=D\,V_{in}\qquad\Rightarrow\qquad V_{in}=\dfrac{V_{o}}{D}=\dfrac{12}{0.8}=15\ \text{V}\)

so statement (a) is correct and (b) is wrong.

Step 2 — the input current. "Lossless" means all the input power reaches the load:

\(P_{in}=P_{out}=20\ \text{W}\)

\(I_{in}=\dfrac{P_{in}}{V_{in}}=\dfrac{20}{15}=\dfrac{4}{3}=1.333\ \text{A}\)

so statement (c) is correct and (d) is wrong. The answer is option 1.

Notice the pattern that confirms the arithmetic. The converter steps the voltage down by the factor D = 0.8, and it steps the current up by exactly the reciprocal:

\(\dfrac{I_{o}}{I_{in}}=\dfrac{1.667}{1.333}=1.25=\dfrac{1}{0.8}\)

A DC-DC converter is a DC transformer with turns ratio set by the duty cycle — and unlike a resistive divider, which would throw away the difference as heat, it conserves power.

Why the duty-cycle relation holds. In the steady state the inductor's average voltage over a cycle must be zero, or its current would ramp away without limit. During the on-time \(DT\) the inductor sees \(V_{in}-V_{o}\); during the off-time \((1-D)T\) it sees \(-V_{o}\). Volt-second balance gives

\(\left(V_{in}-V_{o}\right)DT=V_{o}\left(1-D\right)T\quad\Rightarrow\quad V_{o}=DV_{in}\)

The word "continuous" matters. If the load were light enough for the inductor current to fall to zero each cycle (discontinuous conduction), the output would rise above \(DV_{in}\) and would depend on the load as well as on D — the simple relation used here would no longer apply.

A practical note on the input current : 4/3 A is the average. The actual input current is pulsed, flowing only during the switch's on-time, which is why a buck converter needs a substantial input capacitor to supply those pulses and keep the supply line quiet.

Hence, the correct statements are (a) and (c).