PEA 2025 PEA Q26 | ISI MSQE

PEAMCQHard

2025 PEA Q26

$x \succ y$ iff (i) $x_a + x_b > y_a + y_b$ , or (ii) $x_a + x_b = y_a + y_b$ and $x_a > y_a$ . Which statements are true?

I. $d_a = w/p_a$ if $p_a < p_b$ . \quad II. $d_a = w/p_a$ if $p_a \le p_b$ . \quad III. $d_a = 0$ if $p_b \le p_a$ . \quad IV. $d_b = w/p_b$ if $p_b < p_a$ . \quad V. $d_b = 0$ if $p_b < p_a$ .

Reveal answer and solution

Answer

Solution

1
The consumer first maximises $x_a + x_b$ subject to $p_a x_a + p_b x_b = w$ . To maximise the sum, spend all income on the cheaper good. Tie-breaker: if $p_a = p_b$ , choose the bundle with more apples (rule (ii)), i.e.\ all-apples.
2
\begin{itemize}[leftmargin=*]
3
- If $p_a < p_b$ : cheaper good is $a$ . $d_a = w/p_a$ , $d_b = 0$ .
4
- If $p_a = p_b$ : indifferent in sum; tie-break to apples: $d_a = w/p_a$ , $d_b = 0$ .
5
- If $p_a > p_b$ : cheaper good is $b$ . $d_a = 0$ , $d_b = w/p_b$ .
6
\end{itemize}
7
Hence:
8
\begin{itemize}[leftmargin=*]
9
- I (strict $<$ ) --- TRUE.
10
- II ( $\le$ ) --- TRUE (tie-break gives all apples).
11
- III ( $p_b \le p_a$ ) --- if $p_b = p_a$ , the tie-break gives apples, so $d_a \ne 0$ . FALSE.
12
- IV ( $p_b < p_a$ ) --- TRUE.
13
- V ( $p_b < p_a$ ) --- $d_b = w/p_b > 0$ , so $d_b = 0$ is FALSE.
14
\end{itemize}
15
True statements: II and IV.

Answer structure / marking notes

Wait: re-evaluating III. For III, $p_b \le p_a$ means cheaper is $b$ (or tie). If $p_b < p_a$ , $d_a = 0$ , so III holds in that subcase. But at $p_b = p_a$ , the tie-break gives apples, so $d_a = w/p_a \ne 0$ , breaking III. So III fails on the boundary. II requires $p_a \le p_b$ : if $p_a < p_b$ , $d_a = w/p_a$ holds; if $p_a = p_b$ , tie-break gives apples, so $d_a = w/p_a$ still holds. So II is true. IV: $p_b < p_a$ strict, so $d_b = w/p_b$ . True. Final pair: II and IV $\Rightarrow$ Option (C).

Correction: Final answer (C).

\boxed{\text{Option C}}

Trap

The tie-breaking on equal prices is the subtle point: at $p_a = p_b$ , apples win, so III fails.

Final answer (after correction): Hmm, the printed answer key gives option (B) ``II and III''. Re-examine III. At $p_b = p_a$ in III's condition $p_b \le p_a$ , the sum-maximisation is indifferent and the lexicographic tie-break gives all apples; thus $d_a \ne 0$ . So III is false. The correct pair is II and IV, option (C).

\boxed{\text{Option C}}

%==========================================

Content note

Imported from public/resources/isi/msqe/solutions/pea/2025/ISI_MSQE_PEA_2025_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.