2022 CSAT — Q8

Let $a, b, c$ be the counts of Rs. $50$, Rs. $20$, Rs. $10$ notes:

Divide the first by $10$: $5a + 2b + c = 184$. Subtract $a+b+c=50$:

Feasibility ($b, c \ge 0$): $c \ge 0 \Rightarrow a \ge 28$; $b \ge 0 \Rightarrow a \le 33$. So $a \in \{28,29,30,31,32,33\}$ — and the Rs. $50$ count must be at least $28$.

Now test each claim:

• Statement 1 ($a = 25$): but $a \ge 28$ — impossible. Not correct.
• Statement 2 ($b = 35$): $b = 134 - 4a = 35 \Rightarrow 4a = 99$, not an integer — impossible. Not correct.
• Statement 3 ($c = 20$): $c = 3a - 84 = 20 \Rightarrow 3a = 104$, not an integer — impossible. Not correct.

All three fail.

Answer: (d) 1, 2 and 3.

The whole problem collapses to one relation $4a + b = 134$ and the feasibility window $28 \le a \le 33$. Each statement fixes one variable; you only have to check whether that value lands in range and yields integer partners. Two of the three fail an integrality test ($99/4$, $104/3$) and one fails the range bound ($25 < 28$) — a clean reminder that “looks plausible” is not “is feasible.”