CSAT Solved Papers/ 2021/Q59

2021 CSAT — Q59

Quant Number theory 2.5 marks Hard

Consider the following addition problem:

3P + 4P + PP + PP = RQ2;

where $P$ , $Q$ and $R$ are different digits.

What is the arithmetic mean of all such possible sums?

A 102
B 120
C 202 Answer
D 220

Worked rationale

Read the two-digit numbers by place value: $3P = 30+P$ , $4P = 40+P$ , $PP = 11P$ . The sum is

(30+P) + (40+P) + 11P + 11P = 70 + 24P = \overline{RQ2}.

The result ends in $2$ , so the units digit of $70 + 24P$ must be $2$ : the units digit of $24P$ must be $2$ , i.e. $4P \equiv 2 \pmod{10}$ , giving $P\in\{3, 8\}$ .

$P = 3$ : sum $= 70 + 72 = 142 = \overline{RQ2}$ with $R=1, Q=4, P=3$ — all distinct ✓.
$P = 8$ : sum $= 70 + 192 = 262 = \overline{RQ2}$ with $R=2, Q=6, P=8$ — all distinct ✓.

Both are valid, so the possible sums are $142$ and $262$ . Their arithmetic mean:

\frac{142 + 262}{2} = \frac{404}{2} = 202.

Answer: (c) 202.

Why the other options miss

A

missed a case: finds only $P = 3$ (sum $142$ ) and reports a single value, or averages with a wrong second case.
B

an arithmetic slip: mis-forms one sum (e.g. treats $PP$ once instead of twice) before averaging.
D

missed a case: keeps only $P = 8$ (sum $262$ ) or mishandles the units constraint and averages the wrong pair.

Specialist insight

The skeleton is $70 + 24P = \overline{RQ2}$ , and the units condition $4P \equiv 2 \pmod{10}$ filters $P$ to exactly $\{3,8\}$ — both survive the distinct-digit check, so there are two sums, not one. The phrase “mean of all such possible sums” is the tell that the answer is a number you must aggregate; missing either case (the classic trap) gives a distractor. Enumerate every $P$ that satisfies the units digit, validate distinctness, then average.

The trap, in one line

$70 + 24P = \overline{RQ2}$ needs $4P\equiv2\pmod{10}$ , so $P\in\{3,8\}$ gives sums $142,262$ , mean $202 \Rightarrow$ (c).

← All 2021 CSAT questions