CSAT Solved Papers/ 2022/Q28

2022 CSAT — Q28

Quant Arithmetic & numeracy 2.5 marks Medium

$A$ has some coins. He gives half of the coins and $2$ more to $B$ . $B$ gives half of the coins and $2$ more to $C$ . $C$ gives half of the coins and $2$ more to $D$ . The number of coins $D$ has now, is the smallest two-digit number. How many coins does $A$ have in the beginning?

A 76
B 68
C 60
D 52 Answer

Worked rationale

The smallest two-digit number is $10$ , so $D$ receives $10$ . Each handover gives the recipient $(\text{giver's current half}) + 2$ . Track what each person receives and work backwards.

Let $r_B, r_C, r_D$ be what $B, C, D$ receive. Then each next person gets half-of-the-previous-receipt plus $2$ :

r_D = \tfrac{r_C}{2} + 2,\quad r_C = \tfrac{r_B}{2} + 2,\quad r_B = \tfrac{A}{2} + 2.

Reverse from $r_D = 10$ :

$\tfrac{r_C}{2} + 2 = 10 \Rightarrow r_C = 16$
$\tfrac{r_B}{2} + 2 = 16 \Rightarrow r_B = 28$
$\tfrac{A}{2} + 2 = 28 \Rightarrow A = 52$

Check (forward): $A=52$ gives $\tfrac{52}{2}+2 = 28$ to $B$ ; $B=28$ gives $\tfrac{28}{2}+2 = 16$ to $C$ ; $C=16$ gives $\tfrac{16}{2}+2 = 10$ to $D$ . ✓

Answer: (d) 52.

Why the other options miss

A

an arithmetic slip: works backwards with $(x-2)\times 2$ but slips a sign or an extra step.
B

the wrong inverse: inverts the step as $2(r) + 2$ instead of $2(r - 2)$ once.
C

missed the anchor: uses “half plus $2$ ” forwards from a guessed $A$ that misses the $D = 10$ anchor.

Specialist insight

Reverse a “half-then-add- $2$ ” chain by the inverse operation subtract $2$ , then double: $10 \to 16 \to 28 \to 52$ . Anchoring on “smallest two-digit number $= 10$ ” is the only outside fact needed. Forward-checking the answer is cheap and catches any inversion slip — always run it.

The trap, in one line

Invert "half $+2$ " as " $-2$ then $\times 2$ ": $10 \to 16 \to 28 \to 52 \Rightarrow$ (d).

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