2022 CSAT — Q7

Anchor on a known day and count odd days (days mod $7$).

$1$ January $2000$ was a Saturday. From $1$ Jan $2000$ to $1$ Jan $2099$ is $99$ years; the leap years in $2000\text{–}2098$ are $2000, 2004, \dots, 2096$, i.e. $25$ of them. Odd days $= 99 + 25 = 124 \equiv 5 \pmod 7$. So $1$ Jan $2099 =$ Saturday $+ 5 =$ Thursday.

From $1$ Jan $2099$ to $1$ June $2099$ (and $2099$ is not a leap year, since $99$ is not divisible by $4$):

So $1$ June $2099 =$ Thursday $+ 4 =$ Monday.

Counting forward: $1\to$ Mon, so Sunday is $6$ days later $=$ the $7$th.

Answer: (d) 7.

Two traps decide this item: (1) $2099$ is not a leap year — only divisible-by-$4$ years are, and the century-rule never even comes up here; (2) the leap-year count over the span ($25$, not $24$) hinges on including $2000$ itself (a leap year, divisible by $400$). Under the clock, anchor on $1$ Jan $2000 =$ Saturday and march the odd days — never try to recall the day of a $2099$ date directly.