Complete Syllabus & Gap Fillers

The problems run roughly from easiest to hardest, and every correct answer is worth the same, with no penalty for a wrong one — so never leave a question blank, and do not spend eight minutes on problem $12$ while problems $13$ to $20$ are unread. Forty minutes for twenty-five questions is about $1.6$ minutes each; the first ten usually take less, which buys time for the last five. The sections below fill in the topics the other chapters do not cover in depth: rates and work, averages, and clocks and calendars.

Step 1 — convert times to rates: Ana paints $\tfrac13$ of the fence per hour, Ben $\tfrac16$.

Step 2 — rates add: $\tfrac13 + \tfrac16 = \tfrac12$ of the fence per hour.

Step 3 — the whole fence takes $2$ hours. Check: in $2$ hours Ana paints $\tfrac23$ and Ben $\tfrac13$.

Step 1 — at $10{:}00$ Cara is $4$ km ahead.

Step 2 — Dev gains $12 - 4 = 8$ km every hour (same direction, so the speeds subtract).

Step 3 — closing a $4$ km gap at $8$ km/h takes $\tfrac12$ hour: $10{:}30$. Check: by then Cara has walked $6$ km and Dev has cycled $6$ km.

The mean is really a statement about the total: "the mean of $5$ scores is $84$" means "the scores add to $420$", and almost every mean problem is solved by converting to totals. The mean is also the balance point of the data on a number line, which is why one extreme value pulls it a long way while the median, which only cares about order, barely moves.

Step 1 — mean: total $49$ over $7$ values, $7$.

Step 2 — median: the list is sorted and has $7$ values, so the $4$th, $6$.

Step 3 — mode $5$ (it appears twice); range $13 - 3 = 10$. The value $13$ pulls the mean above the median.

Step 1 — convert to totals: five tests total $5\cdot84 = 420$; six tests at a mean of $86$ total $6\cdot86 = 516$.

Step 2 — the sixth score is the difference, $516 - 420 = 96$.

Step 1 — sorted, the list is $a \le b \le 7 \le d \le e$, with total $5\cdot6 = 30$.

Step 2 — the mode $8$ must appear at least twice, and only $d$ and $e$ can be $8$ or more: $d = e = 8$.

Step 3 — then $a + b = 30 - 7 - 16 = 7$, and $a \ne b$ (otherwise that value would tie with $8$ as a mode).

Step 4 — with $a < b$ and $a + b = 7$, the largest $a$ is $3$: the list $3, 4, 7, 8, 8$.

The hour hand does not jump from number to number; it creeps, and at $7{:}20$ it is a third of the way from $7$ to $8$. That creep is the whole difficulty of clock problems. Calendars run on remainders: the days of the week repeat every $7$ days, so counting days forward means dividing by $7$ and keeping the remainder.

Step 1 — minute hand: $6\cdot20 = 120^\circ$.

Step 2 — hour hand: $30\cdot7 + 0.5\cdot20 = 220^\circ$.

Step 3 — the angle is $220 - 120 = 100^\circ$, which is already less than $180^\circ$.

Step 1 — $m$ minutes after $3{:}00$ the minute hand is at $6m$ and the hour hand at $90 + 0.5m$.

Step 2 — they meet when $6m = 90 + 0.5m$, so $5.5m = 90$ and $m = \tfrac{180}{11} = 16\tfrac{4}{11}$.

Step 3 — at $3{:}16\tfrac{4}{11}$, a little after $3{:}16$. The minute hand gains $5.5^\circ$ per minute and had $90^\circ$ to make up.

Step 1 — December $25$ is day $359$ of the year ($2025$ is not a leap year: $334$ days before December, plus $25$).

Step 2 — it is $358$ days after January $1$.

Step 3 — $358 = 7\cdot51 + 1$, so the day of the week moves forward by $1$: Thursday.

Step 1 — rates: the trip takes $\tfrac{210}{84} = 2.5$ hours, so the train arrives at $12{:}10$.

Step 2 — clocks: minute hand $6\cdot10 = 60^\circ$; hour hand $0.5\cdot10 = 5^\circ$ past $12$.

Step 3 — the angle is $55^\circ$. Each step used one gap-filler; the problem only looks hard because it chains them.