AMC 8 · 2010 · #21

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

fraction-arithmeticlinear-equations-one-varmulti-digit-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: fraction-arithmeticlinear-equations-one-var

📏 Long solution 💡 4 insights

Problem

Hui reads the book over four days. On day 1 she reads

\frac{1}{5}

of the whole book plus 12 more pages. On day 2 she reads

\frac{1}{4}

of what was left plus 15 more. On day 3 she reads

\frac{1}{3}

of what was left plus 18 more. After that there are 62 pages left, which she finishes on day 4. Find the total number of pages.

Pick an answer.

(A)

120

(B)

180

(C)

240

(D)

300

(E)

360

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Work Backwards

The problem describes the book being eaten away day by day, and the last day's leftover (62 pages) is fully known. That is exactly when Tool #11 (Work Backwards) shines: start at the known end and undo each day in reverse. On each day, undoing has two clean moves — add back the "+k extra pages" first, then recognize what's left as a known fraction of that morning's pile and scale up to find the pile. Tool #6 (Guess and Check) is kept as a backup: plug each of the five choices into the forward story and see which one ends with exactly 62 pages left.

1STEP 1

Start from the end: 62 pages remain after day 3. Day 3 was " $\frac{1}{3}$ plus 18," so undo the +18 first, leaving 80 pages just after the $\frac{1}{3}$ chunk.

62 + 18 = 80 pages left after the

\frac{1}{3}

chunk

2STEP 2

Those 80 pages are $\frac{2}{3}$ of day 3's morning pile, so multiply by the reciprocal $\frac{3}{2}$ to get 120 pages that morning.

80 ×

\frac{3}{2}

= 120 pages at the start of day 3

3STEP 3

Now 120 is day 2's leftover. Day 2 was " $\frac{1}{4}$ plus 15," so add the 15 back to get 135 pages just after the $\frac{1}{4}$ chunk.

120 + 15 = 135 pages left after the

\frac{1}{4}

chunk

4STEP 4

Those 135 pages are $\frac{3}{4}$ of day 2's morning pile, so multiply by $\frac{4}{3}$ to get 180 — the start of day 2, which is also day 1's leftover.

135 ×

\frac{4}{3}

= 180 pages at the start of day 2

5STEP 5

Finally 180 is day 1's leftover; add 12 back to get 192, which is $\frac{4}{5}$ of the whole book, so multiply by $\frac{5}{4}$ to get 240 → (C).

192 ×

\frac{5}{4}

= 240 → (C)

Answer

240

Run the story forward with 240 pages to double-check. Day 1:

\frac{1}{5}

× 240 + 12 = 48 + 12 = 60 pages read, 180 left. Day 2:

\frac{1}{4}

× 180 + 15 = 45 + 15 = 60 pages read, 120 left. Day 3:

\frac{1}{3}

× 120 + 18 = 40 + 18 = 58 pages read, 62 left. Day 4: 62. Total = 60 + 60 + 58 + 62 = 240. Everything lines up.

💡Key takeaway

When a problem hands you the very last leftover, walk the story backwards — undo the "+ extras" first, then scale the fraction up to the whole pile. This AMC 8 question only needs Grade 6 ratio reasoning to crack.

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