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2000 AMC 10 Problem 3

2000 AMC 10 Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?

A)
$40$
B)
$50$
C)
$55$
D)
$60$
E)
$75$

Answer

Key insight

Eating 20% leaves 80%, so two days scale the count by (4/5)^2 = 16/25, and the original amount is 32 divided by 16/25, which is 50.

Solution

Each day Jenny keeps $100\% - 20\% = 80\% = \tfrac45$ of whatever is in the jar that morning. Two days in a row multiply the original count $x$ by $\tfrac45$ twice:
$\left(\tfrac45\right)^2 x = \tfrac{16}{25}\,x = 32.$
So $x = 32 \cdot \tfrac{25}{16} = 2 \cdot 25 = 50$ .

Check: $50 \to 40 \to 32$ .

The answer is $\boxed{\textbf{(B)}\ 50}$ .

Why this works

A percentage taken "of what is there" is a multiplicative change, and repeated multiplicative changes compound rather than add. Undoing a factor of $\tfrac45$ means dividing by it, i.e. multiplying by $\tfrac54$ , not adding $20\%$ back.

Alternative approach

Run the days in reverse: the $32$ beans are $\tfrac45$ of the previous evening's count, so that count was $32 \cdot \tfrac54 = 40$ , and one more step back gives $40 \cdot \tfrac54 = 50$ .

The trap

Removing 20% of the original twice (40% total) and solving 0.6x = 32, or adding 20% of 32 twice to go backwards.

Common mistakes

Removing 20% of the original twice (40% total) and solving 0.6x = 32, or adding 20% of 32 twice to go backwards.
Answering $40$ , the count after the first day, instead of the original count.

Techniques

Set up the equation/formula and compute; no special trick needed · Start from the end state / desired conclusion and reverse