AMC 8 · 2008 · #9

Grade 7 arithmetic
percentagefraction-multiplicationmulti-digit-arithmetic identify-subproblems ↑ Prerequisites: percentagefraction-multiplication
📏 Short solution 💡 3 insights

Problem

In 20052005 Tycoon Tammy invested 100100 dollars for two years. During the first year
her investment suffered a 15%15\% loss, but during the second year the remaining
investment showed a 20%20\% gain. Over the two-year period, what was the change
in Tammy's investment?

Pick an answer.

(A)
$5\%\text{ loss}$
(B)
$2\%\text{ loss}$
(C)
$1\%\text{ gain}$
(D)
$2\% \text{ gain}$
(E)
$5\%\text{ gain}$
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Toolkit + CCSS Solution

Understand

Restated: Tycoon Tammy starts with $\$100$. In the first year her investment drops by $15\%$. In the second year, the leftover amount grows by $20\%$. What is the overall percent change after two years?

Givens: Initial investment: $\$100$; Year 1: $15\%$ loss applied to the starting $\$100$; Year 2: $20\%$ gain applied to whatever is left after Year 1; Answer choices: (A) $5\%$ loss, (B) $2\%$ loss, (C) $1\%$ gain, (D) $2\%$ gain, (E) $5\%$ gain

Unknowns: The total percent change from $\$100$ to the value after two years

Understand

Restated: Tycoon Tammy starts with $\$100$. In the first year her investment drops by $15\%$. In the second year, the leftover amount grows by $20\%$. What is the overall percent change after two years?

Givens: Initial investment: $\$100$; Year 1: $15\%$ loss applied to the starting $\$100$; Year 2: $20\%$ gain applied to whatever is left after Year 1; Answer choices: (A) $5\%$ loss, (B) $2\%$ loss, (C) $1\%$ gain, (D) $2\%$ gain, (E) $5\%$ gain

Plan

Primary tool: #7 Identify Subproblems

Secondary: #3 Write an Equation

Two percent changes happen in order, so Tool #7 (Identify Subproblems) splits the problem into two clean steps: first apply the $15\%$ loss, then apply the $20\%$ gain to the result. Tool #3 (Write an Equation) gives the right form for each step — multiply by $0.85$ for a $15\%$ loss, multiply by $1.20$ for a $20\%$ gain — and then one more equation turns the final dollar amount into a percent change. The trap to avoid is adding $-15\%$ and $+20\%$ to get $+5\%$; that ignores the fact that the second percent is taken on a smaller base.

Execute — Answer: D

#3 Write an Equation 6.RP.A.3 Step 1
  • Apply the Year 1 loss.
  • A $15\%$ loss keeps $85\%$ of the money, so multiply the starting $\$100$ by $0.85$.
$\$100 \times 0.85 = \$85$

💡 "Losing $15\%$" is the same as "keeping $85\%$." Multiplying by $0.85$ is the Grade 6 way to find a percent of a number.

#3 Write an Equation 6.RP.A.3 Step 2
  • Apply the Year 2 gain on the new amount.
  • A $20\%$ gain makes the money $120\%$ of what it was at the start of Year 2, so multiply $\$85$ by $1.20$. The base here is $\$85$, not $\$100$.
$\$85 \times 1.20 = \$102$

💡 Sequential percent changes use the running balance as the next base. That is why $+20\%$ on $\$85$ adds only $\$17$, not $\$20$.

#7 Identify Subproblems 7.RP.A.3 Step 3
  • Compare the final amount to the original.
  • The investment went from $\$100$ to $\$102$, so the dollar change is $\$2$, and the percent change uses $\$100$ as the base.
$\dfrac{\$102 - \$100}{\$100} \times 100\% = \dfrac{\$2}{\$100} \times 100\% = 2\% \text{ gain} \;\Rightarrow\; \textbf{(D)}$

💡 A Grade 7 percent-change problem: divide the change by the original, then convert to a percent.

[1] #3 6.RP.A.3 Apply the Year 1 loss. A $15\%$ loss keeps $85\%$ of the money, so multiply the
[2] #3 6.RP.A.3 Apply the Year 2 gain on the new amount. A $20\%$ gain makes the money $120\%$ o
[3] #7 7.RP.A.3 Compare the final amount to the original. The investment went from $\$100$ to $\

Review

Reasonableness: Quick sanity check on the combined factor: a $15\%$ loss followed by a $20\%$ gain multiplies the original by $0.85 \times 1.20 = 1.02$, which is exactly a $2\%$ gain on $\$100$, giving $\$102$. That matches the step-by-step answer. It also explains why the naive $-15\% + 20\% = +5\%$ is wrong: the order and the changing base matter.

Alternative: Tool #11 (Find an Invariant): the combined multiplier $0.85 \times 1.20 = 1.02$ does not depend on the starting amount. So whatever the initial investment, the two-year result is always $1.02$ times the start — a $2\%$ gain. The $\$100$ in the problem is just a convenient number; the percent change would be the same with $\$1$ or $\$1{,}000$.

CCSS standards used (min grade 7)

  • 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, including percent (Translating "$15\%$ loss" into multiplying by $0.85$ and "$20\%$ gain" into multiplying by $1.20$ to get $\$85$ and then $\$102$.)
  • 7.RP.A.3 Use proportional relationships to solve multistep ratio and percent problems (Combining the two sequential percent changes and computing the overall percent change as $(\$102 - \$100) / \$100 = 2\%$ gain.)

⭐ Sequential percent changes do not add up the way they look. A $15\%$ loss then a $20\%$ gain becomes $0.85 \times 1.20 = 1.02$ — a $2\%$ gain — because the second percent is taken on a smaller amount.

⭐ Sequential percent changes do not add up the way they look. A $15\%$ loss then a $20\%$ gain becomes $0.85 \times 1.20 = 1.02$ — a $2\%$ gain — because the second percent is taken on a smaller amount.