When a certain tree was first planted, it was 4 feet tall and the height of the tree increased by...

1. Translate the problem requirements: The tree starts at 4 feet, grows by the same amount each year for 6 years. We need to find this yearly growth amount, given that the height at end of year 6 is \(\frac{1}{5}\) taller than the height at end of year 4.
2. Set up the height timeline: Express the tree's height at key time points (initial, end of year 4, end of year 6) in terms of the unknown yearly growth.
3. Apply the given relationship: Use the fact that year 6 height is \(\frac{1}{5}\) taller than year 4 height to create an equation.
4. Solve for the yearly growth: Algebraically solve the equation to find the constant yearly increase.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in everyday language:

• The tree started at 4 feet tall when first planted
• Each year for the next 6 years, it grew by the same amount (let's call this amount 'x' feet)
• At the end of year 6, the tree was \(\frac{1}{5}\) taller than it was at the end of year 4

We need to find this constant yearly growth amount 'x'.

Process Skill: TRANSLATE - Converting the word problem into mathematical understanding is crucial here

2. Set up the height timeline

Let's think about the tree's height at different points in time:

• Initially (when planted): 4 feet
• After 4 years of growth: \(4 + 4\mathrm{x}\) feet (original height plus 4 years of growth)
• After 6 years of growth: \(4 + 6\mathrm{x}\) feet (original height plus 6 years of growth)

This makes sense because if the tree grows by 'x' feet each year, then after 4 years it has grown \(4\mathrm{x}\) feet total, and after 6 years it has grown \(6\mathrm{x}\) feet total.

So our key heights are:

• Height at end of year 4: \((4 + 4\mathrm{x})\) feet
• Height at end of year 6: \((4 + 6\mathrm{x})\) feet

3. Apply the given relationship

The problem tells us that the height at the end of year 6 is \(\frac{1}{5}\) taller than the height at the end of year 4.

In plain English: if something is \(\frac{1}{5}\) taller than another thing, it means the first thing equals the second thing plus \(\frac{1}{5}\) of the second thing.

So: Height at year 6 = Height at year 4 + \(\frac{1}{5}\) × Height at year 4

This can be written as: Height at year 6 = \((1 + \frac{1}{5})\) × Height at year 4 = \(\frac{6}{5}\) × Height at year 4

Substituting our expressions:

\(4 + 6\mathrm{x} = \frac{6}{5} \times (4 + 4\mathrm{x})\)

4. Solve for the yearly growth

Now we solve the equation: \(4 + 6\mathrm{x} = \frac{6}{5} \times (4 + 4\mathrm{x})\)

First, let's distribute the \(\frac{6}{5}\) on the right side:

\(4 + 6\mathrm{x} = \frac{6}{5} \times 4 + \frac{6}{5} \times 4\mathrm{x}\)

\(4 + 6\mathrm{x} = \frac{24}{5} + \frac{24\mathrm{x}}{5}\)

To make the arithmetic easier, let's multiply everything by 5:

\(5(4 + 6\mathrm{x}) = 5(\frac{24}{5} + \frac{24\mathrm{x}}{5})\)

\(20 + 30\mathrm{x} = 24 + 24\mathrm{x}\)

Now subtract \(24\mathrm{x}\) from both sides:

\(20 + 30\mathrm{x} - 24\mathrm{x} = 24\)

\(20 + 6\mathrm{x} = 24\)

Subtract 20 from both sides:

\(6\mathrm{x} = 4\)

Divide by 6:

\(\mathrm{x} = \frac{4}{6} = \frac{2}{3}\)

Final Answer

The height of the tree increased by \(\frac{2}{3}\) feet each year.

Let's verify: After 4 years, height = \(4 + 4(\frac{2}{3}) = 4 + \frac{8}{3} = \frac{20}{3}\) feet

After 6 years, height = \(4 + 6(\frac{2}{3}) = 4 + 4 = 8\) feet

Check: Is 8 equal to \(\frac{6}{5} \times \frac{20}{3}\)?

\(\frac{6}{5} \times \frac{20}{3} = \frac{120}{15} = 8\) ✓

The answer is D. \(\frac{2}{3}\)

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the timing of growth: Students often confuse when the growth starts. The tree is planted at 4 feet, then grows for 6 years. Some students incorrectly think the tree grows during the planting year, leading to wrong height expressions like "after 4 years: \(4 + 5\mathrm{x}\)" instead of "\(4 + 4\mathrm{x}\)".

2. Misunderstanding "\(\frac{1}{5}\) taller than": Students frequently misinterpret this phrase. Instead of understanding it as "Height at year 6 = Height at year 4 + \(\frac{1}{5}\) × Height at year 4", they might incorrectly set up "Height at year 6 = Height at year 4 + \(\frac{1}{5}\)" (adding just \(\frac{1}{5}\) feet) or "Height at year 6 = \(\frac{1}{5}\) × Height at year 4" (making year 6 height only \(\frac{1}{5}\) of year 4 height).

3. Confusing the reference years: Some students mix up which years to compare. The problem states the tree was \(\frac{1}{5}\) taller "at the end of the 6th year" compared to "at the end of the 4th year", but students might incorrectly compare year 6 to the initial height or year 4 to year 6.

Errors while executing the approach

1. Arithmetic errors with fractions: When distributing \(\frac{6}{5} \times (4 + 4\mathrm{x})\), students often make mistakes like getting "\(\frac{24}{5} + 6\mathrm{x}\)" instead of "\(\frac{24}{5} + \frac{24\mathrm{x}}{5}\)". They forget to apply the fraction to both terms or incorrectly multiply the coefficients.

2. Errors when clearing fractions: When multiplying the entire equation by 5 to eliminate fractions, students sometimes forget to multiply all terms consistently, leading to equations like "\(20 + 30\mathrm{x} = 24 + 4\mathrm{x}\)" instead of "\(20 + 30\mathrm{x} = 24 + 24\mathrm{x}\)".

3. Sign errors during algebraic manipulation: When collecting like terms (\(30\mathrm{x} - 24\mathrm{x} = 6\mathrm{x}\)), students might make sign errors or incorrectly combine coefficients, especially when moving terms from one side of the equation to the other.

Errors while selecting the answer

1. Not simplifying the final fraction: Students might arrive at \(\mathrm{x} = \frac{4}{6}\) but fail to reduce it to \(\frac{2}{3}\), potentially looking for \(\frac{4}{6}\) among the answer choices and getting confused when it's not there.

2. Verification errors leading to answer changes: Some students attempt to verify their answer but make calculation mistakes during verification. When their check doesn't work out due to arithmetic errors, they assume their original answer was wrong and pick a different choice.

Alternate Solutions

Smart Numbers Approach

Key Insight: Instead of working with variables, we can choose a convenient value for the tree's height at the end of year 4, then use the given relationship to find the yearly growth.

Step 1: Choose a smart number for height at end of year 4
Since we need to work with "\(\frac{1}{5}\) taller," let's choose a height that makes this calculation clean.
Let's say the height at end of year 4 = 20 feet
(We choose 20 because \(\frac{1}{5}\) of 20 = 4, giving us nice whole numbers)

Step 2: Find height at end of year 6
Height at end of year 6 = Height at end of year 4 + \(\frac{1}{5}\) × (Height at end of year 4)
= \(20 + \frac{1}{5} \times 20\)
= \(20 + 4 = 24\) feet

Step 3: Find the yearly growth rate
The tree grows for 2 additional years (from end of year 4 to end of year 6)
Growth in 2 years = \(24 - 20 = 4\) feet
Yearly growth = \(4 ÷ 2 = 2\) feet per year

Step 4: Find the initial height
If the tree was 20 feet at end of year 4, and it grew 2 feet per year:
Initial height = \(20 - (4 \times 2) = 20 - 8 = 12\) feet

Step 5: Scale to match the actual problem
In our problem, the initial height is 4 feet, not 12 feet.
Scaling factor = \(4 ÷ 12 = \frac{1}{3}\)
Actual yearly growth = \(2 \times \frac{1}{3} = \frac{2}{3}\) feet per year

Verification:

• Initial height: 4 feet
• Height at end of year 4: \(4 + 4(\frac{2}{3}) = 4 + \frac{8}{3} = \frac{20}{3}\) feet
• Height at end of year 6: \(4 + 6(\frac{2}{3}) = 4 + 4 = 8\) feet
• Check: \(8 = (1 + \frac{1}{5}) \times \frac{20}{3} = \frac{6}{5} \times \frac{20}{3} = \frac{120}{15} = 8\) ✓

Answer: D. \(\frac{2}{3}\)