A tree is at present 9 feet tall. If every year it grows 1/9 th of its

A tree is at present 9 feet tall. If every year it grows 1/9 th of its height, what will be the height of the tree after three years?

12 feet
12.34 feet
13 feet
13.10 feet
This question was previously asked in
UPSC CAPF – 2021
If a tree is 9 feet tall and grows 1/9th of its height every year, its height after three years will be approximately 12.34 feet.
– The current height of the tree is H₀ = 9 feet.
– The growth each year is 1/9th of the height at the beginning of that year. This is a compound growth pattern.
– After 1 year, height H₁ = H₀ + (1/9)H₀ = H₀(1 + 1/9) = H₀(10/9).
– After 2 years, height H₂ = H₁ + (1/9)H₁ = H₁(1 + 1/9) = H₁(10/9) = H₀(10/9)(10/9) = H₀(10/9)².
– After 3 years, height H₃ = H₂ + (1/9)H₂ = H₂(1 + 1/9) = H₂(10/9) = H₀(10/9)³.
– Substitute the initial height H₀ = 9 feet:
– H₃ = 9 * (10/9)³ = 9 * (1000 / 729).
– H₃ = 9000 / 729.
– Simplify the fraction by dividing numerator and denominator by 9:
– H₃ = 1000 / 81.
– Calculate the decimal value: 1000 ÷ 81 ≈ 12.34567…
– Rounding to two decimal places, the height is approximately 12.35 feet. Among the given options, 12.34 feet is the closest value.
The alternative interpretation, where the tree grows 1/9th of the *original* height (1 foot) each year, would result in a height of 9 + 3*1 = 12 feet after 3 years. Since 12.34 is an option and 12 is also an option, the wording “1/9 th of its height” generally implies growth relative to the current height unless otherwise specified, leading to the compounding calculation.
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