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A tree grows only $$\frac{3}{5}$$ as fast as the one beside it. In four years the combined growth of the trees is eight feet. How much does the shorter tree grow in 2 years?
Answer & Solution
Correct Answer:
Option
A
Suppose the taller tree grows x feet in 1 year.
Then the shorter tree grows
$$\eqalign{ & = \frac{{3x}}{5}{\text{ft in 1 year}} \cr & \therefore 4x + 4 \times \frac{3}{5}x = 8 \cr & \Leftrightarrow 4x + \frac{{12}}{5}x = 8 \cr & \Leftrightarrow \frac{{32}}{5}x = 8 \cr & \Leftrightarrow x = \frac{{8 \times 5}}{{32}} = \frac{5}{4} \cr & {\text{Growth of shorter tree in}} \cr & {\text{2 years = }}\left( {2 \times \frac{{3x}}{5}} \right){\text{ft}} \cr & {\text{ = }}\left( {2 \times \frac{3}{5} \times \frac{5}{4}} \right){\text{ft}} \cr & {\text{ = }}\frac{3}{2}{\text{ft}} \cr & {\text{ = 1}}\frac{1}{2}{\text{ft}} \cr} $$
Then the shorter tree grows
$$\eqalign{ & = \frac{{3x}}{5}{\text{ft in 1 year}} \cr & \therefore 4x + 4 \times \frac{3}{5}x = 8 \cr & \Leftrightarrow 4x + \frac{{12}}{5}x = 8 \cr & \Leftrightarrow \frac{{32}}{5}x = 8 \cr & \Leftrightarrow x = \frac{{8 \times 5}}{{32}} = \frac{5}{4} \cr & {\text{Growth of shorter tree in}} \cr & {\text{2 years = }}\left( {2 \times \frac{{3x}}{5}} \right){\text{ft}} \cr & {\text{ = }}\left( {2 \times \frac{3}{5} \times \frac{5}{4}} \right){\text{ft}} \cr & {\text{ = }}\frac{3}{2}{\text{ft}} \cr & {\text{ = 1}}\frac{1}{2}{\text{ft}} \cr} $$
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