We have to read the question carefully to make sure we see that the end of the 6th year is being compared to the end of the 4th year. Also note that the tree grows by a constant amount each year. This is different than the same % each year.
Let x = the constant annual growth in feet.
4ft is the original height of the tree, so after the 6th year, we have
4 + x + x + x + x + x + x….aslo 4 + 6x
After the 4th year we have 4 + x + x + x+ x or 4 + 4x
Lets compare the two values algebraically as the problem does:
6th year is 1/5 taller than the 4th year.
The difference in growth from the 4th to the 6th year will be 1/5 (or 0.2) of the height after the 4th year.
So…(4+6x)-(4+4x) = 1/5(4+4x)…now solve for x
4 + 6x -4 – 4x = 4/5 + 4/5x
2x = 4/5 + 4/5x
Subtract 4/5x from both sides. I’ve changed 2x into 10/5x to easily do the subtraction of fractions.
10/5x – 4/5x = 4/5
6/5x = 4/5
Divide both sides by 6/5, which is the same as multiplying by 5/6
X = 4/5 * 5/6. (Reduce the 5’s or multiply out and) you have 20 / 30 or x = 2/3.
Hope a different approach may help someone else.
When a certain tree was first planted, it was 4 feet tall, and the height of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each year?
There's something particular with the answer of this one which i don't understand, hence the posting.
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
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