Question: 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?
- 3/10
- 2/5
- 1/2
- 2/3
- 6/5
“When a certain tree was first planted, it was 4 feet tall and the height”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.
Solution and Explanation:
Approach Solution 1:
It is given that 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. Since we know that the growth is by a constant amount, we have a linear growth problem. Thus, we can define this constant amount as x = the yearly growth amount, in feet:
Starting height = 4
height after year one = 4 + x
height after year two = 4 + 2x
height after year three = 4 + 3x
height after year four = 4 + 4x
height after year five = 4 + 5x
height after year six = 4 + 6x
We are also given that at the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. This means the height of the tree at the end of the 6th year was 6/5 times as tall as its height at the end of the 4th year. Thus, we can create the following equation:
(6/5)(4 + 4x) = 4 + 6x
To eliminate the fraction of 6/5, we can multiply the entire equation by 5.
6(4 + 4x) = 20 + 30x
24 + 24x = 20 + 30x
6x = 4
x = 4/6 = 2/3 feet
Correct Answer: D
Approach Solution 2:
Height of tree on day 0 = 4
Let d = the height increase each year
Height of tree at the end of the 1st year = 4+d
Height of tree at the end of the 2nd year = 4+d+d = 4 + 2d
Height of tree at the end of the 3rd year = 4+d+d+d = 4 + 3d
Height of tree at the end of the 4th year = 4+d+d+d+d = 4 + 4d
Height of tree at the end of the 5th year = 4+d+d+d+d+d = 4 + 5d
Height of tree at the end of the 6th year = 4+d+d+d+d+d+d = 4 + 6d
At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year
In other words, 6th year height = 4th year height + 1/5(4th year height)
Or we can write 4 + 6d = (4 + 4d) + 1/5(4 + 4d)
Simplify: 4 + 6d = 6/5(4 + 4d)
Multiply both sides by 5 to get: 5(4 + 6d) = 6(4 + 4d)
Expand: 20 + 30d = 24 + 24d
Simplify: 6d = 4
d = 4/6 = 2/3 = D
Correct Answer: D
Approach Solution 3:
In this particular question, the tree was 1/5 taller at the end of 6th year than the end of 4th year.
If the same is taken into consideration, we can solve the same by means of Arithmetic Progression.
Height at the start of 1st Year = 4
The constant increase in the height every year = D
Height At the end of 1st year is = 4 + D
Height At the end of 4th year is = 4 + 4D
Height At the end of 6th year is = 4 + 6D
As per the question => 4 + 4D + 1/5 ( 4 + 4D) = 4 + 6D
Solving the above equation, we get the below mentioned fraction :
4+4D/5 = 2D
Solving that we get the following simplified equation
=> 4 + 4D = 10D => 4 = 6D => D = 2/3
Correct Answer: D
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