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When a certain tree

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When a certain tree

by msbelasco » Thu Sep 06, 2012 6:39 am
When a certain tree is planted, it was 4 ft 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

How should I approach solving this?
Thanks
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by gmatter2012 » Thu Sep 06, 2012 7:17 am
hi

let the tree increase by x feet each year
so initially it is 4 feet tall

at the end of first year = 4+x
at the end of second year = (4+x)+x= 4+2x
....... Third year = 4+3x
...............Fourth year= 4+4x
....... ...... Fifth year= 4+5x
........... sixth year= 4+6x

now at the end six years it is 1/5 taller than it was at the end of the 4th year

so 4+6x -(4+4x)= (1/5)(4+4x)
---> 5(4+6x-4-4x)= (4+4x)
----> 5(2x) =4+4x
----->6x= 4 ---> x= 2/3

ans = D

Hope it helps
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by Brent@GMATPrepNow » Thu Sep 06, 2012 7:53 am
msbelasco wrote:When a certain tree is planted, it was 4 ft 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
Height of tree day 0 = 4
Let d be the increased height 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

We are told that 4+6d is 1/5 greater than 4+4d
In other words 4+6d = (4+4d) + 1/5(4+4d)
or 4+6d = 6/5(4+4d)
Multiply both sides by 5 and (to eliminate fractions) and solve for d to get d=2/3

The answer is D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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