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by Baldini » Fri Mar 06, 2009 3:52 am
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 at the end of the 4th year. By how many feet did the height of the tree increase every year?

a. 3/10
b. 2/5
c. 1/2
d. 2/3
e. 6/5

OA is d
Could someone please explain reasoning behind this question?
Thank you
GMAX
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by bluementor » Fri Mar 06, 2009 4:06 am
This is essentially an arithmetic progression.

Tn = a + (n-1)d

a = 4 (initial value)
d = increment
n = term (in this case its "year", n=1 means when the tree is planted)
Tn = value at the given nth term


T1 = a
T2 = a + d (after 1 year)
T3 = a + 2d (after 2 years)
T4 = a + 3d (after 3 years)
T5 = a + 4d (after 4 years)
T6 = a + 5d (after 5 years)
T7 = a+ 6d (after 6 years)

From the question, we know that:

T7 =(6/5)*T5
a + 6d = (6/5)*(a + 4d)
5*(a + 6d) = 6*(a + 4d)
5a + 30d = 6a + 24d
a = 6d
4 = 6d (we know that a=4)
d=2/3

Choose D

-BM-
Last edited by bluementor on Fri Mar 06, 2009 4:23 am, edited 1 time in total.
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by Baldini » Fri Mar 06, 2009 4:14 am
Thanks Blumentor, but i just don't get one thing:
How do you go from:
T7 =(5/4)*T5

to:
a + 6d = (6/5)*(a + 4d)?
GMAX
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by bluementor » Fri Mar 06, 2009 4:24 am
Baldini wrote:Thanks Blumentor, but i just don't get one thing:
How do you go from:
T7 =(5/4)*T5

to:
a + 6d = (6/5)*(a + 4d)?
I made a careless mistake. Thanks, I've corrected it in my earlier post.

:D

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