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When a certain tree was first planted, it was 4 feet tall,

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Re: 4 feet tall tree ? [#permalink]

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New post 09 Sep 2013, 08:25
End of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year

Given:
4th yr: \(4+4x\)
6th yr: \(4+6x\)
\(\frac{1}{5} = .2\)

in other words, the 6th year it was 1.2 bigger than the 4th year

\(4+6x = 1.2(4+4x)\)

\(4+6x = 4.8+4.8x\)

\(1.2x = .8\)

\(x=\frac{.8}{1.2} ---> x=\frac{8}{12}\)

\(x=\frac{2}{3}\)
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Re: When a certain tree was first planted, it was 4 feet tall, [#permalink]

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New post 09 Sep 2013, 11:15
gameCode wrote:
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?

A. 3/10
B. 2/5
C. 1/2
D. 2/3
E. 6/5


if 4th year is say x
and 6th year is x + x/5, the difference = x/5 for 2 years and hence for each year it will be x/10.

now if 4th year is x then the start of 1st year will be (x-(4x/10)) i.e. x- (2x/5) = 4 i.e. x = 20/3.

Therefore x/10 i.e. for each year = 2/3

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When a certain tree was first planted, it was 4 feet tall, [#permalink]

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New post 24 Aug 2014, 12:10
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Bunuel wrote:
gameCode wrote:
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?

A. 3/10
B. 2/5
C. 1/2
D. 2/3
E. 6/5

There's something particular with the answer of this one which i don't understand, hence the posting.



Let the rate of increase be \(x\) feet per year.

At the end of the 4th year the height was \(4+4x\) and at the of the 6th year the height was \(4+6x\), which was "1/5 taller than it was at the end of the 4th year" --> \((4+4x)+\frac{1}{5}(4+4x)=4+6x\) --> \(\frac{1}{5}(4+4x)=2x\) --> \(x=\frac{2}{3}\).

Answer: D.


Hi Bunuel,

I used a different approach which led me to the wrong answer. I can't seem to figure out why?

1)Difference between EOY 6 and EOY 4 is 1/5, therefore, Y6=(6/5)Y4
2)By the same token, EOY 2 = (6/5) EOY0, therefore EOY2 = (6/5)*4
3)Since it's a constant increase every year, (24/5) / (2) since we are only concerned about the increase per year -- this equals 12/5.

Why is this wrong? If it's a constant increase, it's increasing by the SAME amount each year correct. It's NOT compounding from the previous year. Am I correct? By that theory, shouldn't my method work?

On another note -- if the above was an increase to the previous year, meaning, compounded, how would the equation change?

Thanks!

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When a certain tree was first planted, it was 4 feet tall, [#permalink]

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New post 05 Apr 2016, 19:40
Attached is a visual that should help. Notice that multiplying by \(\frac{6}{5}\) is an easier way to get to "\(\frac{1}{5}\) greater than"
Attachments

Screen Shot 2016-04-05 at 8.34.56 PM.png
Screen Shot 2016-04-05 at 8.34.56 PM.png [ 69.98 KiB | Viewed 1094 times ]


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Re: When a certain tree was first planted, it was 4 feet tall, [#permalink]

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New post 14 Jan 2017, 13:17
Let x be the height of the tree increase each year, then:
[4+6x-(4+4x)]/(4+4x) = 1/5
10x = 4+ 4x
x= 2/3
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Re: When a certain tree was first planted, it was 4 feet tall, [#permalink]

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New post 10 Apr 2017, 07:56
I was surprised to see that everyone has taken an algebraic approach. The problem isn't that complicated and can be solved with no algebra at all.

We will simply run two scenarios based on answer choices (B) and (D). If either of them turns out to be the answer, then we're golden. This will happen 40% of the time. Otherwise, we should be able to see whether the answer lies between (B) and (D) or whether it is an outlier.

To make the scenario easier, I will reclassify the original height for answer choice (B) as 20/5 and for answer choice (D) as 12/3, although I will not write the denominator each time for added speed and simplicity.

Scenario (B)

Year 0—20/5
Year 1—22
Year 2—24
Year 3—26
Year 4—28
Year 5—30
Year 6—32

Since 1/5 of 28 (year 4) is 5 and change, we can see that one fifth more will be 33 and change. So (B) is not the answer—it's too small.

Scenario (D)
Year 0—12/3
Year 1—14
Year 2—16
Year 3—18
Year 4—20
Year 5—22
Year 6—24

Since 1/5 of 20 (year 4) is 4, (D) is the answer. Had (D) not been the answer, we would have been able to determine which of the other three [(A), (C), or (E)] was the answer with no difficulty.
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Re: When a certain tree was first planted, it was 4 feet tall, [#permalink]

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New post 11 Dec 2017, 10:58
gameCode wrote:
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?

A. 3/10
B. 2/5
C. 1/2
D. 2/3
E. 6/5


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 let 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 is 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 6/5, we multiply the entire equation by 5:

6(4 + 4x) = 20 + 30x

24 + 24x = 20 + 30x

6x = 4

x = 4/6 = 2/3 feet

Answer: D
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Re: When a certain tree was first planted, it was 4 feet tall,   [#permalink] 11 Dec 2017, 10:58

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