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by dferm » Mon Mar 24, 2008 9:08 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 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
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by xilef » Mon Mar 24, 2008 11:59 am
1/5 of the height at the end of 4rth + height at the end of 4rth = height at the end of sixth

x - annual increase


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

Answer D.
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by Patrick_GMATFix » Wed Jul 21, 2010 6:44 am
Because the tree grows by the same amount each year (call it x), after 6 years the tree will be 4+6x in height. After 4 years the tree will be just 4+4x in height

since the 6th year height is 1/5 taller (20% more) than the 4th year height, we can say that (4+6x) is 20% more tan (4+4x). Algebraically, 4+6x=(4+4x)+(1/5)(4+4x). Solving for n will give you 2/3. The correct answer is D.

A more involved explanation, take-away lesson and step-by-step video solution can be found at GMATPrep Question 1150. To create drills with similar questions set topic='Algebraic Translations' and difficulty='600-700 & 700+' in the Drill Generator

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by dferm » Sun Jul 25, 2010 8:23 pm
Patrick,

I have a question regarding the tree growing. Is there a way to use a number line to solve these particular questions.

Thanks.

-David
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by dferm » Sun Jul 25, 2010 8:34 pm
Patrick,

I get the same ans when I use 4+6x=6/5(4+4X) as my equation.
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by dinesh19aug » Sun Jul 25, 2010 9:01 pm
dferm wrote:Patrick,

I have a question regarding the tree growing. Is there a way to use a number line to solve these particular questions.

Thanks.

-David
David,
You can plot it on number line

Let's take initial height of the plant as A (A=4 given in the question)
Let's take Y = year and H = Height

Y H
0 A
1 A + d
2 A + 2d
3 A + 3d
4 A + 4d
5 A+5d
6 A + 6d

Now it says that Year 6 - Year 4 = 1/5 (Year 4)
==> (A + 6d) - (A+4d) = 1/5 (A+4d)
Solving the equation would give us ==> 6d = A
Now plug in the values
d = A/6 = 4/6 = 2/3 ANS
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by gmatmachoman » Mon Jul 26, 2010 12:14 am
dferm wrote:Patrick,

I get the same ans when I use 4+6x=6/5(4+4X) as my equation.
Exactly, 1/5 is nothing but 20 %

So 20% more than X will be 1.2 X
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by outreach » Mon Jul 26, 2010 2:42 am
D
method same as @xilef
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by Patrick_GMATFix » Mon Jul 26, 2010 4:38 am
dferm wrote:Patrick,

I get the same ans when I use 4+6x=6/5(4+4X) as my equation.
Hey dferm,

Good equation. The equation you wrote is actually exactly what I would write on my test. In my explanation I instead wrote (4+6x)=(4+4x) + (1/5)(4+4x) because it's easier for most people to understand how that equation relates to "height after 6yrs is 1/5 greater than height after 4 years".

When I write my explanations, I want ppl who have questions to understand so I try to look to put myself in the shoes of someone who doesn't understand the problem.

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by GMATGuruNY » Mon Jul 26, 2010 5:14 am
dferm 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
We also could use the answer choices, which represent the increase in feet each year. Let h= height of tree.

Answer choice C:

After 4 years, h = 4 + 4(1/2) = 6.
After 6 years, h = 6 + 2(1/2) = 7.
7-6=1, but 1/5 * 6 = 6/5. Doesn't work.
1< 6/5, so we need the increase to be greater.

Answer choice D:

After 4 years, h = 4 + 4(2/3) = 20/3.
After 6 years, h = 20/3 + 2(2/3) = 24/3.
24/3 - 20/3 = 4/3, and 1/5 * 20/3 = 4/3. Success!

The correct answer is D.
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