Const increament

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by GMATGuruNY » Mon Feb 14, 2011 9:03 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
We can plug in the answer choices, which represent the increase in feet each year.
Let h = height of tree.

Answer choice C: increase = 1/2 each year.
After 4 years, h = 4 + 4(1/2) = 6.
After 6 years, h = 6 + 2(1/2) = 7.
6 years - 4 years = 7-6 = 1.
1/5 * (4 years) = 1/5 * 6 = 6/5. Doesn't work.
1< 6/5, so the yearly increase must be greater.

Answer choice D: increase = 2/3 each year.
After 4 years, h = 4 + 4(2/3) = 20/3.
After 6 years, h = 20/3 + 2(2/3) = 24/3.
6 years - 4 years = 24/3 - 20/3 = 4/3.
1/5 * (4 years) = 1/5 * 20/3 = 4/3. Success!

The correct answer is D.
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by Night reader » Mon Feb 14, 2011 9:52 am
earnest10 wrote:Image
when first planted tree = 4 OR x=4
each year the tree grows an amount a (const.)
year one --> x+a
year two --> x+2a
...
year six --> x+6a

at the end of the sixth year, tree is 1/5 taller than at the end of the fourth OR (x+6a)= (1+1/5)*(x+4a).
Now if we solve the system of two equations we should get the value of a (the amount of increase every year) :) do not use each in SC part of GMAT for such particular case, as each is used as one out of the two, and every may be used for more than two terms

SO, EQ.1 (x+6a)= (6/5)(x+4a) AND EQ.2 x=4; EQ.1 --> 5x+30a=6x+24a <> 6a=x; EQ.2 x=4 and 6a=4, a=4/6 OR 2/3

answer D