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

[tim]

thanks for sharing!

[AmunaGmat]

This is essentially a sequence problem in disguise. Let x = amount of yearly growth, in feet.

Yr0 = 4
Yr1 = 4+x
Yr2 = 4+x+x=4+2x
Yr3 = 4+x+x+x=4+3x
Yr4 = 4+x+x+x+x=4+4x
Yr5 = 4+x+x+x+x+x=4+5x
Yr6 = 4+x+x+x+x+x+x=4+6x

We are told the amount at the end of Year 6 is 6/5 of the amount at the end of year 4. Thus we can write:

4+6x = 6/5 (4+4x)
5(4+6x) = 6(4+4x)
20+30x = 24+24x
6x=4
x=2/3

I never understood any formula in these stuff, Stacey you nailed it. Thanks!!!

[jnelson0612]

I never understood any formula in these stuff, Stacey you nailed it. Thanks!!!

Great, thanks! Good to hear!

[rkafc81]

I did it this way, and got it wrong because I misinterpreted the yearly-growth wording :

Yr1 = 4
Yr2 = 4+x
Yr3 = 4+x+x=4+2x
Yr4 = 4+x+x+x=4+3x
Yr5 = 4+x+x+x+x=4+4x
Yr6 = 4+x+x+x+x+x=4+5x

instead of

Yr0 = 4
Yr1 = 4+x
Yr2 = 4+x+x=4+2x
Yr3 = 4+x+x+x=4+3x
Yr4 = 4+x+x+x+x=4+4x
Yr5 = 4+x+x+x+x+x=4+5x
Yr6 = 4+x+x+x+x+x+x=4+6x

i really don't understand how i managed to make this mistake! i think I should have read it as 'at the end of the 1st year, = 4 + x".

how did I actually make this error so I know not to do it again?

thanks

[tim]

the sequence starts at 4, and if you are going up to year 6 that means 6 years have elapsed. so if you start at year 1, that means 1 year has already elapsed. that's why you want to start at year 0. just make sure you pay attention to detail in general, and don't throw symbols up on the page unless you have a clear idea in your mind what they stand for..

[rkafc81]

thanks Tim!

[jnelson0612]

:-)

[liarish]

I guess why many non-native English speakers (including me) got stuck at this question is because we read 1/5 as 1 by 5 or 1 over 5, and not 1/5th.
Good take away from this post.
Thanks guys !

[tim]

actually, those all mean the same thing. if you think that is the issue (and especially if that somehow resolved a question of yours), then you need to go back and take another look at this and see if you can understand it properly..

[byuwadd]

Okay, I totally understand this question now (thanks for the help).

However, when I initially read the question, the word 'constant' triggered a successive percents problem in my head. The tree was 4 feet tall and grew at a constant rate: 4(1.0X)(1.0X)(1.0X)(1.0X)(1.0X)(1.0X).

What is it in the wording that makes it so this problem is not a successive rates problem? What would the question have to say to make it into what I have concocted? The tree grew at a compounding rate of x for six years?

Or is it the fact that the problem says 'constant amount' instead of 'constant rate'? Amount and rate are two very different words. I think that's the difference. Please let me know what you think. :)

[tim]

you are right. "constant amount" and "constant rate" are two very different things, and you need to know what each means..

[byuwadd]

Thanks Tim.

Lesson learned.

[jnelson0612]

Thanks Tim.

Lesson learned.

Cool! :-)

[geezer0305]

GMATprep. Exam 1. Question 31

When a certain tree was first planted, it was 4 feet tall, and the heigth 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 increased each year?

Answers: 3/10, 2/5, 1/2, 2/3, 6/5

The correct answer is 2/3 but I have no idea how to solve it. Any help?

thanks

I used the following method:

Because the growth of the tree is constant each year, we can represent the growth as a linear function: y = mx +c
or let us represent it as H= m*T+C
where,
H= height (varies with time)
T = Time (variable)
m = slope of the equation or rate of growth
C= initial constant

To find: "m" or slope or rate of growth
Given:
Initial height of the tree = 4 units
Therefore, C = 4 (this is the "initial" point from where we are concerned with growth)

Now, Height at the end of the 6th year or at T=6 :-
=> H6 = m*(6) + 4-- Eqn (i)

Height at the end of the 4th year or at T=4:-
=> H4 = m*(4) + 4-- Eqn (ii)

We are also given that H6 = (1+1/5)*H4
=> H6 = (6/5) H4--- Eqn (iii)

Simplifying Equations (i), (ii) and (iii), we get

m = 2/3

[jlucero]

Algebraically, this is exactly how to set this problem up. Good work.