Constant growth

[magauovazamat]

I am little confused with these two problems:

First is GMAT prep question:

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: (A) 3/10 (B)2/5 (C)1/2 (D) 2/3 (E) 6/5
OA is D

Here is the second question from GMAT club advanced workshop:

The population of Linterhast was 3,600 people in 1990 and 4,800 people in 1993. If the population growth rate per thousand is constant, then what will be the population in 1996?

Answers: (A) 6,000 (B) 6,400 (C) 7,200 (D) 8,000 (E) 9,600
OA is B but I ended up with A

My interpretation for the first question:
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

Interpretation for the second question:
Let x = amount of yearly growth
1990 = 3600
1991 = 3600+x
1992 = 3600+x+x=3600+2x
1993 = 3600+x+x+x=3600+3x = 4800
1994 = 3600+x+x+x+x=3600+4x
1995 = 3600+x+x+x+x+x=3600+5x
1996 = 3600+x+x+x+x+x+x=3600+6x

My initial solution:
3600+3x = 4800
3x=1200
x=400

So, 1996= 3600+6*400=6000

Then after knowing that I missed the question I solved this way:
(3600+6x) = 4800/3600*(3600+3x)
6(600+3x)=4/3*3(1200+x)
x=600

So, 1996= 3600+6*600=7200

Then,
Found the answer with the logic here is just:

4800/3600*4800=6400

I get the logic of the solution but how my other interpretations are not correct since from the logical sense and GMAT logic my other interpretations are correct. These two problems are identical and the logic is pretty similar.
Why my some interpretations are correct in GMAT prep and incorrect in GMAT club problem.

Thanks!!!

[magauovazamat]

Hi folks!!!

Is there anyone who could help with these questions?
Thanks!

[RonPurewal]

Hi folks!!!

Is there anyone who could help with these questions?
Thanks!

first, don't post messages like this. if you do so, you're actually delaying our response to your question.
see, we answer posts in order from oldest to newest. by posting this kind of thing, you're making your question into the newest thread on the whole forum -- which means it's the last one to be answered again.

--

the problem with your first two solutions to the "linterhast" problem is that you are assuming constant growth -- in other words, you're assuming that the town gains the same number of people each year -- and, well, that's not true.
on that problem, the problem statement says that the growth rate "per thousand" is what's constant. in other words, that's a relative growth rate; it's the same thing as having a constant percentage growth rate.

also, notice that there's no need to consider individual years. the jump from 1990 to 1993 is three years, and the jump from 1993 to the target year 1996 is also three years, so you can just repeat what happens from 1990 to 1993 to figure out what happens from 1993 to 1996.

in the first interval (1990-1993), the town gains 1200 people. that's a growth rate of 1200/3600 = 1/3 = 33 1/3 %. (in literal terms of "per thousand" it's 333.333....... per thousand, but that's pretty icky.)

so, from 1993 to 1996, the town will gain a number of people equivalent to 1/3 of its starting population again.
this time, that's 1/3 of 4800, or 1600 new people, making a total of 4800 + 1600 = 6400.

--

by the way, this is the GMAT PREP folder. even though these two problems are vaguely similar, you can't post the second one here; you need to post it in the General Math folder, per the forum rules.
this thread is now locked. if you have any further questions about that problem, please start a new thread, this time in the correct folder. thanks.