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

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Joined: 23 Mar 2011
Posts: 13

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

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29 Nov 2012, 19:27
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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
[Reveal] Spoiler:
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
[Reveal] Spoiler:
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 GMAT logical sense my other interpretations are correct, even though these two problems are quite identical.
Thus, why my some interpretations are correct in GMAT prep and incorrect in GMAT club problem.

Thanks!!!

Kudos [?]: 1 [0], given: 6

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Joined: 22 Dec 2011
Posts: 295

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Re: Certain tree (GMAT Prep) vs Linterhast (GMAT Club) [#permalink]

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29 Nov 2012, 22:35
GMATtaker777 wrote:
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
[Reveal] Spoiler:
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
[Reveal] Spoiler:
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 GMAT logical sense my other interpretations are correct, even though these two problems are quite identical.
Thus, why my some interpretations are correct in GMAT prep and incorrect in GMAT club problem.

Thanks!!!

In the 2nd problem (GMAT Club), the rate per year is not constant as given in the 1st problem
What's Given in the 2nd problem is "the population growth rate per thousand is constant" - its not mentioned per year or any base.
So the population could have grown by 1000 in the 1st year and 100 each for the subsequent years. All we know for 3 year period population grows by 1200.

On the hand GMAT prep says -> the height of the tree increased by a constant amount each year So in this problem we can use our regular progression formula.

Cheers

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Re: Certain tree (GMAT Prep) vs Linterhast (GMAT Club) [#permalink]

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29 Nov 2012, 22:40
GMATtaker777 wrote:
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
[Reveal] Spoiler:
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
[Reveal] Spoiler:
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 GMAT logical sense my other interpretations are correct, even though these two problems are quite identical.
Thus, why my some interpretations are correct in GMAT prep and incorrect in GMAT club problem.

Thanks!!!

You have considered the second question as an arithmetic progression whereas in reality it is a geometric progression with a=3600 and r=1.1. The increase is not by a certain amount as in the first question but rather is at a certain rate.
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Re: Certain tree (GMAT Prep) vs Linterhast (GMAT Club) [#permalink]

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30 Nov 2012, 03:26
GMATtaker777 wrote:
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
[Reveal] Spoiler:
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
[Reveal] Spoiler:
B

First question is discussed here: when-a-certain-tree-was-first-planted-it-was-4-feet-tall-104128.html
Second question is discussed here: ps-population-m01q09-68558-20.html

Topic is locked.

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Re: Certain tree (GMAT Prep) vs Linterhast (GMAT Club)   [#permalink] 30 Nov 2012, 03:26
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