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ashkrs
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Had to guess on this one since time was running and I was 11 Sep 2007, 20:19
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Had to guess on this one since time was running and I was wasting my time solving this.
Some easy steps?



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b14kumar
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Had to guess on this one since time was running and I was 11 Sep 2007, 20:39
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ashkrs
Had to guess on this one since time was running and I was wasting my time solving this.
Some easy steps?
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Little bit tricky...You need to read the problem carefully.

Ok, my take:

Let the rate of increment is r.
Original height is 4.

At the end of 1st year, height will be (4+r)
....
....

At the end of 4th year, height will be (4+4r)
....
....

At the end of 6th year, height will be (4+6r)

According to the problem, at the end of 6th year height of the tree will be 1/5 taller than what it was at the end of 4th year....

That means,

Height at the end of 6th year = height at the end of 4the year + 1/5(height at the end of 4the year)

=> 4+6r = (4+4r) + 1/5(4+4r)
=> 2r = 4/5(1+r)
=> 10r = 4+4r
=> 6r = 4
=> r = 4/6
=> r = 2/3

- Brajesh
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GMATBLACKBELT
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Had to guess on this one since time was running and I was 11 Sep 2007, 21:15
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ashkrs
Had to guess on this one since time was running and I was wasting my time solving this.
Some easy steps?
Show more


I had this same problem on my GMATPrep 1. Had to guess. I tried using the rate formula to solve this (which I could have), but the rate method made the setup unnessicarily complicated. So I had to move on and guess. Afterwards, I got it in a couple of min w/o tryin the rate formula... owell.


Heres the solution. Make the two years equal eachother in algebraic terms.

Since year 6 is 1/5 taller. Then it is greater than year four by 20%.
X is the growth rate. and then number next to X represents the years.
120/100(4+4x) = 4+6x ---> 6/5(4+4x)=4+6x

24/5+24x/5=4+6x ---> 24/5-20/5=30x/5-24x/5 ---> 4/5=6x/5

x=2/3.
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Had to guess on this one since time was running and I was 11 Sep 2007, 23:23
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Fastest way is to just write down on the scratch pad the following:

Each year grow by x ft.

Start: 4ft
End of year 1: 4+x
End of year 2: 4+2x
End of year 3: 4+3x
End of year 4: 4+4x
End of year 5: 4+5x
End of year 6: 4+6x

1/5(4+4x) + 4+4x = 4+6x
x = 2/3
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ashkrs
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Had to guess on this one since time was running and I was 12 Sep 2007, 07:39
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Thanks all . You made it sound easy.
I was still struggling with verbage.
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lonesome307
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Had to guess on this one since time was running and I was 28 Nov 2007, 22:44
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ywilfred
Fastest way is to just write down on the scratch pad the following:

Each year grow by x ft.

Start: 4ft
End of year 1: 4+x
End of year 2: 4+2x
End of year 3: 4+3x
End of year 4: 4+4x
End of year 5: 4+5x
End of year 6: 4+6x

1/5(4+4x) + 4+4x = 4+6x
x = 2/3
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Thanks guys!!...I recently got it in my gmatprep test too. Had to guess on it and got it wrong. But now it's clear as a crystal! :-)
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ian7777
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Had to guess on this one since time was running and I was 29 Nov 2007, 20:32
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ywilfred
Fastest way is to just write down on the scratch pad the following:

Each year grow by x ft.

Start: 4ft
End of year 1: 4+x
End of year 2: 4+2x
End of year 3: 4+3x
End of year 4: 4+4x
End of year 5: 4+5x
End of year 6: 4+6x

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


I did it the same way. My only change is very minor, but time saving:

1/5 more is the same as 6/5 of... so i immediately wrote:
4 + 6xx = 6/5(4 + 4x)

kinda nit-picky, but applied to multiple problems throughout the test could save a lot of time.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
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Thank you for understanding, and happy exploring!
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