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Source: — Problem Solving |

by Anju@Gurome » Mon Apr 01, 2013 12:45 am
[email protected] wrote:Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period?

A. 1250
B. 1563
C. 2250
D. 2560
E. 2752
Algebraic Approach:
The number of trees in a year is increased by 1/4 of the number of trees in the preceding year. Hence, number of trees in a year = (1 + 1/4) = 5/4 of the number of trees in the preceding year.

Let us assume that number of trees at the beginning of the year = N
So, N*(5/4)*(5/4)*(5/4)*(5/4) = 6250
--> N = 6250*(4/5)*(4/5)*(4/5)*(4/5) = 6250*(4^4)/(5^4) = 6250*256/625 = 2560

The correct answer is D.
Last edited by Anju@Gurome on Mon Apr 01, 2013 12:52 am, edited 1 time in total.
Anju Agarwal
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by cans » Mon Apr 01, 2013 12:46 am
How did you try to solve it?
IMO D
If my post helped you- let me know by pushing the thanks button ;)

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Cans!!
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by Anju@Gurome » Mon Apr 01, 2013 12:51 am
This particular problem can be solved easily by efficiently plugging the answer choices also.

Note that in the 2nd year the number of trees increased by 1/4 of the of the number of of trees in the 1st year. Hence, number of trees in the first year, i.e. the correct answer must be a multiple of 4.

So, the last two digits of the correct answer must be multiple of 4.
Discard A, B, and C.

Check with option D : 2560
2560*(5/4)*(5/4)*(5/4)*(5/4) = 2560*(5^4)/(4^4) = 2560*625/256 = 6250 --> Correct

The correct answer is D.
Anju Agarwal
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Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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by [email protected] » Mon Apr 01, 2013 5:30 am
HI Cans,


I kept the first year's value as x and then did:

x+(x+x/4)+(x+x/4)+(x+x/4)/4)+(x+x/4)+(x+x/4)/4)+(x+x/4)/4=6250
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