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OG 108

Expert replies
Source: — Problem Solving |

by Anju@Gurome » Mon Apr 01, 2013 12:54 am
You have already posted this question here >> https://www.beatthegmat.com/og-108-t213839.html
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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by GMATGuruNY » Mon Apr 01, 2013 3:21 am
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
We can plug in the answers, which represent the original number of trees.
Since the number of trees increases by 1/4 each year, the correct answer must be a multiple of 4.
The last 2 digits of a multiple of 4 must themselves form a multiple of 4.
Eliminate A (1250) and C (2250), since 50 is not a multiple of 4.
Eliminate B (1563), since 63 is not a multiple of 4.

Answer choice D: 2560
After the 1st year, the number of trees = 2560 + (1/4)2560 = 3200.
After the 2nd year, the number of trees = 3200 + (1/4)3200 = 4000.
After the 3rd year, the number of trees = 4000 + (1/4)4000 = 5000.
After the 4th year, the number of trees = 5000 + (1/4)5000 = 6250.
Success!

The correct answer is D.

Note that we had to try only ONE answer choice -- a very efficient way to solve the problem.
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by veenu08 » Wed Apr 03, 2013 9:39 am
These types of qstns can be approached by reverse tracking-

At the end of 4th year no. of tree- 6250
Let n, be the no. of trees at the end of 3rd year.
so n+(1/4)n= 6250
5n/4=6250
n= 5000
similarly let n, be the trees at the end of second year
5n/4= 5000
n=4000
and let n be the tress at the end of first year
5n/4= 4000
n=3200
Therefore trees at the beginning will be : 5n/4= 3200
n= 2560
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