harsh.champ wrote:At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p>0 and q>0. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?

(A)p = q
(B)p < q
(C)p > q
(D)p = q/2
(E)None of the above

The OA is C.

no of goats in the start of 1999 = 108
no of goats bought in the start of 1999 =1.08*p
no of goats sold in the end of 1999 = q (108z+1.08p)/100 = 1.08q+ .0108pq

Now at the end of the year he has got 108 + 1.08p - (1.08q+ .0108pq)
For this to be equal to 108 [because if it is less or more than 108 in one year, the pattern continues and he will either have less than 108 or greater than 108)
108 + 1.08p - (1.08q+ .0108pq) = 108
1.08p - (1.08q+ .0108pq)
1.08p =(1.08q+ .0108pq)
1.08p >(1.08q) (.0108pq Is positive)
p>q

harsh.champ wrote:At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p>0 and q>0. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?

(A)p = q
(B)p < q
(C)p > q
(D)p = q/2
(E)None of the above

Great question for logic instead of complicated algebra.

Let's just look at one year and, instead of using weird numbers, make up our own, since only the trend matters.

We start with 100 goats and end with 100 goats.

At the start of the year we buy 10% more goats, to give us 110 goats.

We want to end up with 100 goats at the end of the year. If we sold 10% of our goats (i.e. 11), we'd only have 99 left. So, we have to sell fewer than 10% of our goats to get back to 100.

Accordingly, the % we buy is higher than the % we sell.

p > q, choose (C).

ajith wrote:

harsh.champ wrote:At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p>0 and q>0. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?

(A)p = q
(B)p < q
(C)p > q
(D)p = q/2
(E)None of the above

The OA is C.

no of goats in the start of 1999 = 108
no of goats bought in the start of 1999 =1.08*p
no of goats sold in the end of 1999 = q (108z+1.08p)/100 = 1.08q+ .0108pq

Now at the end of the year he has got 108 + 1.08p - (1.08q+ .0108pq)
For this to be equal to 108 [because if it is less or more than 108 in one year, the pattern continues and he will either have less than 108 or greater than 108)
108 + 1.08p - (1.08q+ .0108pq) = 108
1.08p - (1.08q+ .0108pq)
1.08p =(1.08q+ .0108pq)
1.08p >(1.08q) (.0108pq Is positive)
p>q

God question and good concept applied by ajith. i appreciate your effort. Harsh will you please post some of similar type questions from the source.