BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Shepherds

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 am
Location: India
Thanked: 64 times
Followed by:6 members
GMAT Score:760

Shepherds

by harsh.champ » Mon Feb 08, 2010 2:52 pm
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.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1275
Joined: Thu Sep 21, 2006 11:13 pm
Location: Arabian Sea
Thanked: 125 times
Followed by:2 members

by ajith » Mon Feb 08, 2010 9:20 pm
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
Always borrow money from a pessimist, he doesn't expect to be paid back.
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Mon Feb 08, 2010 9:47 pm
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).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
User avatar
Legendary Member
Posts: 1022
Joined: Mon Jul 20, 2009 11:49 pm
Location: Gandhinagar
Thanked: 41 times
Followed by:2 members

by shashank.ism » Wed Feb 10, 2010 4:46 am
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.
My Websites:
www.mba.webmaggu.com - India's social Network for MBA Aspirants

www.deal.webmaggu.com -India's online discount, coupon, free stuff informer.

www.dictionary.webmaggu.com - A compact free online dictionary with images.

Nothing is Impossible, even Impossible says I'm possible.
Top
Post new topic Post Reply
• Page 1 of 1