Finance Class

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 304
Joined: Wed Jan 27, 2010 8:35 am
Location: International Space Station
Thanked: 11 times
Followed by:3 members

Finance Class

by Aman verma » Fri Feb 26, 2010 3:32 am
Q: A finance class consists of boys and girls students. If $ 69 was were divided amongst 115 students, the number of girl students in the class is :

I. Each girl gets 50 Cents less than a boy.

II.Each boy received twice the Cents as each girl received.
800. Arjun's-Bird-Eye
Source: — Data Sufficiency |

Senior | Next Rank: 100 Posts
Posts: 40
Joined: Thu Mar 26, 2009 3:02 pm
Thanked: 3 times
GMAT Score:410

by Pedros » Fri Feb 26, 2010 5:28 am
I got C here but I am not sure

User avatar
Master | Next Rank: 500 Posts
Posts: 102
Joined: Sat Feb 20, 2010 5:38 am
Location: IIM Ahmedabad
Thanked: 10 times

by firdaus117 » Fri Feb 26, 2010 7:31 am
Let the number of girls be G and boys be 115-G.
I. Each girl=C cents
Each boy=C+50

GC+(115-G)*(C+50)=6900 Insufficient
II. Each girl=X cents
Each boy=2X cents
GX+(115-G)*2X=6900 ------(A) Insufficient
Combining,
C+50=2C
C=50=X
Substituting in eqn A,we can get G.
Hence,option C
Assumption:Total number of students are 115 only,not stated explicitly in question though.

Senior | Next Rank: 100 Posts
Posts: 84
Joined: Sat Oct 03, 2009 1:13 pm
Thanked: 11 times

by djkvakin » Fri Feb 26, 2010 9:03 am
I think the answer is B.
The answer would not change in DS question if we manipulate the data to make the analysis easier. Consider the following scenario:
we want to divide $9 dollars among the 3 students.
if we take condition 2 we know that total parts is 1+2 and the ratio is 1/2, hence girls would get 1/3 of the pot (9/3*1)=3 dollars, and therefore there is one girl, and boys would get 2 parts (9/3*2)= 6 dollars and there are 2 boys in the class.

condition 1 results in multiple solutions, and therefore is not enough.

Master | Next Rank: 500 Posts
Posts: 304
Joined: Wed Jan 27, 2010 8:35 am
Location: International Space Station
Thanked: 11 times
Followed by:3 members

by Aman verma » Sat Feb 27, 2010 3:29 am
Ans.C.Firdaus117 has adopted a very good approach.I couldn't have possibly solved that algebraically like Firdaus did.
800. Arjun's-Bird-Eye

Senior | Next Rank: 100 Posts
Posts: 60
Joined: Wed Jan 07, 2009 1:56 pm
Thanked: 12 times
GMAT Score:750

by arzanr » Sat Feb 27, 2010 6:18 am
if we take condition 2 we know that total parts is 1+2 and the ratio is 1/2, hence girls would get 1/3 of the pot (9/3*1)=3 dollars, and therefore there is one girl, and boys would get 2 parts (9/3*2)= 6 dollars and there are 2 boys in the class.
Condition 2 states that each boy gets twice that of each girl. However in your scenario above each girl and each boy are getting an equal amount - $3 each

Senior | Next Rank: 100 Posts
Posts: 84
Joined: Sat Oct 03, 2009 1:13 pm
Thanked: 11 times

by djkvakin » Sat Feb 27, 2010 10:50 am
arzanr wrote:
if we take condition 2 we know that total parts is 1+2 and the ratio is 1/2, hence girls would get 1/3 of the pot (9/3*1)=3 dollars, and therefore there is one girl, and boys would get 2 parts (9/3*2)= 6 dollars and there are 2 boys in the class.
Condition 2 states that each boy gets twice that of each girl. However in your scenario above each girl and each boy are getting an equal amount - $3 each
i realize my mistakes now... thanks.