A group of 10 people plan to contribute equally to pay fora friend's gift that costs G dollars. If nadditional people want to contribute to pay for the gift, the required contribution per person will be reduced by how many dollars?
(A)Gn/100+10n
(B)10+n/Gn
(C)10G+Gn/n
(D)Gn+10G/GN-10
(E)10G/n^2+10n :roll:
answer after discussion...[/img]
help me please.!!!
This topic has expert replies
-
- Legendary Member
- Posts: 966
- Joined: Sat Jan 02, 2010 8:06 am
- Thanked: 230 times
- Followed by:21 members
Just substitute easy numbers.
Let the gift cost $100, so initially everyone pays $10 each.
Let n = 10 (10 additional people contribute), the cost per head now will be 100/20 = $5.
Sub G = 100 and n = 10 in the answer choices, you should get 5 as answer. Only A satisfies IMO.
Infact you could even use smaller numbers.
Let the gift cost $100, so initially everyone pays $10 each.
Let n = 10 (10 additional people contribute), the cost per head now will be 100/20 = $5.
Sub G = 100 and n = 10 in the answer choices, you should get 5 as answer. Only A satisfies IMO.
Infact you could even use smaller numbers.
- smackmartine
- Legendary Member
- Posts: 516
- Joined: Fri Jul 31, 2009 3:22 pm
- Thanked: 112 times
- Followed by:13 members
IMO A
Cost of the gift = G
Total people initially = 10
Cost per head = G/10
Extra people who want to contribute = n
So, total # of people = 10+n
Cost per head now= G/(10+n)
Notice that G/10 > G/(10+n) [now each person has to contribute less]
So, contribution per person will be reduced = [G/10]- [G/(10+n)] = Gn/10(10+n)=Gn/100+10n
Cost of the gift = G
Total people initially = 10
Cost per head = G/10
Extra people who want to contribute = n
So, total # of people = 10+n
Cost per head now= G/(10+n)
Notice that G/10 > G/(10+n) [now each person has to contribute less]
So, contribution per person will be reduced = [G/10]- [G/(10+n)] = Gn/10(10+n)=Gn/100+10n
Smack is Back ...
It takes time and effort to explain, so if my comment helped you please press Thanks button
It takes time and effort to explain, so if my comment helped you please press Thanks button
-
- Senior | Next Rank: 100 Posts
- Posts: 92
- Joined: Thu Oct 06, 2011 8:06 am
- Thanked: 18 times