Rare Coins problem

fk352 · Posted: Wed Oct 15, 2008 1:56 pm Post subject: Rare Coins problem

Hi All, I took one look at the following problem and was completely stumped. I looked like a deer in headlights. Anyone want to give it a shot? I'll post the OA later:

Rare Coins

In a rare coin collection, one in six coins is gold. If 10 non-gold coins are subsequently traded for 10 gold coins, the ratio of gold coins to non-gold coins will be 1 to 4. Based on this information, how many gold coins are there now in this collection (after the trade)?

A) 50
B) 60
C) 180
D) 200
E) 300

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dmateer25 · Posted: Wed Oct 15, 2008 2:27 pm Post subject:

Ratio is

1 Gold : 5 Non Gold

Let X = Number of Gold Coins Originally
5X = Number of Non-Gold Originally

That means that:

(x+10)/(5x-10) = 1 Gold/ 4 Non Gold

4X + 40 = 5x -10
50 = X

Originally 50 gold coins.

So now there are 60 gold coins and 240 non gold coins (because of the ratio of 4 non gold for every 1 gold)

I go with B

vishubn · Posted: Wed Oct 15, 2008 5:48 pm Post subject:

Indeed.
i approached the problem in the same way as dmateer25 did.
60 is my call as well

Vishu

oa ???

stop@800 · Posted: Wed Oct 15, 2008 8:23 pm Post subject:

B for me too..

OA?

rohangupta83 · Posted: Thu Oct 16, 2008 12:29 am Post subject:

Total # of coins = 5X
gold coins = X

After trade

Total # of coins = 5X - 10
Gold coins = X + 10

New ratio
Coins: Gold = 1:4

Thus, 5X - 10 = 4X + 40
X = 50 (originally)

New # of gold coins = X + 10 or 50 + 10 = 60

msvmuthu · Posted: Thu Oct 16, 2008 12:54 am Post subject:

let no of gold coins originally be x
and no of non gold coins originally be n

=> one in six coins is gold
no of gold coin originally = 1
no of non gold coins originally = total coins - gold coins = 6-1 =5
there fore forming an equation:
x/n = 1/5
n= 5x
5x-n =0 -> eq1

=>If 10 non-gold coins are subsequently traded for 10 gold coins
current no of gold coins = x+10
current no of non gold coins = n-10

=>the ratio of gold coins to non-gold coins will be 1 to 4
(x+10) / (n-10) = 1/4

4(x+10) = n-10
4x+40 = n-10
4x-n=-50
-4x+n=50 ->eq2

solving eq1 and eq2 will give
x=50

But please remember that x is no of gold coins before the trade.
After the trade 10 gold coins are added to the collection.

There fore no of gold coins in the current collection is 50 + 10 = 60

fk352 · Posted: Thu Oct 16, 2008 1:51 pm Post subject:

haha, sheesh. maybe the problem wasn't as hard as I thought.

Yes, OA is B, 60 gold coins.

Thanks everyone.

vishubn · Posted: Thu Oct 16, 2008 6:30 pm Post subject: