BTGmoderatorLU wrote:In a rare coin collection, one in six coins is gold and all coins are either gold or silver. If 10 silver coins were to be subsequently traded for an additional 10 gold coins, the ratio of gold coins to silver coins would be 1 to 4. Based on this information, how many gold coins would there be in this collection after the proposed trade?

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

Since all of the values in the problem are multiples of 10, the current number of coins is almost certainly a multiple of 10.
Since 1 of every six coins is gold, G : S = 1:5, implying that the number of silver coins is 5 times the number of gold coins.
Options for G and S:
G=10, S=50
G=20, S=100
G=30, S=150
G=40, S=200
G=50, S=250.

In the list above, exchanging 10 silver coins for 10 gold coins must yield G : S = 1:4, implying that the resulting number of silver coins must be 4 times the resulting number of gold coins.
If in each option we increase the value of G by 10 and decrease the value of S by 10, we get:
new G=20, new S=40
new G=30, new S=90
new G=40, new S=140
new G=50, new S=190
new G=60, new S=240.

Only in the green option is the resulting number of silver coins if 4 times the resulting number of gold coins.
Thus, new G = 60.

The correct answer is B.

Algebra:

One in six coins is gold and all coins are either gold or silver.
Since G : S = 1:5, let G = the original number of gold coins and 5G = the original number of silver coins.

If 10 silver coins were to be subsequently traded for an additional 10 gold coins, the ratio of gold coins to silver coins would be 1 to 4.
Thus, increasing the value of G by 10 and decreasing the value of 5G by 10 must yield a 1:4 ratio:
(G+10)/(5G-10) = 1/4
4G+40 = 5G-10
50 = G.

How many gold coins would there be in this collection after the proposed trade?
The proposed trade increases the value of G by 10:
G+10 = 50+10 = 60.

Hi BTGmoderatorLU,

You can try as follows,

Gold . . . . . . . . . . . . . Silver . . . . . . . . . . . Total
X . . . . . . . . . . . . . . . . . . 5X . . . . . . . . . . . . . . 6X (say "X" are the gold coins)
X+10 . . . . . . . . . . . 5X - 10 . . . . . . . . . . (After trade)

(X+10)/(5X-10) = 1/4

X = 50.

Gold coins after trade = 50 + 10 = 60. Hence, Option B is the correct answer.

Regards!

one in six coins is gold, and all coins are either gold or silver.
if x=number of gold coins
number of silver coins= 1-6(x) =5(x)
After the trade, we have;
number of gold= x+10
number of silver= 5x-10
solve for x;
$$\frac{\left(x+10\right)}{5x-10}=\frac{1}{4}\ \left(i.e\ the\ new\ ratio\ of\ \ gold\ to\ silver=1\ to\ 4\right)$$
$$on\ cross\ multiplication,\ we\ have\ 4x+40=5x-10$$
$$5x-4x=40+10$$
$$x=50$$
$$Note\ that\ after\ the\ trade,\ there\ are\ 10\ more\ gold\ coins=50+10=60\ gold\ coins\ after\ proposed\ trade$$