The ratio of the number of one dollar coins in a box was 3:4.six one dollar coins were removed & some fifty cent coins of the same amount were added.the ratio of the number of one dollar coins to the number of fifty cent coins then became 1:3.what was the total amount of money in the box?
if there were x 50-cent coins and y dollar coins, then if 6 dollars were replaced with 12 halves,
x/y = 3/4
(x-6)/(y+12) = 1/3
clearing the fractions, we have
4x = 3y
3(x-6) = y+12
4x-3y = 0
3x-y = 30
now we see that y=3x-30, so
4x-3(3x-30) = 0
4x-9x+90 = 0
5x = 90
x = 18
So, there were 18 dollars and 24 halves.
After the replacements, there were 12 dollars and 36 halves.
I expect you can figure out how much money that was.
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In order to find the answer, we need to set up a system of equations based on the given information.
Let's assume that the initial number of one dollar coins in the box is 3x, where x is the common ratio between the number of one dollar coins and fifty cent coins. Therefore, the initial number of fifty cent coins in the box is 4x.
We are told that six one dollar coins were removed from the box. This means that the updated number of one dollar coins is 3x - 6.
We are also told that an equal amount of fifty cent coins were added. Therefore, the updated number of fifty cent coins is 4x + 6.
Now, we need to set up the second piece of information. The ratio of the updated number of one dollar coins (3x - 6) to the number of fifty cent coins (4x + 6) is 1:3. Therefore, we can write the equation:
(3x - 6)/(4x + 6) = 1/3
To solve this equation, we can cross-multiply:
3(3x - 6) = (4x + 6)
Simplifying, we get:
9x - 18 = 4x + 6
Combining like terms, we have:
9x - 4x = 6 + 18
5x = 24
Dividing both sides by 5, we find:
x = 4
Since x represents the common ratio between the number of one dollar coins and fifty cent coins, we can substitute x=4 into the original expressions to find the initial number of coins:
Initial number of one dollar coins = 3x = 3(4) = 12
Initial number of fifty cent coins = 4x = 4(4) = 16
To find the total amount of money in the box, we need to multiply the number of coins by their respective values. One dollar coins have a value of $1 and fifty cent coins have a value of $0.50:
Total amount of money = (12 * $1) + (16 * $0.50) = $12 + $8 = $20
Therefore, the total amount of money in the box is $20.
Oh, this sounds like a puzzling situation! Let's put on our thinking caps and figure it out together.
Initially, we have a ratio of 3:4 for one dollar coins in the box, which means there were 3x one dollar coins and 4x fifty cent coins. When six one dollar coins are removed, we're left with 3x - 6 one dollar coins.
Now, some fifty cent coins were added, so let's say there were y fifty cent coins added. The ratio of one dollar coins to fifty cent coins becomes (3x - 6):(4x + y) = 1:3.
From the ratio equation, we can establish that (3x - 6) / (4x + y) = 1 / 3. To solve for x and y, we need another equation.
Since we're dealing with coins of the same value, the ratio of the amounts must be equal to the ratio of the coins themselves. One dollar coins are worth double the value of fifty cent coins, so the ratio of their amounts should also be 2:1.
This gives us (3x - 6) / y = 2 / 1. Now we have a system of equations to work with.
To make things simpler, let's solve for x in the second equation: (3x - 6) = 2y.
Now we can substitute this value of (3x - 6) into the first equation: (3x - 6) / (4x + y) = 1 / 3.
After some algebraic fun, we find that x = 12.
Plugging x = 12 back into (3x - 6) = 2y, we find that y = 18.
Now, let's calculate the total amount of money in the box. We already established that initially there were 3x one dollar coins (3 * 12 = 36) and 4x fifty cent coins (4 * 12 = 48), which sums up to $36 + $0.50 * 48 = $60.
So, the total amount of money in the box is $60.
Phew, we made it through! I hope my clown-like explanations lightened the journey.