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The ratio of one-dollar coins to fifty-cent coins in a box was 3:4. 6 one-dollar coins were removed, and some fifty-cent coins of the same amount were added. The ratio of one-dollar coins to fifty-cent coins then became 1:3. What was the total amount of money in the box?

please also show work :)

User TOPKAT
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We can start solving the problem by using algebra. Let x be the original number of one-dollar coins in the box, and y be the original number of fifty-cent coins. Then, based on the given information, we know that:

x:y = 3:4 (original ratio)

x-6:y+z = 1:3 (new ratio)

Where z is the number of fifty-cent coins added.

Now we can use the first equation to find x in terms of y.

x:y = 3:4

x = 3y/4

Now we can substitute x = 3y/4 into the second equation.

(3y/4 - 6) : (y + z) = 1:3

Now we can cross-multiply and simplify the equation to get:

3y/4 - 6 = y + z

3y - 8y = 12 + 4z

-5y = 12 + 4z

y = -12/5 - 4z/5

Now we can use this value of y to find the value of x.

x = 3y/4

x = 3(-12/5 - 4z/5) / 4

x = -9/5 - 3z/5

Now we can use the value of x and y to find the total amount of money in the box.

x + y = (-9/5 - 3z/5) + (-12/5 - 4z/5)

x + y = -21/5 - 7z/5

We know that the number of dollars is equal to the number of one dollar coin, the number of fifty cents is equal to the number of fifty-cent coins, so the total amount of money in the box is:

x + y = -21/5 - 7z/5 = -21/5100 - 7z/550 = -2100-35z

We don't know the value of z so we can't calculate the exact amount of money in the box.

User Jonasespelita
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