Bobby has only pennies, dimes, and quarters. One half of Bobby's coins are pennies. One fourth of coins are quarters. Three of his coins are dimes. How many coins does Bobby have?
To find the number of coins Bobby has, we can use algebra and solve the system of equations that represents the given information.
Let's say the total number of coins Bobby has is "x".
According to the problem, one-half of Bobby's coins are pennies. So, the number of pennies Bobby has is (1/2)x.
Furthermore, one-fourth of the coins are quarters. So, the number of quarters Bobby has is (1/4)x.
We can also determine the number of dimes Bobby has. The problem states that Bobby has three dimes.
Now we can form an equation to represent the total number of coins:
Number of coins = Number of pennies + Number of quarters + Number of dimes
x = (1/2)x + (1/4)x + 3
To simplify, we can apply common denominators:
x = (2/4)x + (1/4)x + 3
Combining like terms:
x = (3/4)x + 3
To eliminate the fraction, we can multiply both sides of the equation by 4:
4x = 3x + 12
Subtract 3x from both sides:
x = 12
Therefore, Bobby has 12 coins.
Let x = total number of coins.
1/2x + 1/4x + 3 = x
3 = x - 3/5x
3 = 1/4x
3 / (1/4) = x
3 * 4 = x
12 = x
let the number of coins he has be x
x/2 + x/4 + 3 = x
times 8
4x + 2x + 24 = 8x
2x = 24
x = 12
12/2 = 6 pennies
12/4 = 3 quarters
+ 3 dimes
= 12 coins
(A better question would have been:
How much money does Bobby have with his coins? )