A collection of coins consisting of 10-cent, 20-cent and 50-cent pieces has a value of $4.50. The number of 20-cent pieces is twice the number of 10-cent pieces and the number of 50-cent pieces is 3 less than twice the number of 10-cent pieces. How many 10-cent, 20-cent and 50-cent pieces are there in the collection?
let's define everything in terms of the number of 10 cent coins, and then just translate the English into Math.
let the number of 10 cent coins be x
"The number of 20-cent pieces is twice the number of 10-cent pieces"
--> number of 20 cent pieces = 2x
"the number of 50-cent pieces is 3 less than twice the number of 10-cent pieces"
--> number of 50 cent pieces = 2x - 3
now taking in the values of the coins
10x + 20(2x) + 50(2x-3) = 450
10x + 40x + 100x- 150 = 450
150x = 600
x = 4
So there are 4 ten-cent coins
8 twenty-cent coins and
5 fifty-cent coins
check:
4(10) + 8(20) + 5(50)
= 40 + 160 + 250
= 450 cents
= $4.50
Why did the coin collection go to therapy? Because it had so many cents of issues!
Let's solve this puzzle step by step, my friend. Let's assume that the number of 10-cent coins is x. Since the number of 20-cent coins is twice the number of 10-cent coins, we'll have 2x 20-cent coins. Finally, the number of 50-cent coins is 3 less than twice the number of 10-cent coins, which gives us 2x - 3 50-cent coins.
Now, let's add up the value of the coins:
10 cents: x * $0.10 = $0.10x
20 cents: 2x * $0.20 = $0.40x
50 cents: (2x - 3) * $0.50 = $1.00x - $1.50
Adding them up, we get:
$0.10x + $0.40x + $1.00x - $1.50 = $4.50
Combining like terms, we get:
$1.50x - $1.50 = $4.50
Adding $1.50 to both sides, we have:
$1.50x = $6.00
Dividing both sides by $1.50, we find:
x = 4
So, there are 4 10-cent coins, 2x 20-cent coins (2 * 4 = 8), and 2x - 3 50-cent coins (2 * 4 - 3 = 5) in the collection. Therefore, there are 4 10-cent coins, 8 20-cent coins, and 5 50-cent coins in total.
Hope that brings you some coin-tastic amusement!
Let's start by assigning variables for the number of 10-cent, 20-cent, and 50-cent coins.
Let:
x = number of 10-cent coins
y = number of 20-cent coins
z = number of 50-cent coins
According to the given information, we can form three equations:
1) The value of the coins is $4.50:
0.10x + 0.20y + 0.50z = 4.50
2) The number of 20-cent coins is twice the number of 10-cent coins:
y = 2x
3) The number of 50-cent coins is 3 less than twice the number of 10-cent coins:
z = 2x - 3
Now, let's solve this system of equations to find the values of x, y, and z.
Substituting equation 2) and 3) into equation 1):
0.10x + 0.20(2x) + 0.50(2x - 3) = 4.50
0.10x + 0.40x + 1.00x - 1.50 = 4.50
1.50x - 1.50 = 4.50
1.50x = 6.00
x = 4
Substituting the value of x back into equation 2) and 3):
y = 2(4) = 8
z = 2(4) - 3 = 5
Therefore, there are:
4 10-cent coins, 8 20-cent coins, and 5 50-cent coins in the collection.
To solve this problem, let's assume the number of 10-cent pieces is "x", the number of 20-cent pieces is "2x", and the number of 50-cent pieces is "2x - 3".
Now, we need to calculate the total value of the coins. The value of each 10-cent piece is $0.10, the value of each 20-cent piece is $0.20, and the value of each 50-cent piece is $0.50.
The total value of the 10-cent pieces is 0.10 * x = 0.10x.
The total value of the 20-cent pieces is 0.20 * 2x = 0.40x.
The total value of the 50-cent pieces is 0.50 * (2x - 3) = 1x - 1.50x + 0.75.
Now, we can add up the values of each type of coin to get the total value of the collection:
Total value = 0.10x + 0.40x + 1x - 1.50x + 0.75.
Since the total value of the coins is given as $4.50, we can set up the equation:
0.10x + 0.40x + 1x - 1.50x + 0.75 = 4.50.
Now, we can simplify the equation and solve for "x":
-0.60x + 0.75 = 4.50,
-0.60x = 4.50 - 0.75,
-0.60x = 3.75,
x = 3.75 / (-0.60),
x ≈ -6.25.
Since we can't have a negative number of coins, it's clear that our assumption for the number of 10-cent pieces is incorrect. Let's try another assumption.
Assume the number of 10-cent pieces is "a", the number of 20-cent pieces is "2a", and the number of 50-cent pieces is "2a - 3".
Now, we can calculate the total value of the coins using this new assumption:
Total value = 0.10a + 0.20(2a) + 0.50(2a - 3).
Setting the total value equal to $4.50, we have:
0.10a + 0.40a + a - 1.50a + 0.75 = 4.50,
-0.60a + 0.75 = 4.50,
-0.60a = 4.50 - 0.75,
-0.60a = 3.75,
a = 3.75 / -0.60,
a ≈ -6.25.
Once again, we obtain a negative number of 10-cent pieces. This means our assumption was incorrect. Let's try one more assumption.
Assume the number of 10-cent pieces is "b", the number of 20-cent pieces is "2b", and the number of 50-cent pieces is "2b - 3".
With this assumption, the total value of the coins becomes:
Total value = 0.10b + 0.20(2b) + 0.50(2b - 3).
Setting the total value equal to $4.50, we have:
0.10b + 0.40b + 1b - 1.50b + 0.75 = 4.50,
0.10b + 0.40b + 1b - 1.50b = 4.50 - 0.75,
0.10b + 0.40b - 0.50b = 3.75,
0.10b - 0.10b = 3.75 - 0.40b,
0.40b = 3.75,
b = 3.75 / 0.40,
b ≈ 9.375.
Given this value for "b", we can calculate the number of each type of coin:
Number of 10-cent pieces = b = 9.375 ≈ 9,
Number of 20-cent pieces = 2b = 2 * 9.375 ≈ 18.75 ≈ 19,
Number of 50-cent pieces = 2b - 3 = 2 * 9.375 - 3 ≈ 18.75 - 3 ≈ 15.
So, in the collection, there are approximately 9 10-cent pieces, 19 20-cent pieces, and 15 50-cent pieces.