Mark has 28 10-cent and 20-cent pieces. How many of each coin does he have if their total value is $4.30?
In a class of 26 students, there are four more than girls. How many boys are there?
x ten cent pieces
(28-x) twenty cent pieces
10 x + 20(28-x) = 430
10 x + 560 - 20 x = 430
10 x = 130
x = 13
28-x = 15
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x= boys
x-4 = girls
2x-4 = 26
x = 15
x-4 = 11
Oh boy, let's solve these conundrums with a touch of humor!
For the first riddle, let's call the number of 10-cent coins "x" and the number of 20-cent coins "y." We know that Mark has a total of 28 coins, so we can express this as the equation: x + y = 28. We also know that the total value of these coins is $4.30, which we can express as another equation: 10x + 20y = 430 (since each 10-cent coin is worth 10 cents and each 20-cent coin is worth 20 cents). Now, let's put our juggling skills to the test and solve these equations!
Multiplying the first equation by 10 on both sides, we get: 10x + 10y = 280. Subtracting this from the second equation, we have: 10x + 20y - (10x + 10y) = 430 - 280. Simplifying the equation gives us: 10y = 150. Dividing both sides by 10, we find that y = 15. So, Mark has 15 shiny 20-cent coins.
Now let's substitute this value into the first equation, x + 15 = 28. Solving for x, we find that x = 13. Therefore, Mark has 13 delightful 10-cent coins.
Whew, now onto the second riddle! We're in a class of 26 students, and we know that there are four more girls than boys. If we let the number of boys be "x," then the number of girls would be "x + 4" since there are four more girls. Adding the number of boys and girls together gives us: x + (x + 4) = 26. Let's flex our math muscles and find out how many boys there are!
Simplifying the equation, we have: 2x + 4 = 26. Subtracting 4 from both sides, we get: 2x = 22. Lastly, dividing both sides by 2, we find that x = 11. So, there are 11 fantastic boys in the class.
I hope my clownish antics brightened up these math problems for you! If you have any more brain-benders or jokes in need of answering, hit me up!
Let's solve these two problems step by step.
Problem 1: Mark's Coins
Let's assume Mark has x 10-cent pieces and y 20-cent pieces. According to the problem, Mark has a total of 28 coins. Therefore, we can form the equation:
x + y = 28 -----------(1)
The total value of these coins is given as $4.30. We can calculate the total value in cents and form another equation:
10x + 20y = 430 -------------(2) [converting dollars to cents]
Now, we have a system of two equations. By solving this system, we can determine the values of x and y.
To solve the system of equations, we can use the substitution method. From equation (1), we can rewrite it as:
x = 28 - y
Substituting this value of x into equation (2), we get:
10(28-y) + 20y = 430
280 - 10y + 20y = 430
10y = 430 - 280
10y = 150
y = 15
Substituting the value of y into equation (1), we get:
x + 15 = 28
x = 28 - 15
x = 13
Therefore, Mark has 13 10-cent pieces and 15 20-cent pieces.
Problem 2: Boys and Girls in the Class
Let's assume the number of boys in the class is x. According to the problem, the number of girls is four less than the total number of students. So, we have:
Number of girls = (Total number of students) - 4
=> Number of girls = 26 - 4
=> Number of girls = 22
Now, we know that there are four more girls than boys. Hence, we can form the equation:
Number of girls = Number of boys + 4
Substituting the value of the number of girls, we get:
22 = x + 4
x = 22 - 4
x = 18
Therefore, there are 18 boys in the class.
To solve the first problem, let's assign variables to the unknown quantities. We can use "x" to represent the number of 10-cent pieces and "y" for the number of 20-cent pieces.
We know that Mark has a total of 28 coins, so we can write the equation:
x + y = 28 (Equation 1)
Next, we know that the total value of the coins is $4.30. In cents, this is equal to 430 cents. Since each 10-cent piece is worth 10 cents and each 20-cent piece is worth 20 cents, we can write another equation to represent the value of the coins:
10x + 20y = 430 (Equation 2)
Now that we have our two equations, we can solve them simultaneously to find the values of x and y.
First, let's solve Equation 1 for x:
x = 28 - y
Now substitute this value of x into Equation 2:
10(28 - y) + 20y = 430
280 - 10y + 20y = 430
10y = 430 - 280
10y = 150
y = 15
Now that we have the value of y, we can substitute it back into Equation 1 to find x:
x + 15 = 28
x = 28 - 15
x = 13
Therefore, Mark has 13 10-cent pieces and 15 20-cent pieces.
To solve the second problem, let's assume that the number of boys is represented by "x" and the number of girls is represented by "y".
We know that the total number of students in the class is 26, so we can write the equation:
x + y = 26 (Equation 1)
We're also told that there are four more boys than girls, which can be expressed as:
x = y + 4 (Equation 2)
Now we can solve these equations simultaneously to find the value of x.
First, let's substitute the value of x from Equation 2 into Equation 1:
(y + 4) + y = 26
2y + 4 = 26
2y = 26 - 4
2y = 22
y = 11
Now substitute the value of y back into Equation 2 to find x:
x = 11 + 4
x = 15
Therefore, there are 15 boys in the class.