Problems involving simultaneous equation.
At a circus, there were twice as many children as there were adults in attendance. Altogether, 1020 attended the circus. How many were children?
c = 2a
c+a=1200
Oh, going to the circus, huh? That sounds like fun! Alright, let's solve this mathemagical problem together.
Let's assume the number of adults in attendance is 'x'. Since there were twice as many children as adults, that means the number of children must be '2x'. And together, they add up to 1020.
So, we can set up a simultaneous equation:
x + 2x = 1020
Now let's simplify that equation:
3x = 1020
To solve for x, we divide each side by 3:
x = 1020 / 3
x = 340
So, there were 340 adults at the circus.
Now that we know the number of adults, we can find the number of children by multiplying it by 2:
2x = 2 * 340
2x = 680
Ta-da! There were 680 children at the circus. Bring on the acrobats and clowns!
To solve this problem, we can use a system of simultaneous equations.
Let's represent the number of adults as x and the number of children as y.
From the given information, we know that:
1) y = 2x (there were twice as many children as adults)
2) x + y = 1020 (altogether, 1020 attended the circus)
Now we can solve this system of equations:
Substitute equation 1 into equation 2:
x + 2x = 1020
3x = 1020
x = 1020 / 3
x = 340
Substitute x = 340 back into equation 1:
y = 2(340)
y = 680
Therefore, there were 680 children at the circus.
To solve this problem involving simultaneous equations, let's start by assigning variables to the unknowns.
Let:
- C represent the number of children in attendance.
- A represent the number of adults in attendance.
From the problem, we know that there were twice as many children as there were adults. We can express this as an equation:
C = 2A [Equation 1]
Additionally, we know that the total number of people who attended the circus is 1020:
C + A = 1020 [Equation 2]
Now we have a system of two simultaneous equations with two unknowns: C and A. We can solve this system using the substitution method or the elimination method. Let's use the elimination method.
First, we'll multiply Equation 1 by 2 to eliminate the coefficients of C:
2C = 4A [Equation 3]
Now we can subtract Equation 3 from Equation 2 to eliminate C:
(C + A) - (2C) = 1020 - 4A
Simplifying the equation:
-A - 3A = 1020 - 4A
-4A = 1020
Dividing both sides by -4:
A = -1020 / -4
A = 255
Now that we know the value of A, we can substitute it back into one of the original equations to find the value of C. Let's use Equation 1:
C = 2A
C = 2(255)
C = 510
So, there were 510 children in attendance at the circus.