At a school concert, the tickets cost \( \$ 20\) per adult and \(\$ 2\) per child. The total paid by the \(100\) people who attended was \(\$ 920\). How many were children?
(A) between 25 and 35 (B) between 35 and 45 (C) between 45 and 55 (D) between 55 and 65 (E) between 65 and 75
Let's solve this problem by considering x as the number of adults and y as the number of children.
Thus, \(x + y = 100\)
From the first statement, we can conclude,
The total number of adults = \(20x\)
The total number of children = \(2y\).
Thus, \(20x + 2y = 920\).
As we know, \(x + y = 100\),
So, \(y = 100 - x\) ----------------------(1)
\(20x + 2y = 920\).
\(\therefore\) \(20x + 2(100 -x)\) [applying (1)]
\(\therefore\)\(20x + 200 - 2x = 920\)
\(\therefore\)\(18x = 920 -200 = 720\)
\(\therefore\)\(x = 720\div 18 = 40\)
Thus, \(x = 40\).
Applying the value of \(x\) in \(eq^n (1)\) we get,
\(y = 100 - x = 100 - 40 = 60\)
So the number of children is \(60\).
The answer is D) between 55 and 65.
The Australian Mathematics Competition (AMC) is one of the largest and oldest annual mathematics competitions in Australia, aimed at fostering interest and excellence in mathematics among students.