A movie theater charges $7 for adults, $5 for school-age children, and $3 for babies. A group of people went to the theater. There were the same number of school-age children as babies and the number of adults was the same as school-age children and babies combined. If the group paid $1562, how many babies were in the group ?

Question

A movie theater charges $7 for adults, $5 for school-age children, and $3 for babies. A group of people went to the theater. There were the same number of school-age children as babies and the number of adults was the same as school-age children and babies combined. If the group paid $1562, how many babies were in the group ?

Answer 1

Please type your subject in the School Subject box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.

Answer 2

same number of school-age children as babies

x=number of school-age children
x=number of babies

number of adults was the same as school-age children and babies combined

y=number of adults
y=x+x
y=2x

$7(2x) + $5(x)+ $3(x) = $1562

similar problem - look at example 1:
http://www.algebra-class.com/algebra-word-problems.html

Answer 3

Let's assume the number of school-age children and babies is 'x'.

Since the number of adults is the same as the school-age children and babies combined, the number of adults is 2x.
The total cost can be calculated as follows:
Cost of adults = 7 * (2x) = 14x
Cost of school-age children = 5 * x = 5x
Cost of babies = 3 * x = 3x
Total cost = 14x + 5x + 3x = 22x
We know that the total cost is $1562, so:
22x = 1562
Dividing both sides of the equation by 22:
x = 1562/22
x = 71
Therefore, there were 71 babies in the group.

Answer 4

To solve this problem, let's break it down step by step.

Let's assume the number of babies in the group is 'b'. Since there were the same number of school-age children as babies, the number of school-age children would also be 'b'.

Now, let's determine the number of adults in the group. The problem states that the number of adults is the same as the school-age children and babies combined. So, the number of adults would be 'b + b = 2b'.

Now, we can calculate the total cost paid by the group. The cost per adult is $7, so the total cost for adults would be '7 * 2b'. The cost per school-age child is $5, so the total cost for school-age children would be '5 * b'. The cost per baby is $3, so the total cost for babies would be '3 * b'.

According to the problem, the total cost paid by the group is $1562. So we can create the equation:
'7 * 2b + 5 * b + 3 * b = 1562'.

Now we can solve this equation to find the value of 'b'.

7 * 2b + 5 * b + 3 * b = 1562
14b + 8b + 3b = 1562
25b = 1562
b = 1562 / 25
b ≈ 62.48

Since we can't have a fraction of a baby, we can round the value of 'b' to the nearest whole number. Therefore, there were approximately 62 babies in the group.