The bongo family has three children, including a pair of twins. The third child is 3 years younger then the twins. The sum of the three ages is 42. How old is each child?
a = age of the twins
b = age of the third child
b = a - 3
a + a + b = 42
2 a + b = 42
2 a + a - 3 = 42
3 a - 3 = 42
3 a = 42 + 3
3 a = 45 Divide both sides by 3
a = 45 / 3
a = 15
b = 15 - 3
b = 12
the first of two films lasted 9 minutes less than twice as long as the second. the combined length of the two films plus a 5-minute break between them was 260 minutes. how long was the first film?
x + 2x - 9 + 5 = 260
3x - 4 = 260
3x = 264
x = 88
To solve this problem, let's assign variables to represent the ages of the children.
Let's call the age of one of the twins 'x'. Since the other twin is the same age, their age is also 'x'. The third child is 3 years younger than the twins, so their age will be 'x - 3'.
Given that the sum of the three ages is 42, we can write an equation:
x + x + (x - 3) = 42
Simplifying the equation, we get:
3x - 3 = 42
Adding 3 to both sides:
3x = 45
Dividing both sides by 3:
x = 15
So, one of the twins is 15 years old. And since the other twin is the same age, the other twin is also 15 years old. The third child is 3 years younger than the twins, so their age is:
15 - 3 = 12
Therefore, the ages of the children in the bongo family are: 15, 15, and 12 years old.