The present ages in years of four cousins are consecutive multiples of 3. Five years ago the sum of their ages was 46. What is their ages now?

if their ages are a,b,c,d, then

a+b+c+d = 3n + 3n+3 + 3n+6 + 3n+9 = 12n + 18
(a-5)+(b-5)+(c-5)+(d-5) = 46
a+b+c+d = 66

so, 12n+18 = 66
n=4

so, the ages are 12,15,18,21
5 years ago, they were 7,10,13,16 and added up to 46

To solve this problem, we can break it down into steps:

Step 1: Set up equations for the problem.
Let's assume the ages of the four cousins are x, x+3, x+6, and x+9.

Step 2: Write the equation for the sum of their ages five years ago.
Five years ago, their ages would be (x-5), (x+3)-5, (x+6)-5, and (x+9)-5.
The sum of their ages five years ago is: (x-5) + (x+3-5) + (x+6-5) + (x+9-5) = 46.

Step 3: Solve the equation.
Combine like terms: 4x + 8 = 46.
Subtract 8 from both sides: 4x = 38.
Divide both sides by 4: x = 9.5.

Step 4: Find their ages now.
Using the value of x, we can find the ages of the cousins.
x = 9.5
x+3 = 9.5+3 = 12.5
x+6 = 9.5+6 = 15.5
x+9 = 9.5+9 = 18.5

Therefore, the ages of the four cousins now are approximately: 9.5, 12.5, 15.5, and 18.5 years.