Adam is 3 times as old as Cynthia and Fred is 16 years younger than Adam. One year ago, Adam's age was twice the sum of Cynthia's and Fred's age. Find their present age.
A=3C
F=A-16
A-1=2(C-1+F-1)
can you take it from here?
To solve this problem, let's assign variables to represent their ages.
Let's say:
Adam's age = A
Cynthia's age = C
Fred's age = F
From the given information, we can form equations:
1) Adam is 3 times as old as Cynthia: A = 3C
2) Fred is 16 years younger than Adam: F = A - 16
3) One year ago, Adam's age was twice the sum of Cynthia's and Fred's age: A - 1 = 2((C - 1) + (F - 1))
Now, let's use these equations to find their present ages.
Substitute equation 1 into equation 2:
F = 3C - 16
Substitute equation 1 and 2 into equation 3:
A - 1 = 2((C - 1) + ((3C - 16) - 1))
Simplify equation 3:
A - 1 = 2(C - 1 + 3C - 17)
A - 1 = 2(4C - 18)
A - 1 = 8C - 36
A = 8C - 35
Now substitute this value of A into equation 1:
8C - 35 = 3C
8C - 3C = 35
5C = 35
C = 7
Now substitute C = 7 back into equation 1 to find A:
A = 3C
A = 3(7)
A = 21
And substitute A = 21 into equation 2 to find F:
F = A - 16
F = 21 - 16
F = 5
Therefore, their present ages are:
Adam is 21 years old.
Cynthia is 7 years old.
Fred is 5 years old.