Question

The sum of the ages of Berma, her mother Rinna and her father Erwin is 80. Two years from now, Rinna’s age will be 13 less than the sum of Erwin’s age and twice Berma’s age. Three years ago, 15 times Berma’s age was 5 less than the age of Rinna. How old are they now?

Answer 1

Answer: Berma is 5 years old

Rinna is 38 years old

Erwin is 37 years old

Step-by-step explanation:

Let x represent Berma's age

Let y represent Rinna's age

Let z represent Erwin's age

Since the sum of their ages is 80,

x + y + z = 80 - - - - - - -1

Two years from now, Rinna’s age will be 13 less than the sum of Erwin’s age and twice Berma’s age. This means that

y +2 = [ (z+2) + 2(x+2) ] - 13

y +2 = z + 2 + 2x + 4 - 13

2x - y + z = 13 + 2 - 4 -2

2x - y + z = 9 - - - - - - -2

Three years ago, 15 times Berma’s age was 5 less than the age of Rinna. It means that

15(x - 3) = (y - 3) - 5

15x - 45 = y - 3 - 5

15x - y = - 8 + 45

15x - y = 37 - - - - - - - -3

From equation 3, y = 15x - 37

Substituting y = 15x - 37 into equation 1 and equation 2, it becomes

x + 15x - 37 + z = 80

16x + z = 80 + 37 = 117 - - - - - - 4

2x - 15x + 37 + z = 9

-13x + 2 = -28 - - - - - - - - -5

subtracting equation 5 from equation 4,

29x = 145

x = 145/29 = 5

y = 15x - 37

y = 15×5 -37

y = 38

Substituting x= 5 and y = 38 into equation 1, it becomes

5 + 38 + z = 80

z = 80 - 43

z = 37