Let's say that 10 years ago Ben was x years old, and Mike was 7x years old.
Now Ben would be x + 10 years old, and Mike 2(x + 10) or 7x + 10 years old.
So we have (Mike age):
2(x + 10) = 7x + 10
2x + 20 = 7x + 10 (subtract 2x and 10 to both sides)
2x - 2x -10 + 20 = 7x - 2x - 10 + 10
10 = 5x (divide by 5 to both sides)
2 = x (Ben's age 10 years ago)
So that Ben is 12 years old now, and Mike is 24 years old (as you see Mike was a very young father, 14 years old).
Mike is 12. Mary is 4. In 4 years, Mike will be 16 and Mary will be 8. So Mike will be twice as old as she is in four years.
9 years
9
Mikey, Mike & Ike
Mike is 30; Bill is 2. m = 15b m - 28 = b 15b - 28 = b 14b = 28 b = 2 m = 15(2) = 30
Mike is 12. Mary is 4. In 4 years, Mike will be 16 and Mary will be 8. So Mike will be twice as old as she is in four years.
9 years
9
12,4...16,8
9
Twice
Mary is 4 years old. [A+]
Leo must be 18. If Leo is twice as old as martin and three times older than mike and the sum of the ages is 33, then martin would be 9 and mike would be 6, making Leo 18.
Yes but only twice.
Write and solve two simultaneous equations. a = Mary's age b = Mike's age Age now: b = 3a Age in 4 years: b + 4 = 2(a + 4)
Twice Once in 1989 and 2nd time in 1996
9