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+]
Yes but only twice.
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.
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