1. Eighteen years ago, a mother was three times as old as her daughter. Now the mother is only twice as old her daughter. What’s the total of the sum of their present ages?
Solution:
Let the daughter’s present age is x.
Let the mother’s present age is 2x.
Eighteen years ago, the mother was three times as old as her daughter.
Therefore,
(2x-18) =
3(x-18)
2x-18 = 3x-54
x = 36
The
present age of the mother is 2x = 2(36) = 72.
The
present age of the daughter is x = 36.
The
sum of their present ages = 72+36 = 108.
2. Neil’s present age is
two-fifth of the age of his father. After 8 years, he’ll be one-half of the age
of his father. How old is the father at present?
Solution:
Let the father’s present age is
x.
As
per the given, Neil’s present age will be (2/5)x.
After 8 years,
(2/5)x + 8 = 1/2 (x + 8)
2(2x + 40) = 5(x+8)
4x+80 =
5x+40
x = 40
The
present age of his father is 40 years.
3. The age of father
10 years ago was thrice the age of his daughter. Ten years hence, father’s age
will be twice of his daughter. What’s the ratio of their present ages?
Solution:
Let the daughter’s age(10 years ago) be x.
The father’s age(10 years ago) will be 3x.
10
years hence,
3x+10+10 = 2(x+10+10)
3x+20 =
2x+20+20
x
= 20
The ratio
of their present ages is (x+10):(3x+10)
30:70
3:7.
Comments
Post a Comment