Age problem

1. The sum of kim’s and kevin’s age is 18. In 3 years, kim will be twice
as old as kevin. What are their ages now?

Sol’n.

Present Future
Kevin X X+3
Kim Y Y+3

x + y = 18
y= 18 – x eq.1
(y+3) = 2(x+3) eq.2

Substitute y in equation 2:

(18-x) + 3 = 2x +6
21 – x =2x +6
X=5

Y= 18-5
Y=13

2. Robert is 15years older than his brother Stan. However “y” years ago, Robert was twice as
old as stan. If stan is now “b” years old b>y, find the value of (b-y)

Sol’n.

Past Present
Robert b+15-y b+15
Stan b-y b
 (b+15 -y ) = 2(b-y)
b+15 – y =2b -2y
2b – b – 2y +y =15
b-y = 15

3. A girl is one- third as old as her brother and 8 years younger than
her sister. The sum of their ages is 38years. How old is the girl?
Sol’n

Let x = age of the girl

y =3x eq.1
z= x+8 eq.2
x + y +z = 38 eq.3

Subs. Eq.1 and 2 in eq.3

x + 3x + (x+8) = 38
x=6

4. A father tell his son, “ I was your age now when you were born.” If
the father is now 38 years old, how old was his son 2years ago?
Sol’n.

Past Present
Father x 38
Son 0 x

38 – x = x -0
x =19
5. If anna was four times as old as belle eight years ago , and if anna
will be twice as old as belle eight years hence, how old is belle
now?
Sol’n.

Let A and B their present ages:

Anna was four times as old as belle 8 years ago

A-8 = 4(B-8) eq.1

Anna will be twice as old as Belle eight years hence

A + 8 = 2(B+8) eq.2

Using Calcu, Mode Equation 2unknown 2 equations.

Therefore
A= 40
B= 16
 6. Noel is 5 years older than Dennis and 10 years younger than Hilda.
In 8 years, their combined ages will be 65. How old is Noel?

7. A boy is one-half as old as his brother and 6 years younger than his
sister. The sum of their ages is 38.
How old is the boy?
8. Twenty-two years ago, Sheryl was twice as old as Pinky and eight
times as old as Roy. That same year, the sum of their ages was 26.
How old is Pinky now?

9. A is 50% older than B and 40% younger than C. If the sum of their
ages is 110, how old is A?

10. The sum of the parent's ages is twice the sum of the children's
ages. Four years ago, the sum of the parent's ages was thrice the
sum of the children's ages. In 16 years, the sum of the ages of the
parents and children will be equal.
How many children are there?

11. The sum of the ages of the parents and the three children is 9
decades over the century. The father is twice as old as the eldest
child. When the eldest child was born, the sum of the parents
ages was 54. When the youngest was born, the sum of the
parents was 70. In 38 years, the sum of the parents ages will be
equal to the sum of the children's ages. How old is the second child?

12. A man lived his life for "x" years. When he was at his midlife,
his first child was born. When he was at his two-thirds of his life, his
second child was born.

When the man died, the sum of the ages of his children is 60. How long did
the man live?
 Solution to 6 -12 Age Problem.

6. Let
x = Noel’s present age
y = Dennis’ present age
z = Hilda’s present age
Noel is 5 years older than Dennis
x=y+5 (1)
Noel is 10 years younger than Hilda
x = z – 10 (2)

In 8 years, their combined ages will be 65.

(x + 8) + (y + 8) + (z + 8) = 65

x + y + z = 41

From (1) and (2), expressing y and z in terms of x then substituting them
into (3), we obtain
x + (x – 5) + (x + 10) = 41
x =12
Noel is now 12 years old

7. Let
x = the boy’s present age
y = the boy’s brother’s present age z = the boy’s sister’s present age.
The boy is one – half as his brother. x = y/2 (1)
The boy is 6 years younger than his sister.
x=z–6 (2)

From (1) and (2), expressing y and z in terms of x then substituting them
into (3),
we obtain
x + 2x + (x + 6) = 38 x = 8
the boy is 8 years old

8. Let
x = Sheryl’s present age y = Pinky’s present age z = Roy’s
present age
Twenty-two years ago, Sheryl was twice as old as Pinky.
x – 22 = 2(y – 22) (1)
In that same year, Sheryl was eight times as old as Roy.
x – 22 = 8(z – 22) (2)
The sum of their ages at that time was 26.
(x – 22) + (y – 22) + (z – 22) = 26 x + y + z = 92 (3)
From (1) and (2), expressing y and z in terms of x then substituting them
into (3),
we obtain
x + [((x-22) / 2 ) + 22 ] + [((x-22) / 8 )+ 22 ] = 92

Solving for x, we get x = 38.

Thus, Pinky is 30 years old

9. Let
x = the present age of A y = the present age of B z = the present
age of C A is 50% older than B. x = 1.5y (1)
A is 40% younger than C. This means A is 60% as old as C.
x = 0.60z (2)
The sum of their ages is 110 x + y + z = 110
From (1) and (2), expressing y and z in terms of x then substituting them
into (3), we obtain
x + (x/1.5) + (x/0.60) = 110 x = 33
A is now 33 years old
10. Let
P = the sum of the parents’ ages C = the sum of the children’s ages n =
the number of children
The sum of the parents’ ages is twice the sum of their children’s ages.
P = 2C (1)
Four years ago, the sum of the parents’ ages was thrice the sum of their
children’s ages.
P – 2(4) = 3 (C – 4n) (2)
In 16 years, the sum of the ages of the parents and their children will be
equal.
P + 2(16) = C + 16n (3)
Substituting P from (1) into (2) then simplying, we obtain
2C – 8 = 3C – 12n
C = 12n – 8 (A)
Substituting P from (1) into (3) then simplying, we obtain
2C + 32 = C + 16n
C = 16n – 32 (B)
Solving (A) and (B) simultaneously, we obtain n = 6.

There are 6 children in the family

11. Let
A = the father’s present age B = the mother’s present age
C = the present age of the first child D = the present age of the second
child
E = the present age of the third child The sum of the ages of the parents
and three children is 9 decades over a century
A + B + C + D + E = 190 (1)
The father is twice as old as the eldest child
A = 2C (2)
When the eldest child was born, the sum of the parents’ ages was 54.
The father was then A – C years old and the mother was B – C years old.
(A – C) + (B – C) = 54
A + B – 2C = 54 (3)
When the youngest child was born, the sum of the parents ages was 70.
The father was the A – E years old and the mother was B – E years
old. (A – E) + (B – E) = 70
A + B – 2E = 70 (4)
In 38 years, the sum of the parents ages will be equal to the sum of the
children’s ages
(A + 38) + (B + 38) = (C + 38) + (D + 38) + (E + 38)
A + B + C + D + E = 38 (5)

The second child is now 24 years old.

12. Let
x = the man’s age when he died y = the first child’s age when the man
died
z = the second child’s age when the man died
When the man was at his midlife, his first child was born. The man at that
time was x – y years old and is equal to half of his life.
x – y = x/2
x = 2y (1)
The man was 2/3 of his life when his second child was born. The man at
that time was x – z years old and is equal to 2/3 of his life.
x – z = (2/3)(x)
x = 3z (2)
When the man died, the sum of his children’s ages is 60.
y + z = 60 (3)
From (1) and (2), expressing y and z in terms of x then substituting them
into (3), we obtain
(x/2) + (x/3) = 60 x = 72
The man lived for 72 years