## Ages Part 4

Problem 1: Ram got married 10 years ago. His present age is $1\frac{1}{3}$ times of the age at the time of marriage. Ram’s sister was 4 years younger to him at the time of his marriage. Find out the present age of Ram’s sister?

Solution:

$1\frac{1}{3}$——->$\frac{4}{3}$

R            M

4             3

1——–>10

4———>?4*10=40

3———>?3*10=30

40-4=36 is  the present age of Ram’s sister

Problem 2: If the two digits of the age of Arun are reversed then the new age so obtained is the age of his wife. 1/11 part of the sum of their ages is equal to the difference between their ages. If Arun is older than his wife then find the difference between their ages?

Solution:

Assume age of Arun =10x+y

Age of his wife = 10y+x

$\frac{1}{11}(11x+11y)=9x-9y$

$10y=8x$

$\frac{x}{y}=\frac{10}{8}=\frac{5}{4}$

A=54     w = 45

Difference between the ages is 9

Problem 3: Arun’s brother is 3 years elder to him. his father was 28 years of age when his sister was born while his mother was 26 years of age when he was born, The ages of Arun’s father when his brother was born?

Solution:

Father age =28——->28+4=32

Mother age = 26——>26+4=30

Sister age=0——–>0+4 = 4

brother age = 0

The age of Arun’s father when the brother born is 32

Problem 4: If 6 years are subtracted from the present age of Randheer and the remainder is divided by 18 then the present age of his grandson Anup is obtained. If the Anup is 2 years younger to Mahesh whose age is 5 years, then whats the age of Randheer?

Solution:

Age of mahesh is= 5

Age of Anup is = 5-3 =2

$\frac{R-6}{18}=3$

R= 54+6 =60

## Ages Part 3

Problem 1: 16 years ago, my uncle was 8 times older than me. After 8 years from today, my uncle will be thrice as old as I will be at that time. Eight years ago, What is the ratio of my age and my uncle’s age?

Solution:

$U^{-16}:M^{-16}=9:1$——->difference is 8 (9-1=8)

$U^{+8}:M^{+8}=3:1$———->difference is (3-1=2)

$U^{-16}:M^{-16}=9:1$ * 1———–>$9:1$

$U^{+8}:M^{+8}=3:1$  *  4————>$12:4$

12-9 = 3          4-1 = 3

8-(-16)=24

3———>24

1———->?$\frac{24}{3}=8$

Age after 8 years

12———>?12*8=96

4———->?4*8=32

Age 8 years ago 8+8=16

96-16=80

32-16=16

M      :       U

16     :        80

1        :         5

Problem 2: If Sita’s mother was 4 times as old as sita 10 years ago. After 10 years the mother will be twice as old as sita. How old is Sita’ s mother at present?

Solution:

$M^{-10}:S^{-10}=4:1$——–>difference is 3 (4-1 = 3)

$M^{+10}:S^{+10}=2:1$——–>difference is 1 (2-1=1)

$M^{-10}:S^{-10}=4:1$ * 1——–>$4:1$

$M^{+10}:S^{+10}=2:1$ *  3——–>$6:3$

6-4=2  ,   3-1=2

10-(-10) = 20

2———->20

1———–>$\frac{20}{2}=10$

Age after 10 years

6———–>?6*10 = 60

3————>?3*10 = 30

Sita’s mother present age = 60-10 = 50

Problem 3: Jayesh is as more younger to Amit as Jayash is older to prashant. If the sum of the ages of Amit and Prashant is 48 years. What is the age of Amit in years?

Solution:

A-J=X

J-P=X

A-J = J-P

A+P = 2J

48 = 2J

J = 24

We can derive only jayaesh age based on the given data, We can not determine the Amith and prasanth age.

## Ages Part 2

Problem 1: A father said to his son, “I was as old as you are at the present at the time of your birth”. If the father age is 38 years now, What was the son’s age five years back?

Solution:

Present age of son=x

Age of son at the time of birth=0

age of father = x

Present age of son=x+0=x

Present age of father =x+x=2x

2x——–>38

x———->?$\frac{38}{2}=19$

Age of son five years ago = 19-5=14

Problem 2: A person present age is two-ninth of the age of his mother. After 10 years, he will be four-eleven of the age of his mother. How old will be her mother after 15 years?

Solution:

P   :   M = 2   :    9———->difference is 7 (9-2 = 7)

$P^{+10}:M^{+10}$=11  :  7——->difference is 7 (11-7 = 7)

4-2=2     ,       7-9=2

2———->10

1———->?$\frac{10}{2}=5$

2———–>?2*5=10

9———–>?9*5=45

Mothers age after 15 years is 45+15 = 60

Problem 3: Ratio of the age of A and B is 5:x. A is 18 years younger to C. After nine years C will be 47 years old. If the difference between the ages of A and B is the same as the age of C, What is the value of X?

Solution:

A   :   B

5x   :   x

C-A=18

$C^{+9}=47\rightarrow&space;C^{0}=47-9=38$

C – A = 18

A = 38-18=20

5X———->20

X————>?$\frac{20}{5}=4$X

$4X-20=38$

$4X=58$

$X=14.5$

## Ages Part 1

Problem 1: The present ratio of the age of A and B is 3:5. If the sum of the present age of A and B is 48 years. Find the age of B before 5 years?

Solution:

A   :   B

3    :     5

8——->48

1———>?$\frac{48}{8}=6$

5——–>6*5=30

Age of B before 5 years = 30-5=25

Problem 2: A man’s age is 125% of what it was 10 years ago, but 83$\frac{1}{3}$% of what it will be after 10 years. What is his present age?

Solution:

125%——–>$\frac{5}{4}$

4 is 10 years ago age and 5 is present age

5-4 = 1

1——–>10

5———->?5*10=50

4———–>?4*10=10

$83\frac{1}{3}$%=$\frac{5}{6}$

5 is the present age and 6 is after 10 years

6-5=1

1——–>10

5———->?5*10=50

6———–>?6*10=60

hence the present age is 50 years

Problem 3: Father is aged three times more than his son. After 8 years, he would be two and half times of son’s age. After 16 years, how many times would he be the son’s age?

Solution:

F = 3*S

$\frac{F}{S}=\frac{3}{1}$

$F:S=3:1$   (3-1=2) *3

$F^{+8}=\frac{5}{2}\times&space;S^{+8}$

$F^{+8}:S^{+8}=5:2$   (5-2=3) *2

F   :   S

9    :   3

10  :    4

10-9=1

1——>8

9——>?9*8=72

3——–>?3*8=24

further 16 years

72+16=88

24+16=40

$\frac{88}{40}=\frac{11}{5}=2.2&space;times$

Fathers age is 2.2 times then the son’s age.