Problems based on ages are generally asked in most of the company examinations, To solve these problems, the knowledge of linear equations is essential. In such problems, there may be three situations:
i. Age some years ago
ii. Present age
iii. Age some years hence
Two of these situations are given and it is required to find the third. The relation between the age of two persons may also be given. Simple linear equations are framed and their solutions are obtained. Sometimes, shortcut methods given below are also helpful in solving such problems.
SOME USEFUL SHORT-CUT METHODS
1. If the age of A, t years ago, was n times the age of B and at present A‘s age is n2 times that of B, then
A‘s present age = (n1-1/n1-n2) n2 t years
and, B‘s present age =(n1-1/n1-n2)t years
Explanation
Let the present age of B be x years.
Then, the present age of A =n2 x years Given, t years ago,
n1(x-t)=n2x-t or, (n1-n2)x = (n1-1) t or, x=(n1-1/n1-n2) t years.
Therefore, B‘s present age = (n1-1/n1-n2) t years.
And, A‘s present age =(n1-1/n1-n2)n2 t years
Illustration 1 The age of father is 4 times the age of his son. If 5 years age father‘s age was 7 times the age of his son at that time, what is father‘s present age?
Solution The father‘s present age = (n1-1/n1-n2)n2t [Here, n1 =7, n2=4 and t=5]
=(7-1/7-4) 4×5 = 6×4×5/3 = 40years.
2. The present age of A is n1 times the present age of B. If t years hence, the age of A would be n2 time that of
B, then
A‘s present age = (n1-1/n1-n2)n2 t years and B‘s present age = (n1-1/n1-n2)t years
Explanation
Let the present age of B be x years. Then, the present age of A= n1x Given, t years hence, (n1x +t)=n2 (x+t)
or (n1-n2)x = (n2-1)t or, x=(n2-1/n1-n2)t
Therefore, B‘s present age =(n2-1/n1-n2)n1t years
and, A‘s present age =(n2-1/n1-n2)n1t years
Illustration 2 The age of Mr. Gupta is four times the age of his son. After ten years, the age of Mr. Gupta will be only twice the age of his son. Find the present age of Mr. Gupta‘s son.
Solution The present age of Mr. Gupta‘s son = (n2-1/n1-n2)t
=(2-1/4-2)10
[Here, n1 = 4, n2 = 2 and t=10]= 5years.
3. The age of A, t1 years ago, was n1 times the age of B. If t2 years hence A‘s age would be n2 times that of B,
then,
A‘s present age =n1(t1+t2)(n2-1)/n1-n2 +t1 years
and, B‘s present age =t2(n2-1)t1(n1-1)/n1-n2 years.
Explanation
Let A‘s present age=x years and B‘s present age=y years. Given:x-t1 = n1(y-t1) and x+t2 =n2(y+t2)
i.e. x-n1y=(1-n1)t1 ....(1)
and x-n2y=(-1+n1)t2...(2)
Solving (1) and (2), we get x=n1(t1+t2)(n2-1)/n1-n2 +t1
Illustration 3 10 years ago Anu‘s mother was 4 times older than her daughter. After 10 years, the mother will be twice older than the daughter. Find the present age of Anu.
Solution Present age of Anu = t1(n2-1)+t1(n1-1)/n1-n2
[Here, n1 =4, n2= 2, t1= 10 and t2 =10]
=10(2-1) +10(4-1)/4-2 = 10-30/2 =20years.
4. The sum of present ages of A and B is S year. If, t years ago of A was n times the age of B, then Present age of A =Sn-t(n-1)/n-1 years, and Present age of B= S+t(n-1)/n+1 years.
Explanation
Let the present ages of A and B be x and y years respectively. Given: x+y = S …(1)
and, x-t =n(y-t)
or x-ny = (1-n)t
Solving (1) and (2), we get x=Sn-t(n-1)/n+1. y= S+t(n-1)/n+1.
Illustration 4 The sum of the ages of A and B is 42 years 3 years back, the age of A was 5 times the age of B. Find the difference between the present ages of A and B.
Solution Here, S=42, n=5 and t =3
Present age of A
= Sn-t(n-1)/n+1= 42×5-3(5-1)/5+1 = 198/6=33 years
and, present age of B
5+t(n+1)/n+1 = 42+3(5-1/5+1 =54/6=9years.
Difference between the present ages of A and B =33-9=24 years.
Note: If, instead of sum(S), difference (D) of their ages is given, replace S by D and in the age denominator (n+1) by (n-1) in the above formula.
5. The sum of present ages of A and B is S years. If, t years hence, the age of A would be n times the age of B, then
Present age of A=Sn+t(n-1)/n+1 years and, present age of B = S-t(n-1)/n+1 years. Explanation
Let the present ages of A and be x and y years, respectively
Given: x+y=S ….(1)
and, x+t=n(y+t)
or, x-ny=t(n-1) ….(2)
Solving(1) and (2), we get
x= Sn+t(n-1)/n+1
and, y= S-t(n-1)/n+1
Illustration 5 The sum of the ages of a son and father is 56 years. After four years, the age of the father will be three times that of the son. Find their respective ages.
Solution The age of father
= Sn+t(n-1)/n+1= 56×3+4(3-1)/3+1 [Here, S=56, t=4 and n=3]
= 176/4= 44years.
The age of son = Sn-t(n-1)/n+1 = 56-4(3-1)/3+1
= 48/4 = 12 years.
6. If the ratio of the present ages of A and B is a: b and t years hence, it will be c: d, then A‘s present age = at(c-d)/ad-bc
Illustration 6 The ratio of the age of father and son at present is 6 : 1. After 5 years, the ratio will become 7:2. Find the present age of the son.
Solution The present age of the son = bt(c-d)/ ad –bc [Here, a =6, b=1, c=7, d=2 and t=5]
=1×5(7-2)/6×2-1×7 = 5years. Note:
If, with the ratio of present ages, the ratio of ages t years age is given, then replace t by(-t) in the above formula. Illustration 7 6 years ago Mahesh was twice as old as Suresh. If the ratio of their present ages is 9 : 5 respectively, what is the difference between their present ages?
Solution Present age Mahesh =--at(c-d)/ad-bc
=-9×6(2-1)/1×9-5×2
[Here, a=9, b=5, c=2, d=1 and t =6] =54 years
Present age of Suresh -bt(c-d)/ad-bc
= -5×6(2-1)/1×9-5×2 =30 years.
Difference of their ages =54-30=24 years.
MULTIPLE CHOICE QUSTIONS
In each of the following questions a number of possible answers are given, out of which one answer is correct. Find out the correct answer.
EXERCISE
1. ten years ago, Mohan was thrice as old as Ram was but 10 years hence, he will be only twice as old. Find Mohan‘s present age.
a) 60 years b) 80 years c) 70 years d) 76 years
2. The ages of Ram and Shyam differ by 16 years. Six years ago, Mohan‘s age was thrice as that of Ram‘s, find their present ages.
a) 14 years, 30 years b) 12 years, 28 years c) 16 years, 34 years d) 18 years, 38 years
3. 15 years hence, Rohit will be just four times as old as he was 15 years ago. How old is Rohit at present? a) 20
b) 25 c) 30 d) 35
4. A man‘s age is 125% of what it was 10 years ago, but 83 1/3 % of what it will be after ten 10 years. What is his present age?
a) 45 years b) 50 years c) 55 years d) 60 years
5. If twice the son‘s age in years be added to the father‘s age, the sum is 70 and if twice the father‘s age is added to the son‘s age, the sum is 95. Father‘s age is
a) 40 years b) 35 years c) 42 years d) 45years
6. Three years ago, the average age of a family of 5 members was 17. A baby having been born the average age of the family is the same today? What is the age of the child?
a) 3 years b) 5 years c) 2years d) 1 year
7. The ratio of A‘s and B‘s ages is 4:5 If the difference between the present age of A and the age of B 5 years hence is 3, then what is the total of present ages of A and B?
a) 68 years b) 72 years c) 76 years d) 64 years
8. The ages of A and B are in the ratio of 6:5 and sum of their ages is 44 years. The ratio of their ages after 8 years will be
a) 4 : 5 b) 3 : 4 c) 3 : 7 d) 8 : 7
9. 5 years ago, the combined age of my mother and mine was 40 years. Now, the ratio of our age is 4:1. How old is my mother? (A) 10 (B) 40 (C) 60 (D) 20 (E) 50
10. Honey was twice as old as Vani 10 years ago. How old is Vani today if Honey will be 40 years old 10 years hence? a) 20 b) 25 c) 15 d) 35 e) 30
11. One year ago, a mother was 4 times older to her son. After 6 years, her age become more than double her son‘s age by 5 years. The present ratio of their age will be?
a) 13 : 12 b) 11 : 13 c) 3 : 1 d) 25 : 7 e) 4 : 3
12. Vandana‘s mother is twice as old as her brother. She is 5 years younger to her brother but 3 years older to her sister. If her sister is 12 years of age, how old is her mother?
a) 30 b) 35 c) 45 d) 40 e) 50
13. Sonu is 4 years younger Manu while Dolly is four years younger to Sumit but 1/5 times as old as Sonu. If Sumit is eight years old, how many times as old is Manu as Dolly?
a) 3 b) ½ c) 2
d) 1 e) ¼
14. Our mother is 3 times as old as my brother and I am 1/3rd times older than my brother. If 4 years ago I was as
old as my brother today, what is the age of my mother. a) 40
b) 36 c) 44 d) 42 e) 48
15. Ruchi‘s age was double that of Niti 2 years ago. If Ruchi was 2 years older to Niti then, try to guess how old she is today. a) 6 b) 4 c) 8 d) 2 e) 20
16. If we add the age of three brothers Sunil, Sanjay and Sonu, then it becomes 60 years today. If 6 years ago the
Sonu was of half the age of Sanjay and 1/3rd to the age of Sunil, then find out the present age of Sanjay.
a) 14 b) 15 c) 16 d) 18 e) 24
17. Sonu‘s age is 2/3rd of Manu‘s. After 5 years Sonu will be 45 years old. Manu‘s present age is
a) 55 b) 56 c) 58 d) 60 e) 64
18. Ratio of Sonu‘s age to Manu‘s is equal to 4:3. If Sonu will be 26 years old after 6 years, the present age of Manu is a) 11 b) 15 c) 14 d) 17 e) 13
19. Binny is born on 1st October. He is younger to Sunny by one week and two days. If on 1st October it was a
Saturday, then Sunny‘s birthday will come on which day this year?
(A) Wednesday (B) Thursday (C) Monday
(D) Saturday (E) Sunday
20. Binny is half as old as Sunny. Chinky is twice old as Sunny. How many times is Chinky as old as Binny?
(A) 6 (B) 4 (C) 8
(D) 3 (E) 2
21. My age becomes half that of my brother‘s if we simply add 2 years to his present age. If I am 25 years old today, my brother will be
a) 46 b) 48 c) 44 d) 36 e) 38
22. My age is 2 years less than twice that of my brother. If I am sixteen years old, how old is my brother? a) 3
b) 18 c) 9 d) 27 e) 6