1. Six years ago Jose was twice as old as Joseph of the ratio of their present age is 9 : 5 respectively, what is the difference between their present ages? [Bank PO 2000]
(d) none of these
Let their present ages be 9x and 5x. six years ago,
(9x-6)=2(5x-6) => x=6
The difference between their present ages = 9x-5x=4×6=24 years
2. Father’s age is 5 yrs more than mother’s age & mother’s age is 3 times of daughter’s age. The present age of daughter is 10 yrs. What was the age of father at daughter’s birth?
(a) 20 yrs
(b) 22 yrs
(c) 25 yrs
(d) 28 yrs
Present age of daughter= 10yrs
Mother= 10 ×3= 30yrs
Father= 30+5= 35yrs
So, father’s age while 10 yrs ago was 25 years.
3. 15 years hence a man will be 4 times as old as he was 15 years ago, His present age is?
Let us take present age of man is x years.
4. Sachin is elder than Rohit by 6 years. If their ages are in the ratio of 6:8, how old is Sachin?
(a) 6 yrs
(b) 8 yrs
(c) 10 yrs
(d) 18 yrs
Let Sachin age be x, Rohit age be x-6
Sachin age= 18 yrs, Rohit age= 12 yrs
5. Present age ratio of x, y is 6:8, 7 years hence, the ages become 4:5. What is x’s present age?
After 7 years, (6m+7)/(8m+7)=4/5
6. A father tells his eldest son, I’m thrice as old as your youngest sister and when I was as old as you, you were as old as youngest sister. If the youngest sister is 17 yrs old, how many years ago was her elder brother born?
(a) 30 yrs ago
(b) 32 yrs ago
(c) 34 yrs ago
(d) 36 yrs ago
Let, father is f, eldest son as s, youngest daughter as d
Given f= 3d, f = 3 x 17 =51;
Fathers age – some years (x) = Sons age; —>1
Sons age- some years(x) = daughters age =17; —>2
54 – x = sons age;
Sons age = 17 + x;
54 – x =17 +x;
Sons age =54-17 =34;
Son’s age is 34.
7. A said to B “I am twice as old as you were when I was as old you are now”. Sum of their ages is 42. Find their present ages. [Delhi police, 2007]
(a) 12, 20
(b) 18, 24
(c) 16, 26
(d) 17, 25
Clearly by choosing option (b) we see that A was equal to B’s age, i. e. 18 years, 6 years ago and at that time B was 12 years old. So, A’s present age, 24 years, is twice of B’s that time of age, i. e. 12 years.
Let Present age of B=x and present age of A=42-x
Then, 18 years ago A’s age=x and B’s age=x-(42+2x)=3x-42
By given condition, A’s present age=2×(B’ s age some years ago)
B’s present age=18 years and A’s present age=24 years.
8. A father says to his son, ‚7 years ago my age was 6 times of your age at that time & after 4 years my age will be four times of your age” Find the present age of the man?
(b) 7. 2
(c) 7. 5
Let us take son’s age as x, father’s age as y
7yrs ago, y-7= 6(x-7)
6x-y = 35 —-1
4x-y= 12 —–2
Solve the two equations,
X= 7. 5, y= 10
Son’s age is 7. 5yrs
9. 10 years ago, the average age of family of 4 members was 26yrs. Two children have been born; the average age of the family is same today. The sum of the ages of the two children is?
Sum of ages of 4 members= 26×4= 104
Ten years ago, 104+(4×10)= 144yrs
Members in the family is now 4+2= 6
Sum of the ages of 6 members= 26×6= 156
10. The average of a father and his two sons is 27 years. Five years ago, the average age of the two sons was 12 years. If the difference between the age of the two sons is 4 years, then the present age of the father is: [SSC Grad. 2000]
(a) 34 years
(b) 47 years
(c) 64 years
(d) 27 years
Five years ago the average age of two sons =12 years
Their present average age = (12+5) = 17 years
Sum of their ages = 17 × 2 =34 years
Average age of three of them = 27 years
Sum of their ages = 27 × 3 =81 years
Father’s ages = 81 – 34 = 47 years