Following Problems on Ages will help you in securing good marks in upcoming SBI PO, SBI Clerk, IBPS PO VI, IBPS Clerk VI, IBPS RRB Exams.
1. Rajeev's age after 15 years will be 5 times his age 5 years back. What is the present age of Rajeev?
Sol. Let Rajeev's present age be x years. Then, Rajeev's age after 15 years = (x + 15) years.
Rajeev's age 5 years back = (x - 5) years.
Therefore x + 15 = 5 (x - 5)
x + 15 = 5x - 25
4x = 40
x = 10.
Hence, Rajeev's present age = 10 years.
2. The ages of two persons differ by 16 years. If 6 years ago, the elder one is 3 times as old as the younger one, find their present ages.
Sol. Let the age of the younger person be x years.
Then, age of the elder person = (x + 16) years.
Therefore 3 (x - 6) = (x + 16 - 6)
3x -18 = x + 10
2x = 28
x = 14.
Hence, their present ages are 14 years and 30 years.
3. The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita is more than Ankit's age by 4 years, what is Nikita's age?
Sol. Let Ankit's age be x years. Then, Nikita's age = 240/x years.
2 * (240 /x ) – x = 4
480 – x2 = 4x
x2 + 4x – 480 = 0
( x+24)(x-20) = 0
x = 20.
Hence, Nikita's age = 240/x = 240/20 years = 12 years.
4. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father.
Sol. Let the son's present age be x years. Then, father's present age = (3x + 3) years
(3x + 3 + 3) = 2 (x + 3) + 10
3x + 6 = 2x + 16
x = 10.
Hence, father's present age = (3x + 3) = ((3 * 10) + 3) years = 33 years.
5. Rohit was 4 times as old as his son 8 years ago. After 8 years, Rohit will be twice as old as his son. What are their present ages?
Sol. Let son's age 8 years ago be x years. Then, Rohit's age 8 years ago = 4x years.
Son's age after 8 years = (x + 8) + 8 = (x + 16) years.
Rohit's age after 8 years = (4x + 8) + 8 = (4x+ 16) years.
2 (x + 16) = 4x + 16
2x = 16 => x = 8.
Hence, son's 'present age = (x + 8) = 16 years.
Rohit's present age = (4x + 8) = 40 years.
6. One year ago, the ratio of Gaurav's and Sachin's age was 6: 7 respectively. Four years hence, this ratio would become 7: 8. How old is Sachin?
Sol. Let Gaurav's and Sachin's ages one year ago be 6x and 7x years respectively.
Then, Gaurav's age 4 years hence = (6x + 1) + 4 = (6x + 5) years.
Sachin's age 4 years hence = (7x + 1) + 4 = (7x + 5) years.
(6x+5) : (7x + 5) = 7 : 8
8(6x+5) = 7 (7x + 5)
48x + 40 = 49x + 35
x = 5.
Hence, Sachin's present age = (7x + 1) = 36 years.
7. Abhay's age after six years will be three-seventh of his father’s age. Ten years ago the ratio of their ages was 1: 5. What is Abhay's father's age at present?
Sol. Let the ages of Abhay and his father 10 years ago be x and 5x years respectively.
Then, Abhay's age after 6 years = (x + 10) + 6 = (x + 16) years.
Father's age after 6 years = (5x + 10) + 6 = (5x + 16) years.
(x + 16): (5x + 16) = 3:7
7(x + 16) = 3 (5x + 16)
7x + 112 = 15x + 48
8x = 64 => x = 8.
Hence, Abhay's father's present age = (5x + 10) = 50 years.
8. The Ratio of Ages of Mona and Sona is 4:5. Twelve Years hence, their ages will be in the ratio of 5:6. What will be Sona's age after 6 years?
Sol. Let their present ages be 4x & 6x
Then (4x + 12)/(5x + 12) = 5/6 or x=12
Sona's age after 6 years = (5x +6) = 66 years
9. Ramu was 4 times as old as his son 8 years ago. After 8 years, Ramu will be twice as old as his son. What their present ages?
Sol. Let son's age 8 years ago be x years
Then Ramu's age at that time = 4x years
Son's age after 8 years = (x +8) + 8 = (x + 16) years
Ramu's age after 8 years = (4x + 8) + 8 = (4x + 16) years
2(x + 16) = 4x + 16 or x=8
Son's present age = (x + 8) = 16 years
Ramu's present age= (4x + 8) = 40 years
10. A man is four times as old as his son. Five years ago, the man was nine times as old his son was at that time. What is the present age of a man?
Sol. Let son's age = x, then man's age =4x.
9(x - 5) = (4x-5) or x=8.
Man's present age = (4x + 7) = 35 years
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