x years ago the age of A was n1 times the age of B, and at present A’s age is n2 times that of B, then;
A’s current years.
and, B’s current age years.
The present age of A is n1 times the present age of B. After x years, age of A becomes n2 times the age of B, then;
t1 years ago, the age of A was X times the age of B and after t2 years age of A becomes Y times the age of B, then;
The sum of present ages of A and B is X years, t years after, the age of A becomes Y times the age of B, then;
The ratio of the present ages of A and B is p: q and after t years, it becomes r: s, then;
The sum of present ages of A and B is X years, t years ago, the age of A was Y times the age of B, then;
Q1: Amitabh said to his friend “If you subtract 18 from my age the two digits of my age will reverse their positions. Also my age is six, less than 8 times the sum of digits of my age”. Find Amitabh’s age.
(a) 46 years
(b) 37 years
(c) 56 years
(d) 42 years
Ans: (d)
Sol: Let Mayank’s age be (10x + y) years
Age by reversing the digits = (10y + x) yrs
Now, 10x + y − 18= 10y + x
9x – 9y = 18
x – y = 2………………(1)
Also,
10x+y = 8(x+y) – 6
2x – 7y = –6……….… (2)
Solving equations (1) and (2),
x = 4 , y = 2
Therefore, Mayank’s age = 10x + y
= 10(4) + 2
= 42 years
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