QUESTION

# The age of a person k years ago was half of what his age would be k years from now. The age of the same person p years from now would be thrice of what his age was p years ago. What is the value of the ratio k:p?(a) 3:2(b) 2:3(c) 1:4(d) 4:1(e) 5:1

Hint: First of all, take the present age of the person to be B years and find the age of the person k years ago and k years from now. Substitute these values in the equation according to the given information and find the value of k from here. Do similar steps to find the value of P. Now divide these two equations to get k and P.

Here, we are given that the age of the person k years ago was half of what his age would be k years from now. The age of the same person p years from now would be thrice of what his age was p years ago. Here we have to find the value of the ratio k:p. Let us assume the present age is B years.
The age of the person k years ago would have been = (B – k) years.
The age of the person k years from now would be = (B + k) years.
Now, we are given that the age of the person k years ago was half of what his age would be k years from now. So, we get,
$\left( B-k \right)=\dfrac{1}{2}\left( B+k \right)$
By multiplying 2 on both the sides of the equation, we get,
$2B-2k=B+k$
$2k+k=2B-B$
$3k=B$
So, we get, $k=\dfrac{B}{3}....\left( i \right)$
Similarly, the age of the person p years ago would have been = (B – p) years.
The age of the person p years from now would be = (B + p) years.
Now, we are given that the age of the person p years from now would be thrice of what his age p years ago. So, we get,
$\left( B+p \right)=3\left( B-p \right)$
$\Rightarrow B+p=3B-3p$
$\Rightarrow p+3p=3B-B$
$\Rightarrow 4p=2B$
So, we get, $p=\dfrac{2B}{4}=\dfrac{B}{2}....\left( ii \right)$
By dividing equation (i) and (ii), we get,
$\dfrac{k}{p}=\dfrac{\dfrac{B}{3}}{\dfrac{B}{2}}$
By canceling the like terms from the above equation, we get,
$\dfrac{k}{p}=\dfrac{2}{3}$
Hence, option (b) is the right answer.

Note: Here, students often make mistakes while forming the equation. Sometimes they may write an equation $\dfrac{1}{2}\left( B-k \right)=\left( B+k \right)$ that is wrong. The correct equation as we have seen was $\left( B-k \right)=\dfrac{1}{2}\left( B+k \right)$. So, this mistake must be taken care of. Also, students must note that if the present age is x years, then the age ‘n’ years ago and ‘n’ years from now is given as (x – n) and (x + n) years respectively.