Answer
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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.
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
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