document.write( "Question 8046: 5 years ago, a man was 7 times as old as his son. 5years hence, father will be 3 times as old as his son. Find their present ages.
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #4415 by kritikapd(4)\"\" \"About 
You can put this solution on YOUR website!
Let present age of son = X
\n" ); document.write( "Age of son before 5 years = X-5
\n" ); document.write( "So, Present age of father = 7(X-5)+5 ……… (i)\r
\n" ); document.write( "\n" ); document.write( "Age of son after 5 years = X+5
\n" ); document.write( "Age of father after 5 years = 3(X+5)
\n" ); document.write( "So, present age of father = 3(X+5)-5 ……….(ii)\r
\n" ); document.write( "\n" ); document.write( "Equating age of father i.e. (i) and (ii)\r
\n" ); document.write( "\n" ); document.write( "7(X- 5)+5 = 3(X+ 5)-5
\n" ); document.write( "7X – 35+5 = 3X + 15-5
\n" ); document.write( "7X-3X = 15 - 5 + 35 - 5
\n" ); document.write( "4X = 40
\n" ); document.write( "X = 40/4 . X (son)= 10 , Father (3(10+5)-5)= 40 yrs\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );