In 1930 the father's age is 8 times of son's age. in 1938 father's age is 10 times of son's age in 1930. in 1940 what is their age's

• In 1930 f=8s; In 1938 I.e 8 years after f+8=10*s;
  8s+8=10s=>s=4
  Therefore f=8*4=32 in 1930;
  In 1940 I.e after 10 years f=42,s=14
• 10 years agoHelpfull: Yes(26) No(1)
• Ans 42, 14
  
  In 1930
  F=8S------1
  
  In 1938
  (F+8)=10S----2
  
  By solving 1 and 2 we get
  Fathers age 32
  Son age 4
  
  After ten years their ages are 42 and 14.
• 10 years agoHelpfull: Yes(3) No(1)
• In 1930 age of son be = y
  In 1930 father age is 8 times of son's age.... so if father age be x
  i.e x=8y-----> 1
  
  in 1938 father age will be
  (x+8)=10 *(y) ----->2
  solve 1 and 2
  
  x+8= 10y---->3
  8y+8=10y
  y=4--->son age in 1938
  put y value in equation 3
  you will get fathers age
  
  x+8=10*4
  x=32 ----> father age in 1938
  
  so in 1940 father's age ll be...x+2
  x=34
  similarly,
  son age in 1940 is y=6
  
  
  
  
  
  
  
• 10 years agoHelpfull: Yes(2) No(5)
• in 1930
  son=x
  f=8x
  in 1938
  8x+8=10x
  x=4
  so age in 1940 are
  son=4+10=14
  father=8*4+10=42
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  8
• 10 years agoHelpfull: Yes(2) No(1)
• let in 1930 son's age be x so his father's age should be 8x
  in 1938 son's age =x+8 & father's age =10x
  so,10x-8x=8,x=4
  so,in 1940 son's age=x+10=4+10=14 & father's age=8x+10=42
• 10 years agoHelpfull: Yes(1) No(0)
• 14,42 because of their age in 1930 ....4,32
• 10 years agoHelpfull: Yes(0) No(0)