there are 48 coins amongst 3 kids A,B,C.A divides half his coins equally among B and C.then B divides half of his coins equally among A and C.Now all of them have equal numbers of coin what is the number of coins each had initially?

• as at last a,b,c each have 16 coins....so
  a/q we may say B HAVE 32 COINS before distributing&after a give his 1/4 th of coin
  LET US SUPPOSE A.B.C HAVE a,b,c cons respectively
  as c have 16 coins at last i.e 16=1/4a+1/4b+c
  B'S COIN AFTER A'S GIVEN 1/4 OF HIS COIN 32=1/4a+b
  16=1/2a+8 i.e a=16 initially A have 16 cons &B HAVE 28 CONS &C HAVE 4 COINS
  
  solve these equ we have
• 10 years agoHelpfull: Yes(1) No(0)
• a=16
  b=28
  c=4
• 10 years agoHelpfull: Yes(0) No(0)
• A = 16
  B = 28
  C = 4
  ...
  (A+4B/8)=(9A+4B/16)=(5A+4B+16C/16)=48/3
  A+4B=128
  9A+4B=256
  5A+4B+16C=256
  BY SOLVING THESE A=16,B=28,C=4......
• 10 years agoHelpfull: Yes(0) No(0)
• you can think from answer,in answer is saying all of them have equal no of coins
  48 is total coins then A=16 B=16 C=16
  half of B divides equally among A and C do that
  16 is half of 32 so each 8 subtract from A and C and add 16 to B
  A=8 B=32 C=8
  half of A divides equally among B and C do that
  8 is half of 16 so each 4 subtract from B and C and 8 add to A
  A=16 B=28 C=4
• 10 years agoHelpfull: Yes(0) No(0)