Latest self Aptitude Question SOLUTION: 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

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(#M40019846) SELF question puzzle Keep an EYE

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?

FAISAL RIZWAN (10 years ago)

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niraj kumar  (10 years ago)

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

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Mukesh Kumar  (10 years ago)

a=16
b=28
c=4

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Udhaya kumar  (10 years ago)

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......

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Jagdish Kaiwaritya  (10 years ago)

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

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