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There are 3 friends X,Y,Z having some chocolates each.
X Gives Y and Z as many chocolates as they already have.
After some days Y gives X and Z as many chocolates as they
have.
After some days Z gives X and Y as many chocolates as they
have.
Finally each has 24 chocolates. What is the original No. of
chocolates each had in the beginning?
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Answer / anjali
solution x = 39, y = 21, z = 12
solve it by backtracking.
we know finally each has 24.
x y z
--- --- ---
final: 24 24 24
12 12 48 before z gives
6 42 24 before y gives
initial: 39 21 12 before x gives
| Is This Answer Correct ? | 54 Yes | 7 No |
Answer / arulmurugan.c
finally
x=24
y=24
z=24
before this
x=24-12=12
y=24-12=12
z=24+12+12=48
before this
x=12-6=6
z=48-24=24
y=12+6+24=42
before this
y=42-21=21
z=24-12=12
x=6+21+12=39
so initially
x=39
y=21
z=12
totally 72 chocolates
| Is This Answer Correct ? | 5 Yes | 2 No |
Answer / ranjith kumar
xgave 8 coclates 2 y and z
y and z has 28 chclate
x has 16 chocloclates
now y gaves 8 choclates to x&z
x has 20 choclates
zhas 32 cloclates
and y has 20 choc
z gives 8 choclates to x znd y
x has 24,
x= 16,+4+4=24
y=28-8+4=24
z=24+4+4-8=24
| Is This Answer Correct ? | 0 Yes | 0 No |
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