Asked by Anonymous

Containers K, L and M contained a total of 684 coins.
The ratio of the number of coins in Container K to the number of coins in
Container L was 1 : 3. After Container L and Container M each had
2/3
their coins taken out, the 3 containers had a total of 300 coins left
How many coins were in Container M at first?
Answered by Anonymous
No. of coins taken out = 6844-300
= 384
2/3 of (L+m) = 384
L + M = 384÷2×3
= 576
K=684-576
= 103
1x =108
L =34
= 3x108
= 324
M= 684-108-374
= 252
Answered by mathhelper
or

original:
K contained x
L contained 3x
M contained 684-4x

Number left after 2/3 taken from L and M:
(1/3)(3x + 684 - 4x) + x = 300
3x + 684 - 4x + 3x = 900
2x = 216
<b>x = 108 <---- K
3x = 324 <---- L
684-4x = 252 <---- M</b>

check: 108+324+252 = 684 , check
1/3 of (324+252) = 192

number left over 108 + 192 = 300, check

The solution of "anonymous" is not correct.
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