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?
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
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
x = 108 <---- K
3x = 324 <---- L
684-4x = 252 <---- M
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.
To solve this problem, we can set up a system of equations to represent the given information.
Let's assume that the number of coins in Container K is "x", the number of coins in Container L is "3x" (since the ratio of the number of coins in K to L is 1:3), and the number of coins in Container M is "y".
According to the given information, the total number of coins in the three containers is 684, so we can create the equation:
x + 3x + y = 684
Next, we are told that after Container L and Container M each had 2/3 of their coins taken out, the three containers had a total of 300 coins left. Let's calculate the number of coins left in Containers L and M:
Container L: 3x - (2/3)(3x) = 3x - 2x = x
Container M: y - (2/3)y = (1/3)y
Now, we can set up another equation using the information that the total number of coins left in the three containers is 300:
x + x + (1/3)y = 300
Simplifying this equation, we get:
2x + (1/3)y = 300
So now we have a system of equations:
x + 3x + y = 684
2x + (1/3)y = 300
To solve this system of equations, we can use substitution or elimination.
Let's solve it using substitution method:
We can rewrite the first equation as:
4x + y = 684
Solving the second equation for y, we get:
(1/3)y = 300 - 2x
y = 900 - 6x
Substituting this value of y in the rewritten first equation, we get:
4x + (900 - 6x) = 684
4x + 900 - 6x = 684
-2x = 684 - 900
-2x = -216
x = -216 / -2
x = 108
Now, substituting the value of x back into the other equation to find y:
y = 900 - 6x
y = 900 - 6 * 108
y = 900 - 648
y = 252
Therefore, at first, Container M had 252 coins.