By trial and error find the smallest number to satisfy (i) and (ii) is 8.
Then go on by trialing numbers of 8 + 15n since 15 is the least common multiple of 3 and 5.
You will find the answer to be 68 within just a few tries.
There should be a better and more official way by modulo arithmetic though so it would be great if someone can also add that to this thread.
The amount of money Andrew starts with we set as X
X - 940 - 5/7(X - 940) + 240 = 2/9X
X- 5/7X -2/9X -940 +4700/7+240 = 0
4/63X = 200/7
4X = 1800
X = 450????
Can someone check my process because either I made a mistake during calculation or there's something wrong with the question. He can't start with 450 and spend 940.
This isn't a question that could be solved very easily.
We try to list equations:
G = 1/7(M+K+J)
M = 3/4(K+J)
J = 2/5K
Let's Plug in:
M = 3/4(7/5K) = 21/20K,
G = 1/7(21/20K+K+2/5K) = 1/7(49/20K) = 7/20K
K = K
J = 2/5K
M = 21/20K
G = 7/20K
Now we split all of that $168 into these four parts. 4 parts combined is 56/20K, and thus K original has 168*20/56 = $60. J original has 168*8/56 = $24
Now you can do the last step: $60 - $X = $24 + $X
Hopefully this proved to be helpful
OK, so I would set all the money he has at first as X, our variable.
X - 1/3X - 120 = 2/3X - 120 is what he has after sentence 1
1/3(2/3X-120)-120 = 2/9X - 160 is what he has after sentence 2
1/3(2/9X - 160) -120 = 2/27X - 520/3 is what he has after sentence 3, but since by this time he doesn't have any money left, we can infer than
2/27X - 520/3 = 0, thus 2/27X = 520/3.
Find X to get the answer.
We set all the marbles Fred has as X.
We can infer from the first sentence that: X - (1/4X+34) = 3/4X - 34 marbles is what Fred has after the first sentence
Now, 3/4X - 3/8(3/4x - 34) = 3/4X-9/32X+51/4 = 15/32X+51/4 is what he has after the second sentence
That suggests that 15/32X + 51/4 = 580, which means 15/32X = 2551/4, and find X to get the answer to the question.