Four statements, One statement is false math problem When trying to recall some facts about the ages of his three aunts, Josh made the
following claims:


*

*Alice is fifteen years younger than twice Catherine’s age.

*Beatrice is twelve years older than half of Alice’s age.

*Catherine is eight years younger than Beatrice.

*The three women’s ages add to exactly one-hundred years.


However, Josh’s memory is not perfect, and in fact only three of these four
claims are true. If each aunt’s age is an integer number of years, how old is
Beatrice?
I have these following equations:
$a=2c-15$ 
$b=\frac12 a +12$ 
$c=b-8$
$a+b+c=100$
How would I solve this problem? Do I assume one statement at a time is false and try the situations one by one?
 A: The approach you suggest would work - as long as you find that only 1 of the situations is possible, and the other 3 aren't (like for example if someone has a non-integer/negative age, or you reach another contradiction of some sort).
On the other hand, you may find that there are several possibilities for which Josh's claims are false. But they may all give the same age for Beatrice which still enables you to answer the question. (Note that I haven't done the question so this may not be true at all)
A: Or simply do it case by case.
Case 1:1 is true.
a=2c-15.
Case 1a:2 is true
b=1/2 a + 12 = c - 7 1/2 + 12.  A contradiction.
Case 1b: 2 is false, 3 and 4 are true
c= b-8
a+b+c =100 so 2c-15 + c + 8 + c =4c - 7 = 100 so 4c = 107.  A contradiction.
Case  2: 1 is false, all others true.  Only case left.
b=1/2 a +12
c= b-8 = 1/2  a +4
a + b+c=2a +16=100
a= 42; b=33;c=25.
A: I think I got it.
I assumed the first one was false because, well, it was the first one on the list. Using substitution where $c=b-8$, the third equation becomes $a+2b=108$, and substituting $a+24$ for $2b$ because of the second equation, I get $2a+24=108$, which simplifies to $a=42$. After doing that, I plug it into the second equation to get $b=33$, and plugging that into the third equation to get $c=25$. So, $a=42$, $b=33$, and $c=25$. $42+33+25=100$, so the answer is $33$.
A: A is false. Suppose A is true. We check case by case. Suppose A, B and C or A, B and D are true. Since we want $b$ to be an integer, $a$ must be an even number by equation B.  However equation A says $a$ is a difference of and even and odd integer, hence $a$ is odd. with this we have that A and B cannot be true at the same time. The only option left for A to be true is if A, C and D are true. If that were the case, we could, by substitution, turn equation D into $4c-7=100$. But again this cannot be the case if c is an integer because $4c-7$ and $100$ have different parities. Thus A is false and the system of equations is B,C and D. Substituing everything in D we get $4b=132$ which implies  $b=33$ $c=25$ and $a=42$.
A: Statement 1 implies a is odd.
Statement 2 implies a is even.
These are contradictory so of them is false so statements 3 and 4 are true.
Statement 3 implies b and c are the same parity.  Since this is true b and c are the same parity so b + c is an even number.
Statement 4 is true so a is even.
So statement 1 is false and statement 2 is true. 
Solving for 2,3,and 4 is straight forward.
