<p>How old was a person exactly 1 year ago, if exactly x years ago the person was y years old?</p>
<p>A) y-1
B) y-x-1
C) x-y-1
D) y+x+1
E) y+x-1</p>
<p>The only way I could think of to solving it was substitution :(</p>
<p>There is literally nothing to solve here. You are just making an equation.</p>
<p>y = his age x years ago;
y + x = his current age;
y + x - 1 = his age last year;</p>
<p>The answer is (E) y + x - 1.</p>
<p>My bad**</p>
<p>If anyone has the blue book it might be easier but</p>
<p>For all numbers x and y, let * (it’s really a square but ill put star in here) be defined by x*y = xy-y. If a and b are positive integers, which of the following can be equal to zero.</p>
<p>I. a<em>b
II. (a+b)</em>b
III. a*(a+b)</p>
<p>A) I Only
B) II Only
C) III Only
D) I and II
E) I and III</p>
<p>I know immediately that I can equal zero if a equals one, with a little bit of work I can find out II can not equal zero, and 3 I get lost on :(</p>
<p>Let’s examine each choice on its own:
I. ab - b; - will equal 0 if a = 1.
II. (a+b)(b) - b; - didn’t ACTUALLY check, but since none of the above is not an option, we can bet that this one WILL NOT equal zero.
III. a(a+b) - (a+b); - will equal 0 if a = 1.</p>
<p>Here is an actual proof for II.
(a+b)(b) - b;
ab + b^2 - b;
b(a + b - 1);
Assuming both a and b are 1, the value is: (1+1-1) = 1;
THEREFORE, because all the terms are either multiplied and divided and we must employ whole, positive integers (see: minimizing the inputs will minimize the output), the MINIMUM of the function is 1. Consequently, it will never equal zero.</p>
<p>Again, the answer is B.
Edit: oops the answer is E. I misread the question itself. The math is still applicable, though.</p>
<p>The answer wouldn’t be B, because it says which of the following can be equal to zero.</p>
<p>But in the blue book explanation, they eventually get to b(a+b-1), but your way is a hell of a lot simpler. 
thanks for your help on that.</p>
<p>I think this is an easy way:</p>
<p>x*y = xy-y = (x-1)y</p>
<p>I) a<em>b=(a-1)b, =0 for a=1
II) (a+b)</em>b=(a+b-1)<em>b can’t be zero for a,b>=1
III) a</em>(a+b)=(a-1)(a+b), =0 for a=1</p>