GHong14 wrote:https://magoosh.question-images.s3.amazo ... nation.png

Can anyone explain to me the how the first statement was simplified?

k-1 < 5 and k-1 > 2
k < 6 and k > 3
k interval { 4, 5 }

I would advise thinking of it logically. If 1/something is between 1/2 and 1/5, and we're only dealing with integers (as we're told in this question), then clearly you're dealing with 1/3 or 1/4. Thus, k-1 has to be 3 or 4. Sometimes that kind of simplistic, non-math thinking is far more efficient on test day than dealing with complex algebra rules.

The algebra is as follows:

Split it into 2 inequalities: 1/5 < 1/(k-1) and 1/(k-1) < 1/2
Take the reciprocal of everything. Like multiplying by a negative, taking the reciprocal of an inequality means you must flip the inequality sign, so you get:

5 > k-1 and k-1 > 2
6 > k and k > 3

Putting it together, we get, of course, 3 < k < 6, or k = {4, 5} since it must be an integer.

Now you get to ask yourself: Would you rather do that algebra, and risk a silly mistake somewhere along the line, or see through the natural counting logic of the problem (1/2, 1/3, 1/4, 1/5) and get an answer faster and more confidently on test day?

Remember this:

x < y AND x and y are both positive, Flip sign -> 1/x > 1/y
x < y AND x and y are both negative, Flip sign -> 1/x > 1/y
x < y AND x is negative and y positive, Dont flip sign -> 1/x < 1/y
x < y AND x we don't know if they are positive or negative, we can't take reciprocals

(1) In this case, they are presenting us 1/x < 1/y < 1/z
Because we know that they are all positive, (1/y is between 2 positive number so it must be positive), then we can take the reciprocal (yeah.. the reciprocal of a "reciprocal" :p), so we get x>y>z (manipulate to get 6>k>3)
Solve and woala!

(2) Manipulate agebraicly, remembering that any operation should be done on each section of the inequality, not just the outside sections.