1000DS Please Help... want to be sure

[rookiez]

19. If x is an integer, what is the value of x?
(1) –2(x + 5) < -1
(2) – 3x > 9
how come OA is C

13. Before play-offs, a certain team had won 80 percent of its games. After play-offs, what percent of all its games had the team won?
(1) The team competed in 4 play-off games.
(2) The team won all of its play-off games.
OA E

15. If x&#8800;0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
OA d

23. Is x less than y?
(1) x-y+1<0
(2) x-y-1<0
OA A

24. Is quadrilateral RSTV a rectangle?
(1) The measure of &#8736;RST is 90o
(2) The measure of &#8736;TVR is 90o
OA E

[DanaJ]

Before writing anything else, I'd like to remind you that you're supposed to post only one question per thread.

19. It's obvious that each statement by itself is not sufficient:
1. –2(x + 5) < -1
x + 5 > 1/2
x > -9/2 --- plenty of values for x: -4, -3, ...

2. – 3x > 9
x < -3 ---- again, lots of possible values: -4, -5, ...

But put both together to get that only -4 is the integer that is in both lists.


13. You've got that the team played x games before the playoffs, out of which they won 80% or 0.8x. You're trying to establish the percentage of wins out of (x + p), with p = total number of playoff games.
1. tells you that p = 4, but you don't know how many of the 4 games the team won. 1 is not enough.

2. tells you that the team won all playoff games, so wins = 100% of playoffs. This brings the total wins to 0.8x + p. However, there's no way of determining the percentage of wins out of (x + p). Consider k = percentage of wins.

k%*(x + p) = 0.8x + p
k% = (0.8x + p)/(x + p) --- there's no way of working around this fraction unless you know the values or some other relationship between x and p.

Put both together and still nothing:

k% = (0.8x + 4)/(x + 4) --- same as with statement 2


15. You've missed something on this question. Let's analyze statement 1 to prove it's enough, but you'll need to edit your post for statement 2:
x^2 < 1
x^2 - 1 < 0
(x - 1)(x + 1) < 0 --- the quadratic equation is negative only between the roots (in this case, the roots, i.e. values for which (x - 1)(x + 1) is zero, are -1 and 1). This means that the inequality holds only if -1 < x < 1. This is why |x| < 1.


23. Take each statement and analyze it:
1. x - y + 1< 0 --- add y to each side
x + 1 < y --- subtract 1 from each side
x < y - 1 --- since x is smaller than y - 1, x will definitely be smaller than y. The answer to the question in the stem will be a clear NO. 1 is sufficient.

2. x - y - 1 < 0 --- add y to each side
x - 1 < y --- add 1 to each side
x < y + 1 --- this does not necessarily mean that y is greater than x. Consider a numerical example:
x = 1
y = 0.8 --- y + 1 = 1.8, which is indeed greater than 1; however, y < x.
2 is insufficient.


24. See the picture for a possible shape of a non-rectangular polygon that has both specified right angles. Please forgive the approximate drawing, the art of MS Paint is a delicate one.

[rookiez]

Thanks DanaJ!
It helps a lot!

Following que needs your attention
15. If x&#8800;0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
OA D

[DanaJ]

2. |x| < 1/x --- when dealing with modulus, you need to break it up into two cases:
a. x is negative, which means that 1/x is also negative. However, |x| will be positive. So, if x is negative, then you get (positive number) < (negative number), which is iimpossible. This means that x cannot be negative.

b. x is positive, meaning that |x| = x. You get that x < 1/x or that x^2 < 1. You're back to statement 1, so 2 is also sufficient.