Data Sufficiency - 53

Answer: D

From statement (1): Square root of a number less than 1 is also less than 1 - sufficient
From statement (2): Reciprocal of any number less than 1 is greater than the number - sufficient

Problem Solving - 50

If x = -|x|, then which one of the following statements could be true?

I. x=0
II. x < 0
III. x > 0

A). none
B). I only
C). III only
D). I and II
E). II and III

Answer: D

Given x = - |x|
We know that |x| >= 0
=> -|x| = x <=0

Thus inequalities represented by I and II

DS Question - 29

If x and y are integers, does IxI = y ?

(1) (y^2 - x^2) = 0
(2) xy/(x+y) = 0

Answer: B

Statement (1) insufficient -- y can take any value i.e can be +ve or -ve. We cannot assume y to be +ve. Therefore y may or may not be equal to x

Statement (2) sufficient -- xy/(x+y) = 0 => xy = 0 => x = 0 or y = 0
In case x = 0, then y cannot be equal to 0. Hence y cannot be equal to x.
In case y = 0, then x cannot be equal to 0. Hence y cannot be equal to x.
Therefore sufficient.

Hence B

Manhattan Challenge Problem of the week! - 04/23/07

Which of the following sets includes ALL of the solutions of x that will satisfy the equation:

Answer: OA - C

OE - One way to solve equations with absolute values is to solve for x over a series of intervals. In each interval of x, the sign of the expressions within each pair of absolute value indicators does not change.

In the equation Ix-2I - Ix-3I = Ix-5I , there are 4 intervals of interest:

x less than 2 - In this interval, the value inside each of the three absolute value expressions is negative.

2 less than x less than 3 - In this interval, the value inside the first absolute value expression is positive, while the value inside the other two absolute value expressions is negative.

3 less than x less than 5 - In this interval, the value inside the first two absolute value expressions is positive, while the value inside the last absolute value expression is negative.

5 less than x - In this interval, the value inside each of the three absolute value expressions is positive.

Use each interval for x to rewrite the equation so that it can be evaluated without absolute value signs.

For the first interval, x less than 2, we can solve the equation by rewriting each of the expressions inside the absolute value signs as negative (and thereby remove the absolute value signs):

- x + 2 - ( -x + 3 ) = - x + 5

- x + 2 + x - 3 = - x + 5

x = 6

Notice that the solution x = 6 is NOT a valid solution since it lies outside the interval x less than 2. (Remember, we are solving the equation for x SUCH THAT x is within the interval of interest).

For the second interval 2 less than x less than 3, we can solve the equation by rewriting the expression inside the first absolute value sign as positive and by rewriting the expressions inside the other absolute values signs as negative:

x - 2 - ( -x + 3 ) = - x + 5

x - 2 + x - 3 = - x + 5

3x = 10

x = 10/3

Notice, again, that the solution x = 10/3 is NOT a valid solution since it lies outside the interval 2 less than x less than 3

For the third interval 3 less than x less than 5, we can solve the equation by rewriting the expressions inside the first two absolute value signs as positive and by rewriting the expression inside the last absolute value sign as negative:

x - 2 - ( x - 3 ) = - x + 5

x - 2 - x + 3 = - x + 5

x = 4

The solution x = 4 is a valid solution since it lies within the interval 3 less than x less than 5

Finally, for the fourth interval 5 less than x, we can solve the equation by rewriting each of the expressions inside the absolute value signs as positive:

x - 2 - ( x - 3 ) = x - 5

x - 2 - x + 3 = x - 5

x = 6

The solution x = 6 is a valid solution since it lies within the interval 5 less than x.

We conclude that the only two solutions of the original equation are x = 4 and x = 6. Only answer choice C contains all of the solutions, both 4 and 6, as part of its set.

Therefore, C is the correct answer.