by Guest Thu Jun 28, 2007 2:31 pm
For the first problem, I really think drawing it our graphically is the way to go. If you try to do using algebra you eventually end up with a system of 3 absolute value equations and one inequality. :shock:
So if the distance between A and B is 8 and C and D is 8, your number line might look like this:
c--------d---a-----------------b
Now let's look at each statement. (2) looks less complicated so we will start with that
A is to the left of D
so somthing line this would work
c------ad----------------b
or somthing like this
ca------d----------b
but in any event (2) is not sufficient and you can rule out B and D.
(1) The distance between C and A is the same as the distance between c and B
This tells us that C is on the mid-point between A and B. However D can be either 8 units to the right of C or eight units to the left...so (1) is not sufficient to answer the question of what the distance is between b and d, and we can rule out answer choice A.
Now consider the two statements together. We know from the two together that C is the mid-point between A and B and that A is to the left of D. However because of the interval between and B (18) is much larger than the interval between C and D (8) we don't know which side of C D is on and still can't answer the question...so the answer is E. Graphically:
AD-------C--------B
or
A--------C------DB
Second problem:
If x is less than 10, then the average of x and 10 is the midpoint on the number line between x and 10:
x-----m-----10
1) on the number line, z is closer to 10 than it is to x.
this means that it is the the right (larger) than the average (m) and so (1) is sufficient. We can rule out B,C,E.
2) z=5x
This one is trying to trap people who don't read the problem carefully. X is a positive number less than 10, not necessarly an integer. consider x=.1. Then the mean of x and 10 is 5.05. if z=5x, then z is .5 and is smaller than the mean. Now consider x=2. The mean of x and 10 is then 6. z=5x=10 and is larger than the mean. So 2 is not sufficient and the answer is A.
I thought the first problem really benefited more from careful graphing than the second problem, which seamed pretty straightforward as long as you read it carefully.
/Jeff