anairamitch1804 wrote:
Whenever see a statement about AVERAGES (of which 'halfway' is one), you should automatically associate averages with SUMS. in other words, if you ever see a statement about an average, you should immediately translate that statement into the language of sums. to do so, just use the following equation:
average = sum / # of data points
or, equivalently,
sum = (average) x (# of data points)
statement (1):
this tells you that 6 is the average of x and y (or, (x + y)/2 = 6).
therefore,
sum of x + y = (average)(# of data points) = 6 x 2 = 12.
you can also do good old fashioned algebra to get this result: multiply both sides of (x + y)/2 = 6 by 2 to yield x + y = 12. in fact, that's probably easier on this problem, but it's important that you learn the average/sum formula so that you can apply it effortlessly to other situations (such as sums of 10, 20, or more numbers) on which an algebraic solution would be awkward or just plain impossible in a reasonable amount of time.
in any case, x + y = 12, so this is sufficient.
statement (2):
clearly insufficient by itself, since x and y could be huge (1 million and 2 million) or tiny (0.0001 and 0.0002).
Hence A.
anairamitch1804 wrote:
Whenever see a statement about AVERAGES (of which 'halfway' is one), you should automatically associate averages with SUMS. in other words, if you ever see a statement about an average, you should immediately translate that statement into the language of sums. to do so, just use the following equation:
average = sum / # of data points
or, equivalently,
sum = (average) x (# of data points)
statement (1):
this tells you that 6 is the average of x and y (or, (x + y)/2 = 6).
therefore,
sum of x + y = (average)(# of data points) = 6 x 2 = 12.
you can also do good old fashioned algebra to get this result: multiply both sides of (x + y)/2 = 6 by 2 to yield x + y = 12. in fact, that's probably easier on this problem, but it's important that you learn the average/sum formula so that you can apply it effortlessly to other situations (such as sums of 10, 20, or more numbers) on which an algebraic solution would be awkward or just plain impossible in a reasonable amount of time.
in any case, x + y = 12, so this is sufficient.
statement (2):
clearly insufficient by itself, since x and y could be huge (1 million and 2 million) or tiny (0.0001 and 0.0002).
Hence A.
What if x=-5 and y = 22 the sum is 17. If x= -10 and y = 32 the sum is 22. Statement never said that the numbers are +ve. Also the stat 1 and 2 speak only about alzebra not about absolute distance.
Please help. I think the answer is C as it gives unique solution ie 4 and 8.