BTGmoderatorDC wrote:Each day at Dan's new job, he earns $10 more than the day before. If he just completed his 15th day of work, how much did he earn on his first day?

(1) His total earnings for the three most recent work days equal $690.

(2) His total earnings for the three most recent work days exceed his total earnings for the first three work days by $360

Since each payday is $10 more than the preceding payday, the list of 15 paydays constitutes an EVENLY SPACED SET.

Statement 1:
For any evenly spaced set:
Median = Sum/Quantity.
Thus:
Pay on the 14th day = median of the 3 most recent days = sum/quantity = 690/3 = 330.
Since the pay on the 14th day is known -- and each payday is $10 more than the preceding payday -- the pay on the first day can be determined.
SUFFICIENT.

Statement 2:
For any evenly spaced set:
nth term = first term + (n-1)d, where d = the difference between successive terms.
Here, each payday is $10 more than the preceding payday, implying that d=10.

Let the first payday = x.
Thus:
2nd payday = x + (2-1)(10) = x + 10
3rd payday = x + (3-1)(10) = x + 20
Sum of the first 3 paydays = x + (x+10) + (x+20) = 3x+30
13th payday = x + (13-1)(10) = x + 120
14th payday = x + (14-1)(10) = x + 130
15th payday = x + (15-1)(10) = x + 140
Sum of the 3 most recent paydays = (x+120) + (x+130) + (x+140) = 3x+390
Difference between the sums = (3x+390) - (3x+30) = 360.

The blue equation indicates that the difference between the sums will be 360 -- for ANY VALUE OF X.
Since Statement 2 will be true for any value of x, the pay on the first day cannot be determined.
INSUFFICIENT.

The correct answer is A.