GmatKiss wrote:ABC Consulting charges more for the first 100 man-hours of a project than for additional hours beyond the first 100. If Acme's fees for using ABC Consulting's services were $14,000, how many man-hours were charged?

(1) ABC Consulting charges $100 per man-hour for the first 100 man-hours and $80 per man-hour for each additional hour or fraction of an hour.
(2) If the charges for the first 100 hours had been $120/man-hour, Acme's total consulting charges would have been $16,000.

Hi!

From the stem, we know that the fee is $x/hour for the first 100 hours and $y/hour for extra hours. We also know that Acme was charged a total of $14000. So, we can create the following formula:

100*$x + (n-100)*$y = $14000

(n = total number of hours charged)

and the question is "what's the value of n?".

Note: we don't know that Acme was charged for more than 100 hours (never make assumptions in DS!), so if n<100 then we just ignore the second term in the equation.

To the statements!

1) x=100 and y=80. So:

100*$100 + (n-100)*$80 = $14000

One variable, bunch of numbers: sufficient... eliminate B, C and E.

2) if x=120, then total = 16000. So:

100*$120 + (n-100)*$y = $16000

Simplifying:

(n-100)*$y = $4000

Two variables, only one equation: insufficient.. eliminate D. Choose A!

Now we could have saved time by using the most powerful DS rule: number of equations vs number of unknowns. Always remember:

To solve for a system of n variables, you require n distinct and linear equations

While there are a number of important exceptions to the rule, looking for opportunities to apply it to DS questions will save you a LOT of valuable time as you avoid calculations.

From the stem: 1 equation, 3 unknowns. Missing: 2 distinct linear equations.

1) 2 distinct linear equations; no new variables: sufficient!
2) 1 distinct linear equation; doesn't get rid of both unwanted variables in one shot: insufficient!

(1) is suff, (2) isn't: choose A!