Each of 60 cars is parked in one of three empty parking lots

Let the number of cars parked in largest lot = L
Middle lot = M & Smallest lot = S
But M=L-8 & S=L-16
So, L + (L - 8) + (L - 16) = 60
3L - 24 = 60
3L = 84
L = 28

IMO Option D

OFFICIAL EXPLANATION

Hi All,

We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...

Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...

Answer D: 28 cars

IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!

Final Answer:
GMAT assassins aren't born, they're made,
Rich

Let x = number of cars in the LARGEST lot

Given: the largest lot holds 8 more cars than the middle lot.
So, x - 8 = number of cars in the MIDDLE lot

Given: the largest lot holds 16 more cars than the smallest lot.
So, x - 16 = number of cars in the SMALLEST lot


Given: Each of 60 cars is parked in one of three empty parking lots.
We can write: (# of cars in LARGEST lot) + (# of cars in MIDDLE lot) + (# of cars in SMALLEST lot) = 60
Substitute values: (x) + (x - 8) + (x - 16) = 60
Simplify: 3x - 24 = 60
Add 24 to both sides: 3x = 84
Divide both sides by 3: x = 28

Answer: D

Cheers,
Brent

The 3 slots A + B + C total to 60 cars

The highest slot C has 8 more than B and 16 more than A

Let us look at the second-largest answer choice which says C=28

As per question A=12 (16 less than C), B=20(8 less than C) ...This gives a total of 60. Hence Correct answer is D
EMPOWERgmatRichC: This is the correct way to do TTA ...right

Hi saukrit,

YES - that's a perfect use of TEST THE ANSWERS. In these sorts of situations, that Tactic (and some basic Arithmetic) is almost always faster and easier that a typical Algebraic approach - so it's something to keep in mind (especially if you have a pacing problem in the Quant section).

GMAT assassins aren't born, they're made,
Rich

\(L = M + 8 = S + 16\)

Again , L + M + S = 60

Now, Plug in the options and check

(A) Can be straightaway rejected as value of L must be > 16 , Since in this case if L = 12 = S + 16 , Or, S = - 4 ( Not Possible )
(B) If L = 20 , M = 12 & S = 4 , L + M + S = 36
(C) If L = 22 , M = 14 & S = 6 , L + M + S = 42
(D) If L = 28 , M = 20 & S = 12 , L + M + S = 60
(E) If L = 30 , M = 22 & S = 14 , L + M + S = 66

Hence, Answer must be (D)

As the middle lot has 8 cars less than the larger it would be X-8
Smaller Lot is having 16 cars less than the larger that means X-16
As all the lots together having 60 it would be X+(X-8)+(X-16)=60
On solving this it becomes X= 28 Cars
Hence Option D would be apt.