Thank you, Bunuel. I thought it was C. Your explanation was clear as always.
A rental car agency purchases fleet vehicles in two sizes: a full-size car costs $10,000, and a compact costs $9,000. How many compact cars does the agency own?
(1) The agency owns 7 total cars.
(2) The agency paid $66,000 for its cars.
Don't have the OA. I think this is an easy question but I am afraid that it might be deceptively simple. I just want to double check.
This is classic C-trap question. C-trap questions are the questions which are obviously sufficient if we take statements together. When we see such questions we should become very suspicious.
Let # of full-size car be F and # of compact cars be C. Question: C=?
(1) The agency owns 7 total cars --> F+C=7. Clearly insufficient to get C.
(2) The agency paid $66,000 for its cars --> 10,000F+9,000C=66,000 --> 10F+9C=66. Here comes the trap: generally such kind of linear equations (ax+by=c) have infinitely many solutions for x and y, and we can not get single numerical values for the variables. But since F and C represent # of cars then they must be non-negative integers and in this case 10F+9C=66 is no longer simple linear equation it's Diophantine equation (equations whose solutions must be integers only) and for such kind on equations there might be only one combination of x and y (F and C in out case) possible to satisfy it. When you encounter such kind of problems you must always check by trial and error whether it's the case.
Now, it's quite easy to check whether 10F+9C=66 has one or more solutions. 9C=66-10F so 66 minus multiple of 10 must be multiple of 9: 66 is not multiple of 9; 56 is not; 46 is not; 36 IS MULTIPLE OF 9 (F=3 and C=4)
; 26 is not; 16 is not and 6 is not. So only one combination of F and C satisfies equation 10F+9C=66, namely F=3 and C=4. Sufficient.
Similar problems: gmat-prep2-92785.html?hilit=linear%20type
Hope it helps.