LGOdream wrote:

A car dealer has 40 cars, each of which is white, red, or blue in color, and large, middle, or small in model. How many white large cars does he have?

(1) All the red and blue cars are either middle or small, and make a total of 19.

(2) The number of large cars is more than the number of the middle and small cars combined.

Correct answer appears to be C but I can't see how.

If all red and blue cars are either middle or small and make a total of 19, that means that all large cars are white, but there could be middle or small white cars also as far as I understand.

If the number of large cars is more than the number of the middle and small cars combined, it means than large cars could be either 20 (L + M/S) or 21 (L).

So far, I'd say E.

Any insight?

So there can be total nine kinds of cars - White large, White middle, White small, Red large,... and so on. Lets name these nine kinds of cars as WL, WM, WS, RL, RM, RS, BL, BM, BS respectively. We need to find the value of WL.

(1) RL=BL=0, and sum RM + RS + BL + BS = 19. This means WL + WM + WS = 40-19 = 21. So white cars in total are 21, but out of which we cant say how many will be large. So

Insufficient.

(2) Large cars > middle and small combined. But this tells us nothing about White or White large cars.

Insufficient.

Combining the two statements, WL + WM + WS = 21. Now only large cars are white cars. Middle and small cars of red/blue colors are 19. Even if we have one more middle or small car, i.e., if WM + WS = 1, then total middle/small cars combined would be 20. And in that case WL = 20 too. Here number of large cars will become equal to number of middle/small cars, and would thus violate statement 2 (large cars have to be more than middle/small combined). So the only way for statement 2 to hold true is if WM + WS = 0, thus WL = 21. White large cars are 21 in number.

Sufficient.

Hence

C answer