A rental car agency purchases fleet vehicles in two sizes: a

Question

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.

Bunuel · Accepted Answer

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 cannot 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 a multiple of 9; 56 is not; 46 is not; 36 IS A 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. Answer: B. For more on this type of questions check: eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html collections-confused-need-a-help-81062.html Hope it helps.

Yalephd · Answer

Thank you, Bunuel. I thought it was C. Your explanation was clear as always.

KarishmaB · Answer

In my blog post given below, I have discussed in detail how to find integer solutions to linear equations in 2 variables (i.e. equations such as this one)
See if it helps.
https://anaprep.com/algebra-integer-sol ... variables/

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an    alternative    solution!

*New project from GMAT Club!!! Check  HERE

mau5 · Answer

F.S 1 clearly insufficient.

F.S 2 states that 10*f+9*c = 66

The only way in which we could get a 6 in the units digit for the RHS is through the factor 9*c --> 9*4 = 36. Thus, the no of compact cars = 4.Sufficient.

B.

shivanigs · Answer

Stmt 1 : f + c = 7 …insufficient since we can not determine a specific number

Stmt 2 : 10000f + 9000c = 66000

Dividing both sides by 1000

10f + 9c = 66

Since cars can only be represented by integers, we can set the following equation :

9c = 66 – 10f

C = (66 – 10f)/9

We are looking for a number that is divisible by 9.We need to plug in numbers and arrive at a multiple of 9.

C = (66 – 10*3)/9 = 36/9 = 4

Hence the number of compact cars owned is 4.

Answer : B

superpus07 · Answer

Alt.:

st. 2)

$10.000   $9000   Total        # of cars   7   0   70.000           6   1   69.000           5   2   68.000           4   3   67.000           3   4   66.000   ---> Bingo it matches

mtk10 · Answer

Your blog posts in comments are always helpful, thankyou.

Eloquently · Answer

We know that X and Y must be integers and prices are 9.000 or 10.000
Regardless the tens of thousand number, the only way to obtain 6 is if we sold 10-4 cars at 9.000 so we have 4 times 9.000 = 36.000 thus we can determine X which is 3.

So B is sufficient while A is not as said in other posts.

Not sure how clear it is but in my head it goes quicker to do 10-4 = 6 so I need 6 * 9.000 than trying to find for which value we have a multiple of 9

Gmatprep550 · Answer

Does C trap question like this does really fall under sub-600 level?  Bunuel

Bunuel · Answer

The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question. So our stats say that it's sub-600 level.

MsInvBanker · Answer

(1) It could be any combination for the two.  INSUFFICIENT

(2) For $66,000, a maximum of 6 full-size cars can be bought but that will leave only $6,000 and a compact car is of $9,000.  This cannot be the combination

If 5 full size cars are bought, that leaves $16,000. 1 compact car can be bought and $7,000  This cannot be the combination

If 4 full size cars are bought, that leaves $26,000. 2 compact cars can be bought and $8,000  This cannot be the combination

If 3 full size cars are bought, that leaves $36,000. 4 compact cars can be bought with no remaining amount    This is the combination we are looking for

Therefore, B is the answer

bumpbot · Answer

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A rental car agency purchases fleet vehicles in two sizes: a

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	$10.000	$9000	Total
# of cars	7	0	70.000
	6	1	69.000
	5	2	68.000
	4	3	67.000
	3	4	66.000	---> Bingo it matches

GMAT Data Sufficiency (DS) Questions

A rental car agency purchases fleet vehicles in two sizes: a

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615 to 715 in 15 Days (Ria's Strategies)

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