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

Yalephd

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02 Dec 2010, 14:14

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Bunuel

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02 Dec 2010, 17:03

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Yalephd

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 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.

General Discussion

Yalephd

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02 Dec 2010, 17:46

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

Bunuel

Yalephd

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.

Answer: B.

Similar problems: gmat-prep2-92785.html?hilit=linear%20type

Hope it helps.

KarishmaB

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Updated on: 19 Sep 2023, 04:58

Originally posted by KarishmaB on 02 Dec 2010, 19:46.
Last edited by KarishmaB on 19 Sep 2023, 04:58, edited 1 time in total.

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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

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31 May 2013, 05:43

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

*New project from GMAT Club!!! Check HERE

mau5

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31 May 2013, 22:16

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Yalephd

F.S 1 clearly insufficient.

F.S 2 states that 10*f+9*c = 66 [f is the # full size cars;c of compact car]

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.

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02 Jun 2013, 14:13

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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

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18 Nov 2013, 11:32

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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

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15 Dec 2017, 08:14

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VeritasPrepKarishma

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://gmatquant.blogspot.com/2010/11/integral-solutions-of-ax-by-c.html

Your blog posts in comments are always helpful, thankyou.

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11 Feb 2018, 09:34

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Bunuel

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

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

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05 Oct 2018, 03:28

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

Bunuel

Yalephd

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:
https://gmatclub.com/forum/eunice-sold-s ... 09602.html
https://gmatclub.com/forum/martha-bought ... 00204.html
https://gmatclub.com/forum/a-rental-car- ... 05682.html
https://gmatclub.com/forum/joe-bought-on ... 06212.html
https://gmatclub.com/forum/a-certain-fru ... 01966.html
https://gmatclub.com/forum/joanna-bought ... 01743.html
https://gmatclub.com/forum/at-an-amuseme ... 26814.html
https://gmatclub.com/forum/collections-c ... 81062.html

Hope it helps.

Bunuel

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05 Oct 2018, 05:45

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parthos

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

Bunuel

Yalephd

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:
https://gmatclub.com/forum/eunice-sold-s ... 09602.html
https://gmatclub.com/forum/martha-bought ... 00204.html
https://gmatclub.com/forum/a-rental-car- ... 05682.html
https://gmatclub.com/forum/joe-bought-on ... 06212.html
https://gmatclub.com/forum/a-certain-fru ... 01966.html
https://gmatclub.com/forum/joanna-bought ... 01743.html
https://gmatclub.com/forum/at-an-amuseme ... 26814.html
https://gmatclub.com/forum/collections-c ... 81062.html

Hope it helps.

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.

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07 Oct 2018, 03:20

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Yalephd

(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

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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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