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A rental car agency purchases fleet vehicles in two sizes: a
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02 Dec 2010, 14:14
Question Stats:
79% (00:47) correct 21% (00:50) wrong based on 1147 sessions
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A rental car agency purchases fleet vehicles in two sizes: a fullsize 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.
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Re: Car Dealer  Data Sufficiency
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02 Dec 2010, 17:03
Yalephd wrote: A rental car agency purchases fleet vehicles in two sizes: a fullsize 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 Ctrap question. Ctrap 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 fullsize 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 nonnegative 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=6610F 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: eunicesoldseveralcakesifeachcakesoldforeither109602.htmlmarthaboughtseveralpencilsifeachpencilwaseithera100204.htmlarentalcaragencypurchasesfleetvehiclesintwosizesa105682.htmljoeboughtonlytwentycentstampsandthirtycentstamps106212.htmlacertainfruitstandsoldapplesfor070eachandbananas101966.htmljoannaboughtonly015stampsand029stampshowmany101743.htmlatanamusementparktomboughtanumberofredtokensand126814.htmlcollectionsconfusedneedahelp81062.htmlHope it helps.
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Re: Car Dealer  Data Sufficiency
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02 Dec 2010, 17:46
Thank you, Bunuel. I thought it was C. Your explanation was clear as always. Bunuel wrote: Yalephd wrote: A rental car agency purchases fleet vehicles in two sizes: a fullsize 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 Ctrap question. Ctrap 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 fullsize 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 nonnegative 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=6610F 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: gmatprep292785.html?hilit=linear%20typeHope it helps.



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Re: Car Dealer  Data Sufficiency
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02 Dec 2010, 19:46
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. http://gmatquant.blogspot.com/2010/11/integralsolutionsofaxbyc.html
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Re: A rental car agency purchases fleet vehicles in two sizes: a
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31 May 2013, 05:43



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Re: A rental car agency purchases fleet vehicles in two sizes: a
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31 May 2013, 22:16
Yalephd wrote: A rental car agency purchases fleet vehicles in two sizes: a fullsize 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. 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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Re: A rental car agency purchases fleet vehicles in two sizes: a
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02 Jun 2013, 14:13
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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Re: A rental car agency purchases fleet vehicles in two sizes: a
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18 Nov 2013, 11:32
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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Re: A rental car agency purchases fleet vehicles in two sizes: a
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15 Dec 2017, 08:14
VeritasPrepKarishma wrote: 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. http://gmatquant.blogspot.com/2010/11/integralsolutionsofaxbyc.htmlYour blog posts in comments are always helpful, thankyou.



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Re: A rental car agency purchases fleet vehicles in two sizes: a
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11 Feb 2018, 09:34
Bunuel wrote: 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 104 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 104 = 6 so I need 6 * 9.000 than trying to find for which value we have a multiple of 9



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Re: A rental car agency purchases fleet vehicles in two sizes: a
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05 Oct 2018, 03:28
Does C trap question like this does really fall under sub600 level? BunuelBunuel wrote: Yalephd wrote: A rental car agency purchases fleet vehicles in two sizes: a fullsize 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 Ctrap question. Ctrap 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 fullsize 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 nonnegative 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=6610F 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: http://gmatclub.com/forum/eunicesolds ... 09602.htmlhttp://gmatclub.com/forum/marthabought ... 00204.htmlhttp://gmatclub.com/forum/arentalcar ... 05682.htmlhttp://gmatclub.com/forum/joeboughton ... 06212.htmlhttp://gmatclub.com/forum/acertainfru ... 01966.htmlhttp://gmatclub.com/forum/joannabought ... 01743.htmlhttp://gmatclub.com/forum/atanamuseme ... 26814.htmlhttp://gmatclub.com/forum/collectionsc ... 81062.htmlHope it helps.
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Re: A rental car agency purchases fleet vehicles in two sizes: a
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05 Oct 2018, 05:45
parthos wrote: Does C trap question like this does really fall under sub600 level? BunuelBunuel wrote: Yalephd wrote: A rental car agency purchases fleet vehicles in two sizes: a fullsize 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 Ctrap question. Ctrap 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 fullsize 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 nonnegative 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=6610F 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: http://gmatclub.com/forum/eunicesolds ... 09602.htmlhttp://gmatclub.com/forum/marthabought ... 00204.htmlhttp://gmatclub.com/forum/arentalcar ... 05682.htmlhttp://gmatclub.com/forum/joeboughton ... 06212.htmlhttp://gmatclub.com/forum/acertainfru ... 01966.htmlhttp://gmatclub.com/forum/joannabought ... 01743.htmlhttp://gmatclub.com/forum/atanamuseme ... 26814.htmlhttp://gmatclub.com/forum/collectionsc ... 81062.htmlHope 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 sub600 level.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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A rental car agency purchases fleet vehicles in two sizes: a
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07 Oct 2018, 03:20
Yalephd wrote: A rental car agency purchases fleet vehicles in two sizes: a fullsize 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. (1) It could be any combination for the two. INSUFFICIENT(2) For $66,000, a maximum of 6 fullsize cars can be bought but that will leave only $6,000 and a compact car is of $9,000. This cannot be the combinationIf 5 full size cars are bought, that leaves $16,000. 1 compact car can be bought and $7,000 This cannot be the combinationIf 4 full size cars are bought, that leaves $26,000. 2 compact cars can be bought and $8,000 This cannot be the combinationIf 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 forTherefore, B is the answer
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