It is currently 20 Nov 2017, 12:29

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

The Carson family will purchase three used cars. There are

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

1 KUDOS received
Intern
Intern
avatar
Joined: 26 Sep 2012
Posts: 38

Kudos [?]: 38 [1], given: 46

Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 01 Aug 2014, 04:06
1
This post received
KUDOS
enigma123 wrote:
The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

A. 24
B. 32
C. 48
D. 60
E. 192


I approached this question as follows:

1) Identified the total number of possible selections (including with two same colors).
We need to select 3 cars out of 8.
8! / 3!*5! = 56 options

2) Deducted the number of options when two cars of the same color are selected:
If we select, for example, green & green => 6 cars for the third option are left => so we need to deduct this 6 options.
We have 4 different colors => 6 * 4 = 24
56 - 24 = 32 (D)

Kudos [?]: 38 [1], given: 46

1 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 13 Jun 2013
Posts: 278

Kudos [?]: 479 [1], given: 13

Premium Member
Re: The Carson family will purchase three used cars. There are two models [#permalink]

Show Tags

New post 08 Nov 2014, 10:27
1
This post received
KUDOS
Avisek47 wrote:
The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

(A) 24
(B) 32
(C) 48
(D) 60
(E) 192


Case 1) all the three selected cars are from model A =4C3=4
Case 2) all the three selected cars are from model B= 4C3=4
Case 3) 1 car from model A and 2 car from model B = 4C1*3C2(because all three cars must have different colors)=12
Case 4) 1 car from model B and 2 car from model A = 4C1*3C2=12

thus total number of combinations= 4+4+12+12=32

Kudos [?]: 479 [1], given: 13

Manager
Manager
avatar
Joined: 27 May 2014
Posts: 87

Kudos [?]: 20 [0], given: 21

Re: The Carson family will purchase three used cars. There are two models [#permalink]

Show Tags

New post 08 Nov 2014, 15:42
Bunuel can you explain why we can't use the following method:

I can choose any 8 cars for my first car
I can choose any 6 cars for my second
I can choose any 4 cars for my third

8x6x4=192

Kudos [?]: 20 [0], given: 21

Manager
Manager
avatar
Joined: 21 Sep 2012
Posts: 218

Kudos [?]: 206 [0], given: 31

Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)
Re: The Carson family will purchase three used cars. There are two models [#permalink]

Show Tags

New post 08 Nov 2014, 21:24
Select 3 colors out of total 4 colors= 4C3= 4
Each color has 2 different models = 2^3=8
Total combinations = 4*8=32

Kudos [?]: 206 [0], given: 31

Senior Manager
Senior Manager
avatar
Joined: 23 Jun 2009
Posts: 360

Kudos [?]: 134 [0], given: 80

Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Re: The Carson family will purchase three used cars. There are two models [#permalink]

Show Tags

New post 09 Nov 2014, 00:06
bankerboy30 wrote:
Bunuel can you explain why we can't use the following method:

I can choose any 8 cars for my first car
I can choose any 6 cars for my second
I can choose any 4 cars for my third

8x6x4=192

Assume that you selected Green A Red B and Black A in one selection and Red B Green A and Black A in the second selection. These two selections are same. So sorting (or permutations) work here. You have to divide 192 by factorial of 3. Thus 192/3!=32 ;)

Kudos [?]: 134 [0], given: 80

Intern
Intern
avatar
Joined: 26 Jul 2015
Posts: 20

Kudos [?]: 6 [0], given: 12

GMAT ToolKit User
Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 23 Aug 2015, 11:53
I found the slot method to be simple to use for this question. There are 8 options for Car 1, 6 options for Car 2, and 4 options for Car 3. For example, if I selected A_Red for Car 1, I cannot select B_Red for Car 2 and so forth. Finally, the order in which I select the cars does not matter so I divide the number by 3!.

The answer then becomes: 8*6*4/3! = 32 (B)

Kudos [?]: 6 [0], given: 12

Intern
Intern
avatar
Joined: 10 May 2015
Posts: 30

Kudos [?]: 3 [0], given: 268

GMAT ToolKit User
Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 26 Oct 2015, 21:45
VeritasPrepKarishma wrote:
deya wrote:
Hello,
I have solved the problem in the following way. From the question I came to know that there are two categories of cars of 4 different colors each. So, in total there are 8 cars. We need to select 3 cars of different colors. So, we can do it in two ways. We can select 2 cars from category A and one from B------> C(4,2)* C(2,1) = 12 ----> C(2,1) for B as we have already selected two colors from A so we are left with the rest 2 colors and we have selected 1 from them.

Secondly, we can select 1 car from A and 2 cars from B-----> C(4,1) * C(3,2) = 12

Therefore, total ways of combinations = 12+12 = 24. But, unfortunately this is not the correct answer. Kindly tell me where I went wrong.

Thank You in advance. :-D


What about selecting all three cars from A or all three cars from B?

All three cars can have the same model. The only constraint is that they need to be of different colors.

Select all from A - In 4C3 = 4 ways
Select all from B - In 4C3 = 4 ways

Total = 24+4+4 = 32 ways



Hi Karishma,
Need some help here.

I tried the method given above as well.But I'm confused regarding this.If I'm selecting 2 cars of Model A in 4c2 ways and correspondingly I select I car from Model B in 2c1 ways(2 colors removed because already selected in A) then why are not taking the cases other way around i.e 2 cars of Model B in 4c2 and 1 car of B in 2c1.After all,they are all different cars.Won't that be added to the combinations as well. In that case, the number of combinations would become (12+12+4)*2=56 ways..Kindly put your thoughts here.

Kudos [?]: 3 [0], given: 268

Expert Post
2 KUDOS received
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7738

Kudos [?]: 17817 [2], given: 235

Location: Pune, India
Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 26 Oct 2015, 22:34
2
This post received
KUDOS
Expert's post
davesinger786 wrote:
VeritasPrepKarishma wrote:
deya wrote:
Hello,
I have solved the problem in the following way. From the question I came to know that there are two categories of cars of 4 different colors each. So, in total there are 8 cars. We need to select 3 cars of different colors. So, we can do it in two ways. We can select 2 cars from category A and one from B------> C(4,2)* C(2,1) = 12 ----> C(2,1) for B as we have already selected two colors from A so we are left with the rest 2 colors and we have selected 1 from them.

Secondly, we can select 1 car from A and 2 cars from B-----> C(4,1) * C(3,2) = 12

Therefore, total ways of combinations = 12+12 = 24. But, unfortunately this is not the correct answer. Kindly tell me where I went wrong.

Thank You in advance. :-D


What about selecting all three cars from A or all three cars from B?

All three cars can have the same model. The only constraint is that they need to be of different colors.

Select all from A - In 4C3 = 4 ways
Select all from B - In 4C3 = 4 ways

Total = 24+4+4 = 32 ways



Hi Karishma,
Need some help here.

I tried the method given above as well.But I'm confused regarding this.If I'm selecting 2 cars of Model A in 4c2 ways and correspondingly I select I car from Model B in 2c1 ways(2 colors removed because already selected in A) then why are not taking the cases other way around i.e 2 cars of Model B in 4c2 and 1 car of B in 2c1.After all,they are all different cars.Won't that be added to the combinations as well. In that case, the number of combinations would become (12+12+4)*2=56 ways..Kindly put your thoughts here.


Right, so 4C2 * 2C1 = 12 (2 cars of model A and 1 car of model B)
From where do we get the other 12? It represents the case where we select 2 cars of model B and 1 car of model A. So these are already accounted for.

Next, you select all from A - In 4C3 = 4 ways
and then select all from B - In 4C3 = 4 ways

Total = 12+12+4+4 = 32 ways
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 17817 [2], given: 235

Intern
Intern
avatar
Joined: 17 Oct 2015
Posts: 15

Kudos [?]: [0], given: 20

Schools: LBS '18 (S)
Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 02 Nov 2015, 08:00
I understand the slot method to solve this problem. i.e 8*6*4/3! but when I saw this problem first I tried to solve it by subtracting the combinations where colours of the cars are the same. I'm still not able to find the solution through this, can some one point out the mistake?

Total Choices: 8*8*8 = 512(unrestricted, I assume you can pic a same colour same model 3 times)

Choices where two cars are Blue: 2*2*8 = 32 (Say Blue for first two choices, then remaining can be any of the 8 remaining choices)
Since 4 colours = 32*4 = 128


So for all cars to have a different colour = total - two-cars-same - three-cars-same = 512 - 128 = 384

I guess I need to divide in each step by 3! since order doesn't matter which would just mean my answer is 512/3! - 128/3! = 64... Which is still not the answer.. .What am I missing?

Kudos [?]: [0], given: 20

Current Student
avatar
Joined: 10 Apr 2015
Posts: 74

Kudos [?]: 11 [0], given: 101

Location: India
Concentration: Healthcare, Strategy
GMAT 1: 680 Q49 V34
GPA: 2.4
WE: Marketing (Health Care)
Reviews Badge
Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 04 Nov 2015, 08:44
Hi Bunuel, VeritasPrepKarishma, Engr2012 , manpreetsingh86 , WoundedTiger

I have approached the problem in this manner.

Total cars are 8.

Let's tag these cars Bl(A), Bk(A), R(A), G(A), Bl(B), Bk(B), R(B), G(B)

Now we have 4 families of cars with similar colors: Bl(A), Bl(B) ; Bk(A), Bk(B); R(A), R(B); G(A), G(B)

As we can't chose 2 cars with same colors, we have to pick 3 different colors from available 4 colors.

So, we have 8 cars and we have to select 3 cars with different cars.

The first car - 1 out of 8 => 8c1
Second car - 1 out of 6 => 6c1
Third car - 1 out of 4 => 4c1

This will be 192.

But guys don't stop here.

Imagine we have picked up Bl(A), Bk(B), G(B)....These cars can be selected in any order.

So the no. of ways will be repeated in the above mentioned 192 arrangements.

So we have to divide the replication.

The replication is 3! ways = 6 ways

So finally the number of ways we can select 3 cars out of 8 cars (of different colors) = 192/6 = 32

Short cut:

(8c1 * 6c1 * 4c1)/ 3! = 32

Regards

Kudos [?]: 11 [0], given: 101

Manager
Manager
User avatar
Joined: 12 Jan 2015
Posts: 222

Kudos [?]: 80 [0], given: 79

Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 01 Feb 2016, 11:50
VeritasPrepKarishma wrote:
ashish8 wrote:
calreg11 wrote:
Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?


Bunuel, Calgreg, if I follow the MGMAT method I don't understand what over counting they are talking about. I understand the 8 for the first choice, 6 for the second, and 4 for the third, but why do the division afterwards?


The method of selecting 8 for the first choice, 6 for the second and 4 for the third is what we call 'basic counting principle'. When you do 8*6*4, you are effectively selecting and ARRANGING the cars: you say 'The FIRST car is selected in 8 ways, the SECOND car in 6 ways etc'. But you don't have a first second third car. You only have a group of 3 cars. So to un-arrange (so to say), you need to divide by 3!

I have explained this concept in this post: http://www.veritasprep.com/blog/2011/11 ... binations/



Hi VeritasPrepKarishma,

In this question I am just doing 8C1*6C1*4C1

As you said, by doing this we are selecting and then arranging the cars. Therefore this approach is wrong.
But here I am applying Combination, which ONLY select the things.

If we would have done 8P1*6P1*4P1--> it means we are first selecting and then arranging

Permutation involves 2 steps- Selection + Arrangement
But combination involves 1 step= only selection

So, I am not able to understand why we need to divide by 3!.

Please assist where is my reasoning wrong.
_________________

Thanks and Regards,
Prakhar

Kudos [?]: 80 [0], given: 79

Expert Post
2 KUDOS received
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7738

Kudos [?]: 17817 [2], given: 235

Location: Pune, India
Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 01 Feb 2016, 21:22
2
This post received
KUDOS
Expert's post
RAHKARP27071989 wrote:


Hi VeritasPrepKarishma,

In this question I am just doing 8C1*6C1*4C1

As you said, by doing this we are selecting and then arranging the cars. Therefore this approach is wrong.
But here I am applying Combination, which ONLY select the things.

If we would have done 8P1*6P1*4P1--> it means we are first selecting and then arranging

Permutation involves 2 steps- Selection + Arrangement
But combination involves 1 step= only selection

So, I am not able to understand why we need to divide by 3!.

Please assist where is my reasoning wrong.


When you "select" using the combinations formula multiple times on the same group, you are arranging by default.

There are 8 cars.
You are selecting one using 8C1. Say you select Red model A.
Next using 6C1 you select Blue model A.
Next, using 4C1 you select Green model B.
This is one way of selection.


In another case,
Using 8C1, say you select Blue model A.
Next using 6C1, you select Red model A.
Next, using 4C1 you select Green model B.
This is another way of selection.

These will be counted as 2 selections even though the cars selected are the same: Blue model A, Red model A and Green model B.

So when the selection is made by selecting multiple times from the same bunch, you do end up arranging. Hence you need to divide by 3!.

As for 8P1, note that this is the number of ways to select 1 car out of 8 and then arrange the selected 1 car. It doesn't make any sense because you cannot "arrange" one object.
Actually 8C1 and 8P1 are the same: 8!/7!

Check out my last week's post. You might find it useful.
http://www.veritasprep.com/blog/2016/01 ... mbination/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 17817 [2], given: 235

Manager
Manager
avatar
B
Joined: 23 Dec 2013
Posts: 235

Kudos [?]: 14 [0], given: 21

Location: United States (CA)
GMAT 1: 760 Q49 V44
GPA: 3.76
Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 21 May 2017, 16:39
enigma123 wrote:
The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

A. 24
B. 32
C. 48
D. 60
E. 192


The total number of cars available to pick is 8 (4 colors and 2 sizes). So for the first choice, there are 8 possible options. After the first choice, cars of that color cannot be picked, so the remaining cars available are 6. Same process for the last choice: 4 options.

8*6*4 = 192 choices.

But the order does not matter, so to eliminate the order, I need to divide by the factorial of the number of decisions (3!).

8*6*4/3*2 = 32.

Kudos [?]: 14 [0], given: 21

Intern
Intern
avatar
B
Joined: 23 Feb 2017
Posts: 38

Kudos [?]: 1 [0], given: 8

Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 07 Jul 2017, 19:13
Hi,
We have A model(B,BL,R,G) and B model(B,BL,R,G)
1. selecting all 3 cars from A none from B= 4C3 = 4
2. selecting all 3 cars from B none from A = 4C3 = 4
3. selecting 2 cars from A and 1 car from B = 4C2*2C1(as we should not select the same color as in A model) = 12
4. selecting 1 car from A and 2 cars from B = 4c1*3C2( as we should not select the same color as in A model) = 12
total = 4+4+12+12 = 32

ans : B

Please correct me if I am wrong.

Kudos [?]: 1 [0], given: 8

Intern
Intern
avatar
B
Joined: 28 Apr 2017
Posts: 38

Kudos [?]: 3 [0], given: 51

Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 08 Jul 2017, 05:53
Hi Bunuel

Help me understand why I am wrong
4C1*3C1*2C1*2*2*2

Kudos [?]: 3 [0], given: 51

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42269

Kudos [?]: 132809 [0], given: 12376

Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 08 Jul 2017, 06:35
deepudiscover wrote:
Hi Bunuel

Help me understand why I am wrong
4C1*3C1*2C1*2*2*2


I guess with 4C1*3C1*2C1 = 4*3*2 you are selecting 3 colors for cars. This will give duplications because the order of the colors does not matter. {blue, black, red} color selection is the same as {red, blue, black}, so you should divide 4*3*2 by 3! to dis-arrange and you'll get 4, which is the same as 4C3 = 4 in my solution above. You could simply list all possibilities to avoid formula:

{blue, black, red}
{blue, black, green}
{black, red, green}
{blue, red, green}
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 132809 [0], given: 12376

Manager
Manager
avatar
S
Joined: 23 Jul 2015
Posts: 148

Kudos [?]: 38 [0], given: 29

GMAT 1: 730 Q50 V40
GMAT ToolKit User
Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 08 Jul 2017, 11:36
In total 4 colors, each with 2 choices (Car A or B)

Select 1 Car from each color = 2C1 * 2C1 * 2C1 = 8
Select 3 colors from total of 4 = 4C3 = 4

4* 8 = 32

Ans. B

Kudos [?]: 38 [0], given: 29

Intern
Intern
avatar
B
Joined: 20 Nov 2016
Posts: 5

Kudos [?]: 1 [0], given: 4

Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 26 Jul 2017, 23:48
Hi There,
Thought I'd share my method of solving this. This is a bit different from the ones mentioned above.

Total Cars we can pick from = 8
i:e;
Model A:
1.Blue
2.Black
3.Red
4.Green

Model B

5.Blue
6.Black
7.Red
8.Green

The question looks at each car differently, so I have ignored which model they belong to.

STEP 1 - FINDING THE TOTAL NO. OF WAYS TO PICK 3 CARS
Total no. of ways we can pick = 8 x 6 x 4 (ie: 192 ways)

Explanation:

1.To start off with, you have 8 ways you can pick a car.
2.For example one you pick 1.blue, you cant pick 5.blue. This drops the total no. of cars you can pick to 6.
3.Now, once you pick car 2.Red, you cant pick 6. Red. This means you have 4 cars to pick from
4.This leaves you with 8 x 6 x 4 total ways.

STEP 2 - ELIMINATING REPEATING VALUES
1. You have 192 ways of picking 3 cars. However, this questions does not need you to maintain a specific sequence:

Let's assume you picked red, blue, black. The value of 192 also takes into account blue, black, red.
We need to eliminate repeating values

2. No. of repeating values = no. of ways the three cars can be arranged.
Lets take RED, BLUE and BLACK. These three cars can be arranged as -
RED BLUE BLACK
RED BLACK BLUE
BLUE BLACK RED
BLUE RED BLACK
BLACK BLUE RED
BLACK RED BLUE

There are a total of 6 ways in which any three colors can be arranged.

STEP 3 - GETTING THE ANSWER

Ways in which cars can be picked = TOTAL NO. OF WAYS OF SELECTING THE CARS / NO. OF REPEATING VALUES
ie- 192/ 6 = 32.

Hope this helps :)

Kudos [?]: 1 [0], given: 4

Director
Director
User avatar
G
Joined: 28 Mar 2017
Posts: 575

Kudos [?]: 145 [0], given: 132

Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 27 Aug 2017, 14:44
Hello VeritasPrepKarishma, Please reply.

Please could you point out where m i going wrong in my following explanation:

We have 2 cars: A and B.

In order to buy 3 cars we can have AAA, AAB,ABB or BBB. --> 4 ways
Out of 4 colors we can have 3 --> 4*3*2=24

Therefore, total ways=24*4=96

What am i missing here. Please reply.

Regards
_________________

Kudos if my post helps!

Helpful links:
1. e-GMAT's ALL SC Compilation

Kudos [?]: 145 [0], given: 132

Intern
Intern
avatar
B
Joined: 21 Sep 2016
Posts: 28

Kudos [?]: 2 [0], given: 261

Re: The Carson family will purchase three used cars. There are [#permalink]

Show Tags

New post 11 Sep 2017, 13:47
I'm not 100% comfortable in combinatorics yet, but I did it this way (not sure, though, if that's the correct one):

There are 2 models: A and B.

Hence, we have 4 combinations of models:
1) AAA;
2) BBB;
3) AAB;
4) BBA.

1)=2)= 4C3*2 = 8. As in, we must choose 3 colours out of 4 for the same model, then we multiply by 2, since we have models A and B.

3)=4) = 4C2*2C1*2 = 24. As in, we must choose 2 colours out of 4 for the model that was bought 2 times, then 1 colour out of the remaining 2 for the other model. Since we have 2 ways of grouping the 2 models, we multiply by 2.

24+8=32.

Kudos [?]: 2 [0], given: 261

Re: The Carson family will purchase three used cars. There are   [#permalink] 11 Sep 2017, 13:47

Go to page   Previous    1   2   3    Next  [ 42 posts ] 

Display posts from previous: Sort by

The Carson family will purchase three used cars. There are

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.