hi - can anyone help me solve this one pls - The Beat The GMAT Forum

hi - can anyone help me solve this one pls

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swati.arunahuja: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Tue Jul 15, 2008 1:53 am

by swati.arunahuja » Tue Jul 15, 2008 2:20 am

Ques 1 - A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

Answers - 4,5,6,12,24

Ques 2 (DS) - If 0≠xy, is x/y < 0?

(1) x = – y
(2) – x = – (– y)

Thanks in advance

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Source: Beat The GMAT — Problem Solving | Tue Jul 15, 2008 2:20 am

saeed: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Tue Nov 27, 2007 9:40 pm

by saeed » Tue Jul 15, 2008 2:59 am

ans-1. I guess the answer could be 5. My logic is that by taking 4 color we can simply make 4 code.then by combining two color we can make 4c2=6 codes. so we can make 10 codes. but our task is to make 12 codes. so, in accord with the question the answer should be 5.
I am not sure about my approach. Hope, some GMAT guru will check it.

ans-2. I will go with E.Statement 1 & statement 2 are the same. so the answer should be D/E. As the question asks to know the sign of X or Y,the given statements cant answer the question sufficiently.

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

by sudhir3127 » Tue Jul 15, 2008 3:02 am

Ques 2 (DS) - If 0≠xy, is x/y < 0?

(1) x = – y
(2) – x = – (– y)

either 1 or 2 is sufficent to solve the problem.

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jsl: Master | Next Rank: 500 Posts; Posts: 203; Joined: Wed Jun 04, 2008 1:01 am; Location: Windsor; Thanked: 5 times; GMAT Score:650

by jsl » Tue Jul 15, 2008 3:12 am

For the first, I believe the answer is 5. I basically wrote out all the combinations until I got to 5 - for this, you just need to understand the question. I'm sure there's a mathematical way of doing this though.

For the second, both statements basically show that you have x/-x and this will always resolve to -1. I.e. 1/-1 = -1 and -1/--1 = -1 Thus -1 is always less than 0. Thus answer is C

Can you quote the official answers? Thx

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swati.arunahuja: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Tue Jul 15, 2008 1:53 am

by swati.arunahuja » Tue Jul 15, 2008 3:38 am

Thanks guys - i still can't understand the solution to ans1 - if anyone can help
Official answers -
Ans 1 - 5
Ans 2 - D

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swati.arunahuja: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Tue Jul 15, 2008 1:53 am

by swati.arunahuja » Tue Jul 15, 2008 3:41 am

sudhir3127 wrote:Ques 2 (DS) - If 0≠xy, is x/y < 0?

(1) x = – y
(2) – x = – (– y)

either 1 or 2 is sufficent to solve the problem.

will you be able to tell how you got the ans?

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

by sudhir3127 » Tue Jul 15, 2008 3:44 am

if X= -Y
assume X to be 1 then it will be 1/-1 = -1<0

simillarly -X=Y ..it will be <0

hope it helps

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swati.arunahuja: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Tue Jul 15, 2008 1:53 am

by swati.arunahuja » Tue Jul 15, 2008 3:48 am

sudhir3127 wrote:if X= -Y
assume X to be 1 then it will be 1/-1 = -1<0

simillarly -X=Y ..it will be <0

hope it helps

Thanks

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parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members; Thanked: 146 times

by parallel_chase » Tue Jul 15, 2008 3:51 am

swati.arunahuja wrote:Thanks guys - i still can't understand the solution to ans1 - if anyone can help
Official answers -
Ans 1 - 5
Ans 2 - D

Ques 1.

It is basically a combination problem, since the order of color does not matter.

You have 12 distribution centers.

If you take 4 colors you can code 4 centers with one unique color.
Since the order does not matter you can code another 4C2 centers with combination of 2 colors.
4C2 = 4!/2!2! = 6, therefore with 4 colors you can code 10 (4+6) centers.

Similarly if we take 5 colors, you can code 5 centers with unique colors.
and remaining 5C2 = 5!/2!3! = 10, therefore with 5 colors you can code 15 (5+10) centers.

Hence you need minimum 5 colors to code 12 centers.

Let me know if you still have any doubts.

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swati.arunahuja: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Tue Jul 15, 2008 1:53 am

by swati.arunahuja » Tue Jul 15, 2008 3:59 am

parallel_chase wrote:
swati.arunahuja wrote:Thanks guys - i still can't understand the solution to ans1 - if anyone can help
Official answers -
Ans 1 - 5
Ans 2 - D
Ques 1.

It is basically a combination problem, since the order of color does not matter.

You have 12 distribution centers.

If you take 4 colors you can code 4 centers with one unique color.
Since the order does not matter you can code another 4C2 centers with combination of 2 colors.
4C2 = 4!/2!2! = 6, therefore with 4 colors you can code 10 (4+6) centers.

Similarly if we take 5 colors, you can code 5 centers with unique colors.
and remaining 5C2 = 5!/2!3! = 10, therefore with 5 colors you can code 15 (5+10) centers.

Hence you need minimum 5 colors to code 12 centers.

Let me know if you still have any doubts.

oh nice one - thanks a tonn

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anksbhandari: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Wed Apr 30, 2008 11:05 am; Location: India

Ano. solution for prob 1

by anksbhandari » Tue Jul 15, 2008 7:43 am

Let go by option

1. 4 for example: a, b, c, d

now we can assign: a - 1, b- 2, c- 3 , d- 4, ab - 5, ac - 6, ad - 7, bc - 8, bd - 9 , and cd - 10.

doesn't solve our problem. so take next option:

2, 5 : a, b, c, d, e,

now repeat the statregy of option 1 we can assign code to 15 items

therefore , 5 is the right answer

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san2009: Master | Next Rank: 500 Posts; Posts: 171; Joined: Fri Apr 16, 2010 1:02 am; Thanked: 1 times

by san2009 » Tue Apr 27, 2010 11:41 pm

can an instructor pls share a good method for solving this
i get to five colors by just thinking through the problem
thanks

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harshavardhanc: Legendary Member; Posts: 526; Joined: Sat Feb 21, 2009 11:47 pm; Location: India; Thanked: 68 times; GMAT Score:680

by harshavardhanc » Wed Apr 28, 2010 12:22 am

san2009 wrote:can an instructor pls share a good method for solving this
i get to five colors by just thinking through the problem
thanks

I'm not an instructor, but can suggest a way. Anyway, when you said "thinking thru the problem"..... that's what expected of us.... right?

Approach:

suppose, you have n colors with you.

Your task: by choosing 1 color at a time OR by choosing 2 colors at a time, you should be able to get unique code which identifies these 12 DCs.

*Choosing translates to combination in counting.
*OR translates to "+ "

mathematically, your task is :

nC1 + nC2 > 12.

if n=3, 3C1 + 3C2 = 6 .....not enough
if n=4, 4C1 + 4C2 = 10 ...not enough
if n=5, 5C1 + 5C2 = 15 ...satisfies the condition.

Hence, we need at least 5 different colors to assign a code which satisfies the condition in the question.

HTH.

Regards,
Harsha

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san2009: Master | Next Rank: 500 Posts; Posts: 171; Joined: Fri Apr 16, 2010 1:02 am; Thanked: 1 times

by san2009 » Wed Apr 28, 2010 1:32 am

sorry - i don't do combinations/permutations.... following MGMAT guides

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