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Sally has five red cards numbered 1 through 5 and four blue

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Sally has five red cards numbered 1 through 5 and four blue  [#permalink]

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New post Updated on: 05 Oct 2013, 04:27
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Question Stats:

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Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?

A. 8
B. 9
C. 10
D. 11
E. 12

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Originally posted by fozzzy on 05 Oct 2013, 04:08.
Last edited by Bunuel on 05 Oct 2013, 04:27, edited 2 times in total.
Edited the question.
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Re: Sally has five red cards numbered 1 through 5 and four blue  [#permalink]

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New post 05 Oct 2013, 04:26
5
3
fozzzy wrote:
Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards??

A) 8
B) 9
C) 10
D) 11
E) 12


The cards are stacked so that the colors alternate: R-B-R-B-R-B-R-B-R.

We are also told that the number on each red card divides evenly into the number on each neighboring blue card.

There are two primes in Blue cards: 3 and 5. Their divisors are 1 and 3 AND 1 and 5, respectively. Thus R1 must be between R3 and R5, R3 must be by B3 and R5 must be by B5: 3-3-1-5-5.

Next, add multiple of 3 to the left of 3: 6-3-3-1-5-5.

Add factor of 6 to the left of 6: 2-6-3-3-1-5-5.

Add multiple of 2 to the left of 2: 4-2-6-3-3-1-5-5.

And finally add factor of 4 to the left of 4: 4-4-2-6-3-3-1-5-5.

The sum of the numbers on the middle three cards is 6+3+3=12.

Answer: E.

Hope it's clear.
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Re: Sally has five red cards numbered 1 through 5 and four blue  [#permalink]

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New post 17 Oct 2013, 00:10
Bunuel wrote:
fozzzy wrote:
Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards??

A) 8
B) 9
C) 10
D) 11
E) 12


The cards are stacked so that the colors alternate: R-B-R-B-R-B-R-B-R.

We are also told that the number on each red card divides evenly into the number on each neighboring blue card.

There are two primes in Blue cards: 3 and 5. Their divisors are 1 and 3 AND 1 and 5, respectively. Thus R1 must be between R3 and R5, R3 must be by B3 and R5 must be by B5: 3-3-1-5-5.

Next, add multiple of 3 to the left of 3: 6-3-3-1-5-5.

Add factor of 6 to the left of 6: 2-6-3-3-1-5-5.

Add multiple of 2 to the left of 2: 4-2-6-3-3-1-5-5.

And finally add factor of 4 to the left of 4: 4-4-2-6-3-3-1-5-5.

The sum of the numbers on the middle three cards is 6+3+3=12.

Answer: E.

Hope it's clear.


Hi Bunuel, Does such questions actualy appear on GMAT..?? its really tough and makes my mind wooof..!!
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Re: Sally has five red cards numbered 1 through 5 and four blue  [#permalink]

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New post 17 Oct 2013, 03:11
ishdeep18 wrote:
Bunuel wrote:
fozzzy wrote:
Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards??

A) 8
B) 9
C) 10
D) 11
E) 12


The cards are stacked so that the colors alternate: R-B-R-B-R-B-R-B-R.

We are also told that the number on each red card divides evenly into the number on each neighboring blue card.

There are two primes in Blue cards: 3 and 5. Their divisors are 1 and 3 AND 1 and 5, respectively. Thus R1 must be between R3 and R5, R3 must be by B3 and R5 must be by B5: 3-3-1-5-5.

Next, add multiple of 3 to the left of 3: 6-3-3-1-5-5.

Add factor of 6 to the left of 6: 2-6-3-3-1-5-5.

Add multiple of 2 to the left of 2: 4-2-6-3-3-1-5-5.

And finally add factor of 4 to the left of 4: 4-4-2-6-3-3-1-5-5.

The sum of the numbers on the middle three cards is 6+3+3=12.

Answer: E.

Hope it's clear.


Hi Bunuel, Does such questions actualy appear on GMAT..?? its really tough and makes my mind wooof..!!


Yes, the question is hard, though I think you can expect such questions if you are doing well on the test.
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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

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Re: Sally has five red cards numbered 1 through 5 and four blue  [#permalink]

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New post 12 Mar 2014, 08:53
Bunuel wrote:
fozzzy wrote:
Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards??

A) 8
B) 9
C) 10
D) 11
E) 12


The cards are stacked so that the colors alternate: R-B-R-B-R-B-R-B-R.

We are also told that the number on each red card divides evenly into the number on each neighboring blue card.

There are two primes in Blue cards: 3 and 5. Their divisors are 1 and 3 AND 1 and 5, respectively. Thus R1 must be between R3 and R5, R3 must be by B3 and R5 must be by B5: 3-3-1-5-5.

Next, add multiple of 3 to the left of 3: 6-3-3-1-5-5.

Add factor of 6 to the left of 6: 2-6-3-3-1-5-5.

Add multiple of 2 to the left of 2: 4-2-6-3-3-1-5-5.

And finally add factor of 4 to the left of 4: 4-4-2-6-3-3-1-5-5.

The sum of the numbers on the middle three cards is 6+3+3=12.

Answer: E.

Hope it's clear.




Hi bunuel,

I am not able to understand the logic behind the sequence.
Why it can't be like this
2-6-3-3-4-4-1-5-5

And have sum as 3+4+4 = 11

Or any other combination
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Re: Sally has five red cards numbered 1 through 5 and four blue  [#permalink]

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New post 12 Mar 2014, 09:30
riteshgmat wrote:
Bunuel wrote:
fozzzy wrote:
Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards??

A) 8
B) 9
C) 10
D) 11
E) 12


The cards are stacked so that the colors alternate: R-B-R-B-R-B-R-B-R.

We are also told that the number on each red card divides evenly into the number on each neighboring blue card.

There are two primes in Blue cards: 3 and 5. Their divisors are 1 and 3 AND 1 and 5, respectively. Thus R1 must be between R3 and R5, R3 must be by B3 and R5 must be by B5: 3-3-1-5-5.

Next, add multiple of 3 to the left of 3: 6-3-3-1-5-5.

Add factor of 6 to the left of 6: 2-6-3-3-1-5-5.

Add multiple of 2 to the left of 2: 4-2-6-3-3-1-5-5.

And finally add factor of 4 to the left of 4: 4-4-2-6-3-3-1-5-5.

The sum of the numbers on the middle three cards is 6+3+3=12.

Answer: E.

Hope it's clear.




Hi bunuel,

I am not able to understand the logic behind the sequence.
Why it can't be like this
2-6-3-3-4-4-1-5-5

And have sum as 3+4+4 = 11

Or any other combination


Your sequence is not possible because it violates the condition that the number on each red card divides evenly into the number on each neighboring blue card.

2-6-3-3-4-4-1-5-5 --> 4 there is NOT a factor of neighboring 3.

While in the following sequence 4-4-2-6-3-3-1-5-5 the number on EACH red card is a factor of EACH neighboring blue card.

Hope it's clear.
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Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

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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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Sally has five red cards numbered 1 through 5 and four blue  [#permalink]

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New post 30 Jul 2014, 02:06
1
fozzzy wrote:
Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?

A. 8
B. 9
C. 10
D. 11
E. 12


1,2,3,4,5
3,4,5,6

Rule states that the red card must be a multiple of the 2 blue cards next to it. Therefore, we should start off with the most multiple (6)
2-6-3

1,2,3,4,5
3,4,5,6
We should then attempt to connect 2 & 3 with a multiple among the remaining red cards
4-2-6-3-3

1,2,3,4,5
3,4,5,6
We then need to look for a factor of 4 & 3.
4-4-2-6-3-3-1

1,2,3,4,5
3,4,5,6
Fill in missing Blue 5 & Red 5 on the far right side
4-4-2-6-3-3-1-5-5

We are looking for the sum of the 3 middle digits.
4-4-2-6-3-3-1-5-5

6+3+3=12

Answer is E
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Re: Sally has five red cards numbered 1 through 5 and four blue  [#permalink]

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New post 05 Dec 2017, 00:09
Why can't it be 3-3-6-2-4-4-1-5-5

Sum here equals 10.
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Re: Sally has five red cards numbered 1 through 5 and four blue  [#permalink]

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New post 05 Dec 2017, 00:13
strawhat316 wrote:
Why can't it be 3-3-6-2-4-4-1-5-5

Sum here equals 10.


You should read a question and solution more carefully.

Your sequence is not possible because it violates the condition that the number on each red card divides evenly into the number on each neighboring blue card.

3-3-6-2-4-4-1-5-5 --> 6, 4, and 1 are NOT factor of neighbouring cards.
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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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Re: Sally has five red cards numbered 1 through 5 and four blue  [#permalink]

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New post 05 Dec 2017, 00:17
Got it. I got confused about the order. Thanks Bunuel

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Re: Sally has five red cards numbered 1 through 5 and four blue &nbs [#permalink] 05 Dec 2017, 00:17
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Sally has five red cards numbered 1 through 5 and four blue

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