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Sally has five red cards numbered 1 through 5 and four blue
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Updated on: 05 Oct 2013, 03:27
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75% (hard)
Question Stats:
60% (02:43) correct 40% (03:09) wrong based on 185 sessions
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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?
Re: Sally has five red cards numbered 1 through 5 and four blue
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05 Oct 2013, 03:26
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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.
Re: Sally has five red cards numbered 1 through 5 and four blue
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16 Oct 2013, 23:10
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..!!
Re: Sally has five red cards numbered 1 through 5 and four blue
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17 Oct 2013, 02:11
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.
_________________
Re: Sally has five red cards numbered 1 through 5 and four blue
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12 Mar 2014, 07:53
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
Re: Sally has five red cards numbered 1 through 5 and four blue
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12 Mar 2014, 08:30
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.
Sally has five red cards numbered 1 through 5 and four blue
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30 Jul 2014, 01:06
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
Re: Sally has five red cards numbered 1 through 5 and four blue
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04 Dec 2017, 23:13
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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