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M31-06

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M31-06  [#permalink]

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New post 07 Jun 2015, 08:01
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A
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C
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\(t_1, \ t_2, \ t_3, \ ..., \ t_n, \ ...\)

In the sequence above, each term after the first term is equal to the preceding term plus the constant \(k\). If \(t_1+ t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\)?


A. 8
B. 12
C. 24
D. 32
E. 72

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Re M31-06  [#permalink]

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New post 07 Jun 2015, 08:01
Official Solution:


\(t_1, \ t_2, \ t_3, \ ..., \ t_n, \ ...\)

In the sequence above, each term after the first term is equal to the preceding term plus the constant \(k\). If \(t_1+ t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\)?


A. 8
B. 12
C. 24
D. 32
E. 72


According to the stem:

\(t_2=t_1+k\);

\(t_3=t_2+k=t_1+2k\);

\(t_4=t_3+k=t_1+3k\);

...

\(t_n=t_1+(n-1)k\);

Since \(t_1+t_3+t_5+t_7=32\), then \(t_1+(t_1+2k)+(t_1+4k)+(t_1+6k)=32\). So, \(t_1+3k=8\)

We need to find the of \(t_2+t_4+t_6=(t_1+k)+(t_1+3k)+(t_1+5k)=3(t_1+3k)\). From above we know that \(t_1+3k=8\), thus \(3(t_1+3k)=3*8=24\).


Answer: C
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Re: M31-06  [#permalink]

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New post 08 Jun 2015, 04:38
t2 = t1 + k
t3 = t2 + K = t2 = 2K

tn = t1 + (n-1)K

equation1 --- > t1+t3+t5+t7 = 32

Replacing individual component with corresponding t1 , equation 1 becomes
4t1 + 12K = 32
t1 + 3K = 8

Now t2 + t4 + t6 = ?
replacing individual component with corresponding t1 + (n-1)K, above equation becomes

3t1 + 9K = 3( t1 + 3K) = 3 * 8 = 24
option C
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Re: M31-06  [#permalink]

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New post 29 Oct 2015, 05:56
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. Where is it stated that k is a positive integer? Why can't k be -1/2?
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Re: M31-06  [#permalink]

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New post 29 Oct 2015, 10:40
danjbon wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. Where is it stated that k is a positive integer? Why can't k be -1/2?


How does it matter whether k is an integer or not? Where in the solution it's mentioned that k is an integer?
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Re: M31-06  [#permalink]

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New post 21 Jul 2016, 05:08
I think this is a high-quality question and I agree with explanation. Nice manipulative question
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Re: M31-06  [#permalink]

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New post 19 Jun 2018, 14:50
Possible alternative method?: plugging in values such that the rules are fulfilled in the original problem.

Set K=2, and t1=2.

Sequence (starting with t1) = 2,4,6,8,10,12,14,16,(...) each term is the previous +2, so given constraint in the problem is fulfilled.

t1+t3+t5+t7=2+6+10+14 = 32 (condition is fulfilled).

Then, t2+t4+t6 = 4+8+12 = 24, therefore (C) is the answer.
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Re: M31-06  [#permalink]

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New post 12 Jul 2018, 22:08
Could you help me understand why there's a square box in front of t3, t5 and 32? I assumed this was to indicate that the 'square box' has a certain value but I see that when t3 is converted to t1 + 2k or t5 is converted to t1 + 4k the box is simply removed. What's the point of the box?
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M31-06  [#permalink]

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New post 12 Jul 2018, 22:28
varun72 wrote:
Could you help me understand why there's a square box in front of t3, t5 and 32? I assumed this was to indicate that the 'square box' has a certain value but I see that when t3 is converted to t1 + 2k or t5 is converted to t1 + 4k the box is simply removed. What's the point of the box?


Those boxes are just formatting errors. Edited. Thank you.
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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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M31-06 &nbs [#permalink] 12 Jul 2018, 22:28
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