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M08-35

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New post 16 Sep 2014, 00:38
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If operation \(@X\) is defined as \(@X = X + 2\) if \(X\) is even and \(@X = X - 1\) if \(X\) is odd, what is \(@(...@(@(@(15)))...)\) 99 times?

A. 120
B. 180
C. 210
D. 225
E. 250
[Reveal] Spoiler: OA

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New post 16 Sep 2014, 00:38
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Official Solution:

If operation \(@X\) is defined as \(@X = X + 2\) if \(X\) is even and \(@X = X - 1\) if \(X\) is odd, what is \(@(...@(@(@(15)))...)\) 99 times?

A. 120
B. 180
C. 210
D. 225
E. 250

\(@(15) = 14\). We have to compute \(@(...@(@(@(14)))...)\) 98 times. Each \(@\) just adds 2 to the previous result. Therefore, \(@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210\).

Answer: C
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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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M08-35 [#permalink]

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New post 03 Feb 2015, 08:30
Bunuel wrote:
Official Solution:

If operation \(@X\) is defined as \(@X = X + 2\) if \(X\) is even and \(@X = X - 1\) if \(X\) is odd, what is \(@(...@(@(@(15)))...)\) 99 times?

A. 120
B. 180
C. 210
D. 225
E. 250

\(@(15) = 14\). We have to compute \(@(...@(@(@(14)))...)\) 98 times. Each \(@\) just adds 2 to the previous result. Therefore, \(@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210\).

Answer: C



if 14 is the first term. common difference (d) = 2 and there are 98 terms then
why Last term is not = 14 + (98-1)(2) ??
why 99 used instead of 98 in the formula : T(last term) = a + (n-1) d

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New post 03 Feb 2015, 10:10
earnit wrote:
Bunuel wrote:
Official Solution:

If operation \(@X\) is defined as \(@X = X + 2\) if \(X\) is even and \(@X = X - 1\) if \(X\) is odd, what is \(@(...@(@(@(15)))...)\) 99 times?

A. 120
B. 180
C. 210
D. 225
E. 250

\(@(15) = 14\). We have to compute \(@(...@(@(@(14)))...)\) 98 times. Each \(@\) just adds 2 to the previous result. Therefore, \(@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210\).

Answer: C



if 14 is the first term. common difference (d) = 2 and there are 98 terms then
why Last term is not = 14 + (98-1)(2) ??
why 99 used instead of 98 in the formula : T(last term) = a + (n-1) d


1. \(@(15) = 14\)
2. \(@(14) = 14 +2=16\)
3. \(@(16) = 14 +2+2=14+2*2=18\)
4. \(@(18) = 14 +2+2+2=14+2*3=20\)
...
99. 14 + 2*98 = 210.
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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

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Re: M08-35 [#permalink]

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New post 14 Aug 2016, 08:40
Bunuel,

Are we sure that this is a 500 level question? Generally, 500 level questions do not involve calculations trick- they are pretty straight forward. I was taking GMAT club test (the 11th set) and found several similar questions to be ranked at 500 level and probably that is the reason the score is not upto the mark.

Just a thought!

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Yash

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New post 09 Dec 2016, 18:18
If 14 is the first term. common difference (d) = 2 and there are 98 terms then
why is the Last term not = 14 + (98-1)(2) ?? n=98.. and 14 is the 99th term.

T(last term) = a + (n-1) d

Please help me understand what am I missing here?

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M08-35 [#permalink]

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New post 21 May 2017, 09:35
HI earnit

I too had the same doubt. It took me some time to realise that the first term of the AP is not 14. In fact, the first term is @(14) = 16. After this, everything made sense.

We have \(Tn = a + (n-1)d\). Where, Tn = Nth term. a = First term, n = number of terms in the sequence, d = common difference of AP
\(Tn = 16+(98-1)*2 = 16+97*2 = 16+ 194 = 210\).

Lesson Learnt : I now think its better to write the series as Bunuel did here. Just so that the "short cut" of using nth term does not lead me into a trap! :?

earnit wrote:
Bunuel wrote:
Official Solution:

If operation \(@X\) is defined as \(@X = X + 2\) if \(X\) is even and \(@X = X - 1\) if \(X\) is odd, what is \(@(...@(@(@(15)))...)\) 99 times?

A. 120
B. 180
C. 210
D. 225
E. 250

\(@(15) = 14\). We have to compute \(@(...@(@(@(14)))...)\) 98 times. Each \(@\) just adds 2 to the previous result. Therefore, \(@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210\).

Answer: C



if 14 is the first term. common difference (d) = 2 and there are 98 terms then
why Last term is not = 14 + (98-1)(2) ??
why 99 used instead of 98 in the formula : T(last term) = a + (n-1) d

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Re: M08-35 [#permalink]

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New post 14 Sep 2017, 20:38
I think by A.P. formula T(last term) = a + (n-1)d
the last term should be = 14 + (98-1)(2)=208.
210 is not correct.

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New post 14 Sep 2017, 22:12

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Re: M08-35   [#permalink] 14 Sep 2017, 22:12
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