GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Jun 2018, 17:08

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M08-35

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 46167

### Show Tags

16 Sep 2014, 00:38
2
2
00:00

Difficulty:

35% (medium)

Question Stats:

81% (01:13) correct 19% (01:54) wrong based on 90 sessions

### HideShow timer Statistics

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 46167

### Show Tags

16 Sep 2014, 00:38
1
1
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, $$@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210$$.

_________________
Current Student
Joined: 06 Mar 2014
Posts: 257
Location: India
GMAT Date: 04-30-2015

### Show Tags

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, $$@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210$$.

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
Math Expert
Joined: 02 Sep 2009
Posts: 46167

### Show Tags

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, $$@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210$$.

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.
_________________
Intern
Joined: 02 Nov 2015
Posts: 21

### Show Tags

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!

Regards
Yash
Manager
Joined: 30 Jun 2015
Posts: 61
Location: Malaysia
Schools: Babson '20
GPA: 3.6

### Show Tags

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

Manager
Joined: 12 Jun 2016
Posts: 221
Location: India
WE: Sales (Telecommunications)

### Show Tags

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, $$@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210$$.

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

_________________

My Best is yet to come!

Intern
Joined: 24 Oct 2016
Posts: 29

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 46167

### Show Tags

14 Sep 2017, 22:12
kishor1 wrote:
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.

_________________
Re: M08-35   [#permalink] 14 Sep 2017, 22:12
Display posts from previous: Sort by

# M08-35

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.