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# If the average of a sequence of consecutive multiple of 18 is 153, and

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If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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17 Sep 2015, 01:30
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If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average?

A. 5
B. 6
C. 7
D. 8
E. 9

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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17 Sep 2015, 02:11
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Expert's post
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average?

A. 5
B. 6
C. 7
D. 8
E. 9

Let the sequence have N terms starting 18*K
--> 18K, 18(K+1), 18(K+2), .... , 18(K+N-1) is the sequence.
The sum of the sequence is
---> 18K + 18(K+1) + 18(K+2) + .... + 18(K+N-1) =18*(K+(K+1)+...+(K+N-1)) = 18*(N*(2K+N-1)/2).
So the average is
153=[18*(N*(2K+N-1)/2)]/N = 18*(2K+N-1)/2 = 9 *(2K+N-1) ---> by cancelling 9 we get 2K+N-1=17 --> 2K+N=18 ---*).

On the other hand, since the greatest term is 270, 18(K+N-1)=270 --> by cancelling 18 we get K+N-1=15 --> K+N=16 ---**).
By *), **) we have K=2, N=14.

Since 153/18 = 8.5, 18*8<153<18*9. So the number of terms which are smaller than the average is 18*2, 18*3, ..... ,18*8 ---> 7. The answer is C.
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"Only $79 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Director Joined: 21 May 2013 Posts: 613 Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink] ### Show Tags 17 Sep 2015, 02:30 7 This post received KUDOS Bunuel wrote: If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average? A. 5 B. 6 C. 7 D. 8 E. 9 Kudos for a correct solution. Average of consecutive terms=(Largest+Smallest/2) Now, 153=(270+Smallest)/2 Smallest Value=18 Since the sequence has consecutive multiples of 18, series is 18,36,54,72,90,108,126,144 Answer=7 C Manager Joined: 29 Jul 2015 Posts: 159 Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink] ### Show Tags 17 Sep 2015, 15:55 Bunuel wrote: If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average? A. 5 B. 6 C. 7 D. 8 E. 9 Kudos for a correct solution. Average in an evenly spaced sequence is given by (first term + Last term)/2 Let x be the first term. Then (x+270)/2=153 or x= 36 Now, next multiple greater than the average is 162 So the number of multiples less then 162 will be (last-first)/increment or (162-36)/18 = 126/18 = 7 Answer:-C Manager Joined: 13 Mar 2013 Posts: 176 Location: United States Concentration: Leadership, Technology GPA: 3.5 WE: Engineering (Telecommunications) Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink] ### Show Tags 17 Sep 2015, 22:09 1 This post received KUDOS If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average? A. 5 B. 6 C. 7 D. 8 E. 9 Kudos for a correct solution. given that sequence of consecutive multiple of 18 and avg is 153 therefore median = 153 greatest 270 therefore visualize this X------- 153--------270 Where x is min and 270 is 15th term of the sequence as odd number so 8th term will the median ---- and therefore there are 7 terms less than the median ans is c _________________ Regards , Director Status: I don't stop when I'm Tired,I stop when I'm done Joined: 11 May 2014 Posts: 562 Location: Bangladesh Concentration: Finance, Leadership GPA: 2.81 WE: Business Development (Real Estate) Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink] ### Show Tags 19 Sep 2015, 14:34 1 This post received KUDOS KS15 wrote: Bunuel wrote: If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average? A. 5 B. 6 C. 7 D. 8 E. 9 Kudos for a correct solution. Average of consecutive terms=(Largest+Smallest/2) Now, 153=(270+Smallest)/2 Smallest Value=18 Since the sequence has consecutive multiples of 18, series is 18,36,54,72,90,108,126,144 Answer=7 C IMO,According to your calculation Smallest value Should be 36,NOT 18 _________________ Md. Abdur Rakib Please Press +1 Kudos,If it helps Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges Director Status: I don't stop when I'm Tired,I stop when I'm done Joined: 11 May 2014 Posts: 562 Location: Bangladesh Concentration: Finance, Leadership GPA: 2.81 WE: Business Development (Real Estate) Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink] ### Show Tags 19 Sep 2015, 14:36 2 This post received KUDOS Bunuel wrote: If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average? A. 5 B. 6 C. 7 D. 8 E. 9 Kudos for a correct solution. Since the sequence is in a evenly spaced set,the average of the sequence of consecutive integer formula is mean=$$\frac{First Term +Last Term}{2}$$ ,or 153=$$\frac{First Term+270}{2}$$ ,or First Term=36 So the consecutive numbers less than mean of the sequence are 36,54,72,90,108,126,&144.So the total number if integer below the mean of the sequence is 7 So the Correct Answer is C _________________ Md. Abdur Rakib Please Press +1 Kudos,If it helps Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11270 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink] ### Show Tags 19 Sep 2015, 16:55 2 This post received KUDOS Expert's post Hi All, This question can be approached in a number of different ways. Since we're given so many facts to work with, you can actually use some math 'logic', a bit of arithmetic and some 'brute force' to PROVE how many terms are below the average. First, we're given a series of facts: 1) We're dealing with CONSECUTIVE multiples of 18 2) The average of all those multiples of 18 is 153 3) The LARGEST of those multiples is 270 We're asked to for the number of terms in that sequence that are BELOW the average. To start, since we're dealing with consecutive terms, the SAME number of terms will be above the average as below the average. Since the largest term is 270, and the average is 153, we can determine the number of terms that are ABOVE the average (we just have to keep subtracting 18 to find all of those terms....): 270 252 234 216 198 180 162 At this point, we can stop. If we subtract another 18, we'll end up with 144 - which is BELOW the average. Since we have 7 terms that are ABOVE the average, and we're dealing with CONSECUTIVE multiples of 18, there has to be the SAME NUMBER of terms below the average (to 'balance out' the calculation). Final Answer: [Reveal] Spoiler: C GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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19 Sep 2015, 19:38
1
KUDOS
Start by identifying the last term:
a -> First term
L -> Last term
$$Average = (a + L)/2$$
$$=> 153 = (a+270)/2$$
=> a = 306 - 270
=> a = 36 (or) 18*2

L = 270 (or) 18*15

$$153/ 18 = 8.5$$

So, we have all multiples of 18 between 18*2 and 18*8 (both inclusive) that will be lesser than the average. Which is 6+1 = 7
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Joined: 17 May 2015
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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09 Nov 2015, 03:06
kunal555 wrote:
Bunuel wrote:
If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average?

A. 5
B. 6
C. 7
D. 8
E. 9

Kudos for a correct solution.

Average in an evenly spaced sequence is given by (first term + Last term)/2
Let x be the first term.
Then (x+270)/2=153 or x= 36

Now,
next multiple greater than the average is 162
So the number of multiples less then 162 will be
(last-first)/increment or (162-36)/18 = 126/18 = 7

dont we need to consider 18 too in the list?
Math Expert
Joined: 02 Sep 2009
Posts: 44319
Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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09 Nov 2015, 04:14
kunal555 wrote:
Bunuel wrote:
If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average?

A. 5
B. 6
C. 7
D. 8
E. 9

Kudos for a correct solution.

Average in an evenly spaced sequence is given by (first term + Last term)/2
Let x be the first term.
Then (x+270)/2=153 or x= 36

Now,
next multiple greater than the average is 162
So the number of multiples less then 162 will be
(last-first)/increment or (162-36)/18 = 126/18 = 7

dont we need to consider 18 too in the list?

18 is indeed a multiple of 18 but in this particular sequence it turned out that the smallest term is 36.
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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11 Dec 2015, 04:59
abhisheknandy08 wrote:
If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average?

A. 5
B. 6
C. 7
D. 8
E. 9

Kudos for a correct solution.

given that sequence of consecutive multiple of 18 and avg is 153
therefore median = 153
greatest 270

therefore visualize this
X------- 153--------270
Where x is min

and 270 is 15th term of the sequence
as odd number so 8th term will the median ---- and therefore there are 7 terms less than the median

ans is c

Hi,
abhisheknandy08,
the coloured portion is not correct..
you cannot say that 270 is the 15th term since you do not know the first term....
you have to find the first number through average and last number given...
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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25 Jun 2016, 23:29
In series of consecutive number avg is is defined as avg of two middle terms , if numbers of terms are even
and avg is defined as middle term , if number of terms are odd.
153 is odd number , which cannot be the multiple of 18 , which is even number; hence this has to be the avg of middle even terms.

Number of terms above is equal to number of term below the avg
Next multiple of 18 is 162 last multiple given is 270
270-162/18 +1 = number of terms.
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If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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26 Jun 2016, 06:20

my method-
first find the lowest term=270+x/2=153
x=36

now find the number od terms 270 - 36/18 +1=14

as the number os terms is even the median or average will be 7th +8th term /2.this concludes the 7th term is less than 153

so there are total 7 terms less than 153 and 7 terms greater than 153.
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If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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26 Jun 2016, 15:54
let x=number of terms in sequence
[270+18(15-x+1)]/2=153
x=14 terms
153/18=8.5
mean (153) occurs halfway between 7th term (8*18=144) and 8th term (9*18=162)
there are 7 terms smaller than mean
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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01 May 2017, 08:06
No of numbers less than average = { {270 / 18} / 2 } ; I don't think that 153 is required. Since its a normal distribution. always half are less than avg and another half are more than average. .
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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01 May 2017, 09:07
Bunuel wrote:
If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average?

A. 5
B. 6
C. 7
D. 8
E. 9

Kudos for a correct solution.

let 18x=lowest term
(18x+270)/2=153
x=2
18*2=36=lowest term
let n=number of terms
36+18(n-1)=270
n=14 terms
14/2=7 terms less than average
C
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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04 Sep 2017, 02:30
Clearly the solution is C
see attached image
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and   [#permalink] 04 Sep 2017, 02:30
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