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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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New post 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


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

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New post 17 Sep 2015, 02:30
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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
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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New post 17 Sep 2015, 02:11
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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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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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

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New post 17 Sep 2015, 22:09
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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.

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

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New post 19 Sep 2015, 14:34
1
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
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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New post 19 Sep 2015, 14:36
2
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
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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New post 19 Sep 2015, 16:55
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1
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).

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

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New post 19 Sep 2015, 19:38
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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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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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New post 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

Answer:-C



dont we need to consider 18 too in the list?
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Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

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New post 09 Nov 2015, 04:14
akadmin wrote:
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

Answer:-C



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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New post 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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New post 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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New post 26 Jun 2016, 06:20
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answer is 7

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

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New post 04 Sep 2017, 02:30
Clearly the solution is C
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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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