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If Q is an odd number and the median of Q consecutive

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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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New post 01 May 2017, 18:49
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2

My 2 cents.
Plug in.

Assume that Q = 3
119,120,121, the largest is 121

Plug in.
a) (3-1)/2 + 120 = 121 check.
all others, just by briefly looking at them, you know that they are out.

A.
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If Q is an odd number and the median of Q consecutive  [#permalink]

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New post 15 Oct 2017, 09:50
You need to study the arithmetic sequence formulas , which should get it straight forward

Xn=a+(n-1)d , n=numbers in sequence , a=first number and d=differnce
Xq=a+(q-1)-->1
(Xq+a)/2 =120 -->2 Median=Arithmetic mean equation

from 2&1 , x=2(120)-X+Q-1 , 2X=2(120)+Q-1

X=120 + (Q-1)/2

most of the replies tell to test numbers , but in the real exam testing number could be very hard and confusing , remember the formulas
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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New post 20 Apr 2018, 15:08
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Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2


A very fast solution is to see what happens when Q = 1.
This means that there's only ONE integer in the set.
So, if the median of the set is 120, then the set is {120}, which means the greatest value in the set is 120

So the correct answer choice should yield 120 when Q = 1.

a) (1-1)/2 + 120 = 120 PERFECT!
b) 1/2 + 119 = some non-integer
c) 1/2 + 120 = some non-integer
d) (1+119)/2 = 60
e) (1+120)/2 = some non-integer

Since only answer choice A yield the correct output, it is the correct answer.

Cheers,
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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New post 06 Jun 2018, 06:28
(120-n)<---------->(120+n) this is the Greatest number equation for series
...............120.............,
Consider,
integers to the left of 120= n
intergers to the right of 120= n
Therefore, Q= n+1+n, (*note that 120 is also another integer in the series of Q Odd numbers)
Q= 2n+1
Q-1= 2n
(Q-1)/2= n,

Therefore the greatest number = (Q-1)/2+ 120 Which is same as 120+n..
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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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New post 31 Jan 2019, 13:59
Hi,

My two cents for this question.

So given series is arithmetic series where say first term is \(a_1\) and last term is \(a_n\).

We know that AP is given by \(a_n\)= \(a_1\) +(n-1)d----(1) where n is number of terms and d is common difference .

Also if we recall that in AP we can find mean = \(\frac{a_1+a_n}{2}\).

It is always true of evenly spaced sets that the median – the middle term of the set – is equal to the mean, or arithmetic average, of the set.

So in AP mean = median.

Let median M, and Mean =m
M=m= \(\frac{a_1+a_n}{2}\)

so \(a_1\)= 2m-\(a_n\) which is also same as 2M-\(a_n\) -------(2)( Becuase M=m)

substituting 2 in 1 we get

\(a_n\) =2M-\(a_n\) +(n-1)d

Now since the given series is of consecutive of integers we have d =1

so \(a_n\) =2M-\(a_n\) +(n-1)

\(2a_n\)=2M +(n-1)

\(a_n\)=M +\(\frac{n-1}{2}\)

Here we are given that M=120 and n= Q

So we have

\(a_n\)= 120 +\(\frac{Q-1}{2}\)

Probably logic behind this question is above .

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Re: If Q is an odd number and the median of Q consecutive  [#permalink]

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New post 06 Aug 2019, 17:40
Walkabout wrote:
If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?

(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2


Consecutive integers means median = avg = 120 = First+Last/2 = 120
So, F+L = 240
Also, Q = (Last - First)/1 + 1 = L-F+1

Combine:
F+L = 240
L-F + 1 = Q
2L + 1 = Q + 240
2L = Q - 1 + 240
L = (Q-1)/2 + 120
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Re: If Q is an odd number and the median of Q consecutive   [#permalink] 06 Aug 2019, 17:40

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