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555-605 (Medium)|   Statistics and Sets Problems|                                 
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Walkabout
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,
Brent
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Walkabout
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

Let Q = 3
Set = {119,120,121} which makes 121 the largest integer

(A) (3-1)/2 + 120 = 121
(B) and (C) yields a decimal. ELIMATE!
(D) 122/2 = 61 ELIMINATE!
(E) yields a decimal. ELIMINATE!

Answer: A
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Really nice method Bunuel!
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Bunuel
Walkabout
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

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.

Sorry, if this is a really stupid question. I am probably missing something, but I do not understand the logic of this approach:

If 120 is the median (i.e. in my mind the middle number of the set) how can 121 then be the largest number? I am asking, since this does not make sense to me, yet and thus I would have never come up with such a good shortcut.

Many thanks
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Revenge2013
Bunuel
Walkabout
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

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.

Sorry, if this is a really stupid question. I am probably missing something, but I do not understand the logic of this approach:

If 120 is the median (i.e. in my mind the middle number of the set) how can 121 then be the largest number? I am asking, since this does not make sense to me, yet and thus I would have never come up with such a good shortcut.

Many thanks

We are told that there are Q consecutive integers in a set and Q is odd. We are also told that the median of the set is 120.

Now, say Q=3=odd. So, we have that the median of 3 consecutive integers is 120.
Question: what is the largest of these 3 integers? The set in this case must be {119, 120, 121} (3 consecutive integers with median of 120), so the largest of these 3 integers is 121.

Hope it's clear.
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I came across an alternate method.
Since Q is odd, therefore Q/2 is a fraction this options B and C are eliminated.( questions talks about integers only).E option is nullified since Q+120 is odd and the result will be a fraction on division with respect to 2. Left A and D just put Q =1,3 or any odd number D gives a value less than 120 therefore it cannot be the largest.

Posted from my mobile device
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Walkabout
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

Average of Q consecutive ints in a list = average of first and the last ints in the list
also for consecutive int mean = median
F = First Number
l = Last Number

avg = \frac{(F + L)}{2}

now L = F + (Q-1)

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

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

L = 120 - \frac{(Q-1)}{2} + (Q-1) = 120 + \frac{(Q-1)}{2}
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Bunuel
Walkabout
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

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.

Little eager....might be stupid too but can you assume q=1. This brings me to an important question can you have a median in a set that has only one element. I know this might sound stupid but is this possible. Further if that being so my largest number through option 1 becomes 120. But that is wrong. So am I missing something or the question has been set up assuming odd number starts with 3.

Bunuel request you to please clarify this.
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Walkabout
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

Let us say that the numbers are {119, 120, 121}
Q = 3

(A) (3 - 1)/2 + 120 = 121
(B) 119 + 1.5
(C) 1.5 + 120
(D) (122)/2
(E) (123)/2

Hence option A is the answer
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davidfrank
Bunuel
Walkabout
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

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.

Little eager....might be stupid too but can you assume q=1. This brings me to an important question can you have a median in a set that has only one element. I know this might sound stupid but is this possible. Further if that being so my largest number through option 1 becomes 120. But that is wrong. So am I missing something or the question has been set up assuming odd number starts with 3.

Bunuel request you to please clarify this.

The median of a single element set is that number itself. For example, the median of {-11} is -11.

Next, you can consider Q to be 1, in this case the set is {120} and the largest integer is 120. Substituting Q=1 into the options gives A as the answer.

Hope it's clear.
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Bunuel
Walkabout
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

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.

Dear Bunuel
Is this question properly formulated? "If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?"
Shouldn't it say "If Q is a set of consecutive numbers and Q is is odd, then wat is the largest of..."?
I find the wording a little confusing.
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Bunuel
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reto
Bunuel
Walkabout
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

Consider the easiest case, say Q=3, then;
Set = {119, 120, 121};
The largest integer = 121.

Now, plug Q=3 into the answers to see which yields 121. Only answer choice A works. Notice that we don't really need to plug for B, C, or E, since these options do not yield an integer value for any odd value of Q.

Answer: A.

Dear Bunuel
Is this question properly formulated? "If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?"
Shouldn't it say "If Q is a set of consecutive numbers and Q is is odd, then wat is the largest of..."?
I find the wording a little confusing.

No.

The set given is a set of "Q consecutive integers", where Q is an odd number. For example, if Q=5=odd, then the set can be {118, 119, 120, 121, 122} (set of 5 consecutive integers).
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Hi reto,

You'll likely find it best to break Quant prompts down into 'pieces', so that you can deal with each piece one-at-a-time and simplify any complex-looking situations.

Here, the first thing we're told is "Q is an ODD number..." Now what does THAT mean? Here are some examples...

Q COULD be -5, -3, -1, 1, 3, 5, 7, etc.

Next, we're told "the median of Q consecutive integers is 120." 'Q' clearly refers to the 'Q' in the earlier part of the sentence; we're also dealing with the statistical term "median" (which means 'middle number of a group, when the numbers are ordered from least to greatest').

So let's list out some possibilities (it's worth noting that you can't have "negative consecutive integers", so the negative examples that I listed are not possible here):

IF...Q = 1....the median of 1 consecutive number is 120......so the 'group' is {120}
IF...Q = 3....the median of 3 consecutive numbers is 120....so the 'group' is {119, 120, 121}
IF...Q = 5....the median of 5 consecutive numbers is 120....so the 'group' is {118, 119, 120, 121, 122}
Etc.

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Walkabout
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

We are given that Q is an ODD NUMBER and that the median of Q CONSECUTIVE INTEGERS is 120. Let's choose a convenient number for Q, such as 3. We can now say:

The median of 3 consecutive integers is 120. Since 120 is the MEDIAN, or middle number of these integers, our 3 integers are the following:

119, 120, 121

The answer choices present us with formulas for the value of the largest integer in the sequence. To determine the correct formula, we therefore will plug 3 in for Q in each answer choice until we get 121, which is the largest value in our sample data set.

A) (Q-1)/2 + 120

(3-1)/2 + 120 = 1 + 120 = 121

This IS equal to 121.


B) Q/2 + 119

3/2 + 119 = 1.5 + 119 = 120.5

This IS NOT equal to 121.

C) Q/2 + 120

3/2 + 120 = 1.5 + 120 = 121.5

This IS NOT equal to 121.

D) (Q+119)/2

(3+119)/2 = 122/2 = 61

This IS NOT equal to 121.

E) (Q+120)/2

(3+120)/2 = 123/2 = 61.5

This IS NOT equal to 121.

The only answer that is equal to 121 is that given in answer choice A.

Answer: A
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Hi, I know it's given that Q has to be an integer and basis that we eliminate B, C and D. But C gives us 121.5 cant we round this to 122 and this would be greater than A: 121?
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Hi, I know it's given that Q has to be an integer and basis that we eliminate B, C and D. But C gives us 121.5 cant we round this to 122 and this would be greater than A: 121?

Hi Kritisood,

We can certainly use Number Properties to eliminate three of the answers - since Q is an ODD integer and we're looking for an answer that is ALSO an integer (re: the largest integer in the group of CONSECUTIVE INTEGERS that the question describes) - Answers B, C and E can be eliminated because those answers will all be NON-integers.

The question asks us for the largest integer in the SEQUENCE of consecutive integers - NOT the largest result among the 5 answer choices. That's ultimately why Answer C is incorrect (even though we've already eliminated it for being a non-integer).

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I have a problem with plugging in numbers , i always think that it won't work if i try different values. How do i make sure that it would be the same result everytime ?
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