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Expert replies
Source: — Problem Solving |

by Brent@GMATPrepNow » Sun Jun 21, 2015 9:07 am
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 answer.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Brent@GMATPrepNow » Sun Jun 21, 2015 9:10 am
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
Here's another approach.

If q is odd, then the median of the q integers will be the middle number.
So, of the q integers, the MIDDLE number is 120
Of the remaining q-1 integers (not including 120), HALF of them are greater than 120 and HALF are less than 120.
In other words, (q-1)/2 of the integers are greater than 120.

So, to find the biggest number, we'll take the median (120) and add (q-1)/2 to get ... 120 + (q-1)/2

Answer = A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by [email protected] » Sun Jun 21, 2015 4:13 pm
Hi tapanmittal,

Brent's approach (TESTing VALUES) is a great way to answer this question; it's exactly how I would have approached this prompt.

This prompt is built around an interesting Number Property, which you could have used to answer the question without doing much math at all.

We're told that Q is an ODD number, we're dealing with CONSECUTIVE INTEGERS and we're asked for the LARGEST INTEGER..

Take a look at answer B. Since Q is ODD, will answer B EVER be an integer???
The answer is NO (we'd end up with a fraction, so B can't be the answer.

This same rule holds true for answer C and E. So, we've eliminated B, C and E.

Now look at D; we're taking (Q+119) and cutting it IN HALF. That math doesn't seem like it would produce the LARGEST INTEGER that we're looking for. Eliminate D.

The answer must be A. It would certainly give us an integer: (ODD - 1)/2 = INTEGER. Add THAT integer to 120 and you either get 120 or a bigger integer.

Final Answer: A

GMAT assassins aren't born, they're made,
Rich
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by Scott@TargetTestPrep » Tue Jun 23, 2015 3:16 pm
tapanmittal wrote:If Q is an odd number and the median of Q consecutive intgers 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

OA is A
Solution:

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

Scott Woodbury-Stewart
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by Matt@VeritasPrep » Tue Jun 23, 2015 5:51 pm
tapanmittal wrote:If Q is an odd number and the median of Q consecutive intgers 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

OA is A
Another idea:

Let's take out the median itself. That gives us (Q - 1) integers. Half of those are smaller, half of those are larger, so the set is

(Q - 1)/2 integers, then the median, then (Q - 1)/2 more integers

So the LARGEST number in the set = median + (Q - 1)/2, or 120 + (Q - 1)/2.

This is more abstract but much faster than plugging in values.
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