Median is More Than Just Midpoint

by on August 29th, 2012

A whale goes on a feeding frenzy that lasts for 9 hours. For the first hour it catches and eats x kilos of plankton. In every hour after the first, it consumes 3 kilos of plankton more than it consumed in the previous hour. If by the end of the frenzy the whale will have consumed a whopping accumulated total 450 kilos of plankton, how many kilos did it consume on the sixth hour?

This is a seemingly tough question, requiring several steps. Fortunately, this question, and others like it, can be solved with the use of the properties of arithmetic sequences. An arithmetic sequence has several properties which are tested in GMAT problems. One of them is the average property:

For all sets of integers with a constant difference between adjacent terms, Average of the set = Median of the set.

Take a look at this simple set of consecutive integers: {1, 2, 3}.

The median of the set is easily recognizable – since the set has an odd number of terms, the median is the number in the middle, or 2. The average of the set = Sum / Number of items = (1+2+3) / 3 = 6/3 = 2.

This stays true even if the set includes an even number of consecutive integers, as in the set {1, 2, 3, 4}. Since there is an even number of members in the set, the median is calculated as the average of the two middle terms: (2+3) / 2 = 5 / 2 = 2.5. Surprisingly enough, the average of the set is also (1+2+3+4) / 4 = 10 / 4 = 2.5. This example shows the most common occurrence of this property, which is sets of consecutive integers.

However, this property is not confined to consecutive integers alone; In any set of integers which constitute an arithmetic sequence, regardless of the number of terms in the sequence, Average = Median. This holds true for all of the following categories:

  1. Consecutive integers: (For example: 1, 2, 3, 4, 5)
  2. Consecutive odd or even integers (For example: 1, 3, 5, 7, 9 or 2, 4, 6, 8, 10)
  3. Multiples of an integer (For example: 7, 14, 21, 28, 35 or 8, 16, 24, 32)
  4. Arithmetic series (For example: 2, 11, 20, 29, 38 or 63, 75, 87)

Let’s try tackling the ravenous whale question again now.

A whale goes on a feeding frenzy that lasts for 9 hours. For the first hour it catches and eats x kilos of plankton. In every hour after the first, it consumes 3 kilos of plankton more than it consumed in the previous hour. If by the end of the frenzy the whale will have consumed a whopping accumulated total 450 kilos of plankton, how many kilos did it consume on the sixth hour?

38
47
50
53
62

The question describes an arithmetic sequence with a difference of 3: in the first hour our whale consumes x kilos, in the second (x+3), in the third (x+6), etc. Adding these together will give a total of 450, from which we can find x, but that is not an easy calculation. By the time you’re done with that, you might easily forget that the question does not ask for x, but rather for the consumption in the sixth hour, which is actually x+15.

Instead, recall the average property of arithmetic sequences: Average = Median. Since the question kindly provides the total kilos of Plankton (450) and the number of hours (9), the average hourly consumption of Plankton can be easily calculated: 450 / 9 = 50.

Therefore, the Median of our set of consecutive integers is also 50. Since the set has an odd number of members, the median is the number in the middle, or the 5th hour. If the whale consumes 50 kilos of Plankton in the 5th hour, he will consume 50+3 = 53 kilos in the sixth hour. Quick and easy – with the right approach.

Remember:

In a set of integers with a constant difference between them (arithmetic sequence), Average = Median.

Now, try the following question using this approach.

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?
5
6
7
8
9

18 comments

  • Opps first I solved the question with the traditional method..I got 9x+108 = 450..got the value of x and then calculated for x+15..I took 2.5 mins to solve this problem..:(
    using your method I think anyone can solve it under 1 min..:)..Thanks
    What is the answer for the 2nd question?
    hmm..There are 15 numbers as 15*18 =270 ..so odd numbers of integers.right?the 8th term should be 153 so there are 7 terms smaller than the avg..right ?

  • ans. is 7 Clearly, 153 in not whole multiple of 18 and also 270 is 16th multiple of 18... Going by the logic expressed in previous ques., there must be even no. of terms in series and that would be total 16 terms. So, the avg. would be of 8th and 9th term, thus there are 7 terms smaller than the avg. value..

    • Prasoon,
      Are you considering the zeroth term in the series too? If that is the case then the answer should be 9 (E). Since, there will be 16 terms in total (inclusive of zero) and the average value is the average for the 8th and 9th term. I calculated it mechanically and this does give a value of 153. However, in this case the number of terms smaller than the average becomes 9 and not 7. The eighth term for the series is 144 which is < 153 and the zeroth term should also be included.

      @ Faruk,
      I think the approach you have taken might be incorrect. Since for the odd numbers the median = average = middle term will constitute the 8th term in the series. This 8th term is 144. However, the question mentions that the average is 153.

    • iI got same results as you. is it possible the average in question is wrong. I keep getting 144 as average. its in 8th term out of a total of 15 terms. If there were 16 terms then average would be 153, but even then when you solve for the averge its not 153. so i got 16 terms counting 0, average and median = 135, so there is seven smaller terms. 15 terms = 144 smaller terms =7 , and 14 terms =153 = 6 smaller terms. but since question say 18*15 =270 and 1-15 is odd number the meadian and averge is 8. there are seven smaller numbers. i am going with 7.

  • @Rohan,
    No, I'm not considering the zeroth term....

    • Could you please list the details of how you solved and arrived at the answer? I am still a bit confused about this one.

  • Answer is 7.

    Largest term is 270 and average is 153.

    You can easily find the first term.

    (270+x)/2 = 153 x = 36

    So, we have multiple 18*2 to 18*8 less than 153 i.e 7 numbers.

    Hope this helps.

  • This is perfect ! Thanks for the explanation SN. I went too far out of the range and thought of everything possible under the sun ! Your explanation cleared the fog out totally. Thanks again.

    PS: I guess I was a bit overworked and did not think of the average of consecutive numbers as (last term + first term) / 2.

  • Here's how I would do it:
    (first term + last term)/2 = average; therefore first term is 36
    {(last term - first term)/consecutive multiple }+1 = number of terms, which is 14 in this case
    The median is the average of the 7th and 8th term. Thus, there are 7 numbers smaller than the median.

    • Thanks for your explanation!!

    • Thanks... This is really helpful.

  • C. 7

    For me the question was pretty simple after reading the above article.. it says multiples of 18 upto 270.. meaning 15 multiples of 18.. and the average of these 15 multiples is 153, meaning median is 153, which is the 8th term as the series is odd.

    Therefore, 7 terms in the sequence will be smaller than the average/median, 153. 

    • The series is not odd as 153 is not a part of the series. The series is even.

  • Oh that is such a shame.. yes you are right Siva. Thanks for correcting me. 

    So if I re-think about the problem, it says multiples of 18 upto 270 have an average of 153, meaning (x+270)/2 =153, where x is the first term and 270 is the last term. This gives us the first term as 36. Therefore, there are 14 terms in the series.
    Median will be the average of 7th and 8th term. Hence, there will be 7 terms less than the median or the average. 

    Is this correct?

  • Finding the first term = 153*2=306-270=36
    and the median is 270/36*1/2=7.5
    Therefore, there are 7 terms below the median

    • by median i meant the median term

  • 7 terms. But I forgot to consider the 0 in the list and got stuck figuring out how is 153 the average when the median was the 8th term but now its clear.

  • GOOD ONE....

Ask a Question or Leave a Reply

The author Economist GMAT Tutor gets email notifications for all questions or replies to this post.

Some HTML allowed. Keep your comments above the belt or risk having them deleted. Signup for a Gravatar to have your pictures show up by your comment.