The range of a certain data set is 12, and the median is 10 more than

Question

The range of a certain data set is 12, and the median is 10 more than the smallest number in the data set. If the average (arithmetic mean) of the data set is greater than the median of the data set, what is the smallest possible number of terms that the data set could contain?

A. 11
B. 13
C. 15
D. 17
E. 19

A. 11
B. 13
C. 15
D. 17
E. 19

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KarishmaB · Accepted Answer

Range is 12 so if smallest term is x, the greatest is x+12. Median is x+10

Terms: x, x+10, x+12

The average here will be less than x + 10. It will be x + 22/3 = x + 7 approx.

We need to make it more than x + 10 so we should add greater values to the data set, but the median  should remain x + 10 only and the range should remain 12 only (so the greatest value must remain x + 12 only)

So what do we do? We add a term at x+12 (the greatest possible value) but to balance it to keep the median x+10, we must add another x+10.

x, x+ 10, x+10, x+12, x+12

Now the average has become x + 44/5 = x + 8.8 (greater than before)
So every time we add 2 terms x+10 and x+12 to keep all conditions same we move the average to the right.

Check for 11: If we add another 6 terms (to make total terms = 11), the average will become x + 110/11 = x + 10
But we need the average to be more than x + 10 so the next time we add 2 more terms, the average will become more than x+10.

Hence answer is 13 terms.

Answer (B)

gmatophobia · Answer

Few observations before we begin to solve  
 
 Each answer choice is an odd number. This information tells us that the number of terms on the right side of the median equals to the number of terms on the left side of the median. 
 As the range of the entire set is 12, the maximum difference between the value of the terms to the right of the median and the median is 2. 
 As average is greater than the median, the difference between the average and the smallest term is greater than 10. 
 To minimize the number of terms and to ensure that the average is greater than the median, each term to the right of the median must be equal to its maximum possible value (imagine this as tug-of-war, the heavier side pulls the center towards it. Hence, each person should have the maximum weight possible). 
 All values to the left of the median should also have maximum weight possible. Hence, except the smallest term, each term should be equal to the median.

With these observations in mind, let's delve into the answers.

Answer Choice Elimination

A. 11

There are 5 terms to the right of the median and 5 terms to the left of the median. The median is the sixth term. Let's assume that the lowest value = 0. The value of the median therefore is 10.

The value of the 1st to the 5th term will therefore be equal to 10, and the value of the 7th to the 11th term will be equal to 12.

0 10 10 10 10 10 12 12 12 12 12

The average of this set =  \frac{60 + 50 }{ 11} = 10

The average = median. However, we need average > median. We can eliminate A.

B. 13

There are 6 terms to the right of the median and 6 terms to the left of the median. The median is the sevnth term. Let's assume that the lowest value = 0. The value of the median therefore is 10.

The value of the 1st to the 6th term will therefore be equal to 10, and the value of the 8th to the 13th term will be equal to 12.

0 10 10 10 10 10 10 12 12 12 12 12 12

The average of this set =  \frac{72 + 60 }{ 13} = \frac{132}{13}= 10.XX

Hence, average > median.

Option B

GMATinsight · Answer

smallest number in set = x
The range of a certain data set is 12 i.e. Biggest number in set = x+12
Median = 10+x
Average > Median
i.e. Average > 10+x

So terms in set could be

x _ x+10 _ x+12

For n terms, one term is 1, another (Median) is 11, and biggest is 13

For maximum average, (n-3)/2 terms should be 11 and
For maximum average, (n-3)/2 terms should be 13

i.e. Average =  /n =  /n = (12n-11)/n

Now, (12n-11)/n > 11
i.e. n > 11
i.e. n = 13

Answer: Option B

Validation: 
{1, 11, 11, 11, 11, 11, 11, 13, 13, 13, 13, 13, 13}

Average = 11.15
Median = 11
SATISFIED

sayan640 · Answer

GMATinsight  ,

can you please explain this part of your solution ?

For maximum average, (n-3)/2 terms should be 11 and
For maximum average, (n-3)/2 terms should be 13

i.e. Average =  /n =  /n = (12n-11)/n

Nidzo · Answer

We are given 3 algebraic values: the smallest number in the set, the middle value and the largest value.

Let the smallest value =  x

( x )...   ...( x+10 )...   ...( x+12 )

As of right now our average is  \frac{3x+22}{3}≈ x+7  which is lower than the median.To increase the average and make it greater than the median, with the fewest numbers possible, one needs to maximise the sum of the values in both  blank space 1  and  blank space 2 . The greatest value that can occupy  blank space 1  will be equal to the median so  x+10  and for the  blank space 2  will be equal to the greatest value in the set  x+12 .

Now as we have three values currently, and all of the answer choices are odd, we know that the number of values that fill in  blank space 1  and  blank space 2  are equal and have to sum to an even number. In other words, the average of the sum that will be added, will be  \frac{x+10+x+12}{2}=x+11

We are essentially mixing  \frac{3x+22}{3}  with  x+11  and wanting to get a number greater than  x+10 .

Treating this as a mixture, and using alligation to get a 'solution' of   x+10 :

As we have  \frac{3x+22}{3} , to make the alligation process simpler, multiply the other two values through by  \frac{3}{3}  . Now we are mixing  \frac{3x+22}{3}  with  \frac{3x+33}{3}  to get  \frac{3x+30}{3} .

Mixtures:                  \frac{3x+22}{3}                         \frac{3x+33}{3} 
                                                                                         
 Desired mixture out:                    \frac{3x+30}{3}  
                                                                                           
 Ratio required:              3                                   8

This shows us that if we add an additional four  x+12  and four  x+10  to the three numbers we already have, then we will have an average of  x+10 . However, as we want the average to be greater than the median we need the next biggest odd number which is 13,  (B) .

ANSWER B

GMATinsight · Answer

Hi sayan640 Since we need to meet teh condition that mean is greater than the median, so we need to maximize the mean for lowest possible number of terms. To maximize the average, we need to maximize the sum of teh values hence we need to maximize each value that is unexplained. Suppose there are n terms, then we already know three terms out of those n terms which are first term, median and last term so assuming first term is 1, median becomes 11 and last term becomes 13 now to maximize the terms before median, those terms all (i.e. (n-3)/2 terms) can be 11 and similary, the terms after median (i.e. (n-3)/2 terms) can all be 13 I hope now it makes more sense.

MartyMurray · Answer

The range of a certain data set is 12, and the median is 10 more than the smallest number in the data set. If the average (arithmetic mean) of the data set is greater than the median of the data set, what is the smallest possible number of terms that the data set could contain?

A. 11
B. 13
C. 15
D. 17
E. 19

The information we have is that the data set has a range of 12 and a median 10 above the smallest number. So, let's use the easiest possible numbers that fit:

0, 10, 12

To get the mean to exceed the median with the fewest possible elements, we maximize the values of the numbers in the data set. So, every number other than 0 that's not above the median will be 10, and every number above the median will be 12.

Since all the answer choices are odd, we know there are equal numbers of elements above and below the median. So, the elements of the set will be the following:

1 * 0, x * 10, x * 12

The mean is sum/number =  \frac{0 + 10x + 12x}{1 + 2x} 
 
We need x to be such that  \frac{0 + 10x + 12x}{1 + 2x} > 10

22x > 10 + 20x

2x > 10

x > 5

Since elements are counted in whole numbers, the minimum x = 6.

The total minimum number of elements is 1 + 6 + 6 = 13

A. 11
B. 13
C. 15
D. 17
E. 19

Correct answer:  B

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