A list of numbers has six positive integers. Three of those integers a

Hi all those looking for a short method..

points to NOTE..
1) the median of 6 integers will be the centre value of two MIDDLE integers, here it will be 3rd and 4th in ascending/descending order.
2) when you are looking for centre of two integers, the centre value will be either an integer or a number with .5 in decimals. Here it is given between 7 and 8, the value is 7.5
3) sum of two of the three unknowns is 7.5*2=15

Let's work further..
Total of unknowns=10*6-(24+5+4)=27..
The third unknown =27-15=12..

for other two totalling 15, the lower one cannot be less than 5, as the two lowest are 4 and 5..
So values can be 5 and 10,... or 6 and 9... or 7 and 8...
So A and C are not possible...
Bunuel , we may have to change the median from ' between 7 and 8(exclusive) to from 7 to 8(exclusive)..

So now the Q says that the median could be 7 also,
total of two is 2*7=14 and third unknown becomes 27-14=13....
And other two values can be 5 and 9...... Or 6 and 8.....

Now only 11 is not possible..
Ans C

If the mean is 10, the sum is 60. The three known numbers add to 33.
If the median is between 7 and 8, that means two digits were averaged to the median. (7<(x+y)/2<8 is the equation form).
x+y must be 15 (between 14 and 16). The other number is 12 (to equal 27).

I chose to solve using the answer choices. 10 and 5 add up to 15, so certainly both cannot be the answer.

For 11 to be the answer, the other number is 4.

4,4,5,11,12,27 is the new set. The median is 8. The question said 7 and 8 exclusive. Therefore, 11 is the answer.
C

Set of terms is {4, 5, x, y, z, 24}

Since x, y and z are distinct so let's assume Let, x < y < z

Mean = 10 i.e. Sum = Mean*no. of terms = 10*6 = 60

i.e. Sum of {4, 5, 24, x, y, z} = 60
i.e. x+y+z = 27

median = Average of two middle terms (for even no. of terms in set) = between 7 and 8 which can only be 7.5
if x+y = 7.5*2 = 15
i.e. z = 27 - 15 = 12

Case 1: x = 5 and y = 10 and z = 12
Case 2: x = 6 and y = 9 and z = 12
Case 3: x = 7 and y = 8 and z = 12

OR if z = 13 then x+y = 14
i.e. x and y can be 4 and 10 respectively and the set becomes {4, 4, 5, 10, 13, 24} i.e. Mean = 7.5

So x, y and z can be anything but 11

Answer: Option C

VERITAS PREP OFFICIAL SOLUTION:

The question gives us concrete information about mean – it is 10 – but not about median – it is between 7 and 8 (exclusive). What can we say about median from this? That it cannot be 7 or 8 but anything in between. But we know that the list has all integers. When we have even number of integers, we know that the median is the average of the middle two numbers – when all are placed in increasing order. So can the average of the two middle numbers be, say, 7.1? Which two positive integers can average to give 7.1? None! Note that if the average of two integers is a decimal, the decimal must be (some number).5 such as 7.5 or 9.5 or 22.5 etc. This happens in case one number is odd and the other is even. In all other cases, the average would be an integer.

Since the median is given to be between 7 and 8, the median of the list of the six positive integers must be 7.5 only.

Now we know that the mean = 10 and median = 7.5

Method 1: Algebra/Logic

Let’s try to solve the question algebraically/logically first.

There are 6 elements in the list. The average of the list is 10 which means the sum of all 6 elements = 6*10 = 60

4 + 5 + 24 + x + y + z = 60

x + y + z = 27

Median of the list = 7.5

So sum of third and fourth elements must be 7.5 * 2 = 15

There are two cases possible:

Case 1: Two of the three integers x, y and z could be the third and the fourth numbers. In that case, since already 4 and 5 are less than 7.5, one of the unknown number would be less than 7.5 (the third number) and the other two would be greater than 7.5.

The sum of the third and fourth elements of the list is 15 so

15 + z = 27

z = 12

So, two numbers whose sum is 15 such that one is less than 7.5 and the other greater than 7.5 could be

5 and 10

6 and 9

7 and 8

x, y and z could take values 5, 6, 7, 8, 9, 10 and 12.

Case 2: The known 5 could be the third number in which case one of the unknown numbers is less than 5 and two of the unknown numbers would be more than 7.5.

If the third number is 5, the fourth number has to be 10 to get a median of 7.5. Hence, 10 must be one of the unknown numbers.

The sum of the other two unknown numbers would be 27 – 10 = 17.

One of them must be less than 5 and the other greater than 10. So possible options are

4 and 13

3 and 14

2 and 15

1 and 16

x, y and z could take various values but none of them could be 11

Answer (C)

Method 2: Process of Elimination

Let’s now try to look at the process of elimination here and see if we can find an easier way.

The three unknowns need to add up to 10*6 – 4 – 5 – 24 = 27.

Two of the given options are 5 and 10. They have a median of 7.5 so lets assume that two of the unknown numbers are 5 and 10 (5 can be one of the unknowns since we are not given that all six integers need to be distinct). If two unknowns make up third and fourth numbers in the list and have a median of 7.5, their sum would be 15 and the third unknown will be 12 (to get the mean of 10). This case (5, 10, 12) satisfies all conditions so options (B), (D) and (E) are out of play.

Now we are left with two options 13 and 11. Check any one of them and you will know which one is not possible. Let’s check 13.

From the given options, any number greater than 7.5 must be either the fourth number or the fifth number. 13 cannot be the fourth number since the third number would need to be 2 in that case to get median 7.5. But we have 4 and 5 more than 2 so it cannot be the third number. So 13 must be the fifth number of the list. We saw in the case above that if two unknowns are third and fourth numbers then the fifth number HAS TO BE 12. So the already present 5 must be the third number and the fourth number must be 10. In that case, the leftover unknown would be 4 (to get a sum of 27). So the three unknowns would be 4, 10 and 13. This satisfies all conditions and is possible. Hence answer must be (C). 11 will not be possible.

Let’s see what would have happened had you picked 11 to try out. If 11 were the fourth number, to get a median of 7.5, we would need 4 as the third number. That is not possible since we already have a 5 given. So 11 must have been the fifth number. This would mean that the already present 5 and one unknown 10 would make the median of 7.5. So the third unknown in this case would be 6 (to get a sum of 27). But 6 would be the third number and the median in this case would be (6 + 10)/2 = 8. So one of the numbers cannot be 11.

Answer (C)

The sum of 3 numbers is 33
The sum of remaining numbers is 27 since mean is 10
The fourth number is 10 since the third number is 5 and the median is between 7 and 8
The sum of the remaining 2 numbers has to be 17. If we try to fit to the choices we can see they can be 13 and 4 or 12 and 5
11 does not figure in the list.
So C.

How could A give result of set with median of 7.5? It is 2, 4, 5, 12, 13, 24 then the median becomes 8.5 is it not?

1. The set would look like \(x = {4, 5, 24, x, y, z}\) not in any specific order

2. We are given that the mean of set \(x\) is \(10\) i.e. \(\frac{4 + 5 + 24 + x + y + z}{6} = \frac{33 + x + y + z}{6} = x + y + z = 27\)

3. We are also given that median is between \(7\) and \(8\) i.e. \(7.5\). Hence, the sum of the middle \(2\) terms will be \(15\)

4. The first combination that I can think of is \({4, 5, 5, 10, 12, 24}\). The mean of this set is \(10\) and median is \(7.5\). B, D and E can be eliminated

5. The second combination that I can think of is \({4, 4, 5, 10, 13, 24}\). We can eliminate A

Ans. C

Known integers: 4, 5 and 24
Median is the avg of the middle two integers. The avg of two integers will be Integer/2 and hence will either be an integer or integer.5
Since median lies between 7 and 8, it must be 7.5. So the middle two numbers could be (7, 8) or (6, 9) or (5, 10). The middle two numbers cannot be (4, 11) because if 4 is the 3rd number, the 4th number will be 5. No other values of middle two numbers are possible.

Mean = 10 so sum of all numbers is 60.
If the middle two numbers are (7, 8) or (6, 9), we are left with only one unknown and it must be 12 ( = 60 - 4 -5 - 24 - 15)
If the middle two numbers are (5, 10), the list could be 4, 5, 5, 10, 24 in which case only one unknown is left which must be 12.
or the list could be 4, 5, 10, 24 with two unknowns such that one of them is smaller than 5. If another unknown is 4, the last unknown becomes 13.

So our unknowns can take the values of 5, 10, 12 and 13, but not 11.

Answer (C)

Now try this question on mean, median and range: https://anaprep.com/sets-statistics-mea ... -question/

The 3 unknowns add up to 27.

If one of the numbers is less than 5, then 5 becomes the highest of the 3 and is matched with the next highest number and divided by 2 to be between 7&8.

That means the next highest number could only be 10.

Choosing 4 as the lower number and 10 being the required other number, means that

13 could be one unknown, along with 10

Choosing 3 or lower isn't necessary since it only increases from 13 and those numbers aren't among the choices.

Now, if instead all 3 numbers are >5, trying 6 as the next highest number means that only 9 can be the next, leaving

27-9-6= 12 as a possibility

Finally, choosing 5 as a number means again that 10 has to be another, leaving 27-5-10=12.

So by elimination, only 11 can be the correct choice.

Posted from my mobile device

Question about median could be a really tricky one... I wanted to visualize it.

Condition 1) Median is 7.5 (numbers are all integers)
Condition 2) x + y + z = 27. So, two numbers should be added up to 15 and remaining number should be 12.

------3---4---5---6---7-(median)-8---9---10---11---12---13-----------------------------24

1) Draw a number line & Eliminate answer choices quickly ----> 12 is already eliminated(above), 5 & 10 can easily eliminated.
2) 11 vs. 13 ---> Number line tells me to eliminate 11. (because we already have "5", so can't match 4-11 )
But to verify, try 13. ---> 13, 10, 4 could be matched because we already have "5" !!!

Therefore answer is C.

­If we solve via this way, then how are you eliminating Option A? for {2,13,12} the number in ascending order would look like {2,4,5,12,13,24}
Medium here will be 5 + 12 = 17/2 = 8.5(out of scope)
 

­What does  "the median lies between 7 and 8 (exclusive)" mean?

is it that median has to lie between 7 and 8 ?

 

­
Yes, ­ "the median lies between 7 and 8 (exclusive)" means that the median lies between 7 and 8 (exclusive).