A list of 11 positive integers has a mean of 10, a median of 9, and a

Question

A list of 11 positive integers has a mean of 10, a median of 9, and a unique mode of 8. What is the largest possible value of an integer in the list?

(A) 24
(B) 30
(C) 31
(D) 33
(E) 35

firas92 · Accepted Answer

We can back-work to find the set.

In order to maximize one element, we need to minimize the rest

Now we know that sum of elements of the set will be 11*10 = 110 and all elements are positive integers. So the minimum value an element can take is 1.

8 is the mode which means that 8 appears the maximum number of times and 9 is the median so 9 will be the 6th element

So we can set this all up as

{1,1,8,8,8,9,9,10,10,11,x} - Keep in mind that 1, 9 or 10 cannot appear a third time since 8 is the unique mode.

Any other setup than the above will not satisfy the conditions or will not maximize x

So now 1+1+8+8+8+9+9+10+10+11+x=110

x=35

Answer is  (E)

Kudos if this helped !

neha283 · Answer

Hi  mohitranjan05  ,

on putting x=35 in above equation, n = (110-44)/ 8 
which is 66/8 and does not make n a multiple of 8.

ScottTargetTestPrep · Answer

We see that the sum of the 11 integers is 11 x 10 = 110.

Since the median is the 6th number in the list (assuming the list has been sorted in ascending order) and the mode (8) is less than the median (9), then the mode can appear (at most) 5 times. If that is the case, we can let the first 5 numbers be 8, the next 4 numbers 9, and the second largest number 10, so the largest number must be 110 - (8 x 5 + 4 x 9 + 10) = 110 - 86 = 24.

On the other hand, if the mode repeats only 2 times, then, we can let the first 10 numbers be 1, 2, 3, 8, 8, 9, 10, 11, 12, and 13, totaling 77, so the largest number must be 110 - 77 = 33.

If the mode repeats 3 times, then we can let the first 10 numbers be 1, 1, 8, 8, 8, 9, 9, 10, 10 and 11; totaling 75, so the largest number in this case is 110 - 75 = 35. Since 35 is the greatest answer choice, there is no need to consider any other case.

Answer: E

thakurarun85 · Answer

The sixth biggest number (median) must be 9. The numbers immediately smaller than the 9 must be 8s, and there must be more of those than any other number. all the other numbers except the biggest must be as small as possible as the total must be 110 and we want the maximize biggest.

Looking at the number of 8s…

Two 8s gives 1, 2, 3, 8, 8, 9, 10, 11, 12, 13, 33
Three 8s gives 1, 1, 8, 8, 8, 9, 9, 10, 10, 11, 35
Four 8s gives 1, 8, 8, 8, 8, 9, 9, 9, 10, 10, 30
Five 8s gives 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 24

curtsy:- Quora

samarpan.g28 · Answer

I have gone through all the solutions provided above. However to conclude that mode 8 repeats 3 times, first 2 terms will be 1, 10th term will be 11.. this is time consuming. It took me so long to figure out what the possible values could be.

bumpbot · Answer

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