A set of 7 different integers has a median of 9, range of 11

Question

A set of 7 different integers has a median of 9, a range of 11, and an average of 10. What is 
the highest possible value of the lowest integer in the set?

A. 3
B. 4
C. 5
D. 6
E. 8

Source: Experts Global

alldaytestprep · Accepted Answer

A set of 7 different integers has a median of 9, a range of 11, and an average of 10. What is 
the highest possible value of the lowest integer in the set?

A. 3
B. 4
C. 5
D. 6
E. 8

Let's call our numbers in ascending order a, b, c, d, e, f, g.

d = median = 9 ----------> a, b, c, 9, e, f, g

We are looking for max "a" can be.

Given that each number is different the absolute max "a" could possibly be is 6. Therefore eliminate E) as an answer choice.

To have an average of 10 the numbers must sum to 70(70/7 = 10). Start by plugging in highest number remaining for "a" to see if it works.

If "a" = 6 then b = 7, c = 8, and g = 17(since range must be 11) ----> 6, 7, 8, 9, e, f, 17 
------> to satisfy our requirements that the set sums to 70, e & f need to sum to 23 and each be between 9 - 17. e = 11, f = 12 works to give us a set of:  6, 7, 8, 9, 11, 12, 17

median = 9 
range = 11
mean = 10

Answer: D

Hero8888 · Answer

If integers are all different, then max (1st) will be if terms 1,2,3 and  4 (which is median)  will be consecutive, so

6,7,8, 9 ,x,y, 6+11.

We have to check, since we know that sum of the set = 7*10=70, hence 6+7+8+9+x+y+17=70, x+y = 23, x can be 10 and y=13 - it works. Answer (D)

seed · Answer

A. 3
B. 4
C. 5
D. 6
E. 8

Let numbers be:

a, b, c, 9, d, e, f

f-a=11
Avg=10 -> Sum of numbers = 70.

What is the maximum value of a?

a, b and c will be < 9 and d, e and f will be more than 9.

The maximum value of 'a' will be when the decrease of each term less than 9 is minimum.

So, 6, 7, 8, 9, d, e, f

Note: We cannot have all numbers less than 9 be 8 because its given that the numbers are different.

We need to confirm that the remaining 70 - (6+7+8+9) = 40 is sufficient for d, e and f to be different. 
We are also given f=a+11=6+11=17 so, now we just need to confirm if d and e can be different and greater than 9.

d+e = 40-f= 40-17=23 clearly, we can have d, e values different and greater than 9 (d+e = 11+12 or 10 + 13).

Answer: (D).

akadiyan · Answer

A set of 7 different integers has a median of 9, a range of 11, and an average of 10. What is 
the highest possible value of the lowest integer in the set?

A. 3
B. 4
C. 5
D. 6
E. 8

Step 1: The median of odd set of integers is always the middle number - In this if median is 9 , then 4th digit is 9.

Step 2: If 4th digit is 9 and all digits are different, then numbers smaller than median can be 6,7 and 8.

Step 3: The range is 11 then one option for highest number could be 17

Step 4: The average of 7 numbers are 10 , so sum of 7 digits is 70. From the steps above , we know the numbers can be 6,7,8,9 and 17. The sum of 5 digits is equal to 47.

Now we know that sum is equal to 70 and sum of first 5 digits is 43. From this we get the sum of digits 5 and 6 is equal to 23

the fifth and sixth digits can be 10+13 or 11+12. Both are greater than 9 and lesser than 17 and matches our criteria.

Ans:   6

ScottTargetTestPrep · Answer

Since the average is 10, we have a sum of 7 x 10 = 70.

Also, since we need to determine the highest possible value of the lowest integer in the set, we can let that value be 6 so that the 3 numbers less than the median are 6, 7, 8 (recall that all the integers are different and the median is 9). Since the range is 11, the largest number is 6 + 11 = 17. Therefore, so far we have a sum of 6 + 7 + 8 + 9 + 17 = 47. This leaves us a sum of 70 - 47 = 23 for the remaining two numbers. So we can let them to be 11 and 12 (or, 10 and 13). Therefore, we see that it is totally feasible to have 6 as the smallest integer and it will also be the highest possible value for the smallest integer since if we take any integer higher than that, either the median will not be 9 or the numbers will not be all different.

Answer: D

gracie · Answer

not really conceptual, but here's how I do it:
start with 9-9-9-9-9-9-9 
then do range:
(9+x)-(9-x)=10→
x=5
subtract 5 from 1st term; add 5 to last term:
4-9-9-9-9-9-14
that works, but we can still go lower
subtract 1 from 1st and last terms, and add 2 to 6th:
3-9-9-9-9-11-13
that works too, and 3 is as low as we can go for 1st term
to get down to 2, we'd have to add 2 to 6th term, which would make it higher than last term, 12
3
D

pushpitkc · Answer

We need 7 different integers with a range of 10

First integer in the set: x
Last integer in the set: x+10
Since the median is 9, the 4th integer is 9

(If the set has an odd number of elements, the median is the middle value)

The lowest value that the 5th and 6th integers can take is 10 and 11. In order to have
the lowest possible value for the smallest integer, the largest integer will have a value
of 12. So, the set will look like this - 2,b,c,9,10,11,12 
In this set, the maximum value of b & c is 15. As the sum is 7*9 = 63(average - 9), b+c = 19.

If the lowest integer were slightly bigger, the set could be 3,7,8,9,11,12,13 whose sum is 63.

Therefore, the lowest possible value of the smallest integer in the set is  3(Option D)

KanishkM · Answer

Set of 7 different integers

a b c d e f g

d = 9

now a can be 6

D

EgmatQuantExpert · Answer

Given

• A set of 7 different integers has a median of 9.
• It has a range of 11 and an average of 10.

To Find

• The highest possible value of the lowest integer in the set

Approach and Working Out

• Average is 10 so the sum is 7*10 = 70
 o Median is 9 and hence three numbers should be less than that.
o 6, 7, 8, 9, a, b, c are the best possible scenario we have.
o Let us check it’s feasible or not.
o Sum is 30 + a + b + c = 70
o a + b + c = 40
o The highest one needs to be 17 (as the range is 11)
o So b + c = 23
o It can be 10, 13 or 11, 12

Correct Answer: Option D

griffith7 · Answer

Not sure if this method is any good ~

Say we have 7 integers in order, and we know the middle term is 9 while the final term is (first term + 11):

a, b, c, 9 ,d, e, a+11 ~ all of which sum to 70 (mean * #terms).

a + b + c + 9 + d + e + a + 11 = 70
2a + b + c + d + e = 50

Hereon, we're trying to solve for the maximum value of (a). For (a) to be maximized, the other terms must all be minimized using all known constraints.

We know the minimum values of (d) and (e) are 10 and 11, because they're ordered integers greater than 9.

2a + b + c + 10 + 11 = 50
2a + b + c = 29

We know the minimum values of (b) and (c) are (a+1) and (a+2).

2a + a + 1 + a + 2 = 29
4a + 3 = 29
4a = 26
2a = 13
a = 6.5

Hence the maximum integer value of the first term must be 6.

Gemmie · Answer

A set of 7 integers; Avg = 10 => Sum = 70

a_1: 6 (this is the highest possible based on the median)
a_2: 7
a_3: 8
a_4: 9 (median)
a_5: y
a_6: x
a_7: 17

Check: 6 + 7 + 8 + 9 + x + y + 17 = 47 + x + y = 70
=> x + y = 23
9 < x, y < 17
=> (x, y) can be (10, 13) or (11, 12)

Kinshook · Answer

pushpitkc 
Given: A set of 7 different integers has a median of 9, a range of 11, and an average of 10.
Asked: What is the highest possible value of the lowest integer in the set?
Let the lowest integer be x.
Highest integer = x + 11

Sum of 7 integers = 10*7 = 70

To maximize lowest integer in the set, we have to minimize all others
Let the set be {x,x+1,x+2,9,10,11,x+11}
4x + 44 = 70; 4x = 26; x = 6.5: Since x is an integer x = 6 since x can not be 7 which is greater than 6.5

IMO D

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A set of 7 different integers has a median of 9, range of 11

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