List A contains 5 positive integers, and the average (arithmetic mean)

List A contains 5 positive integers, and the average (arithmetic mean) of the integers in the list is 7. If the integers 6, 7, and 8 are in list A, what is the range of list A?

(1) The integer 3 is in list A.
(2) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term.

Kudos for a correct solution.

First put everything together: 5 Integers * Average of 7 = Sum of 35 for the whole List A. 6, 7 and 8 included equals 21. Therefore 2 integers left with a value of 14.

Statement 1: Sufficient
The list contains 3, 6, 7, 8 and 11 as the last integer (all adding up to 35). Range = 11-3 = 8

Statement 2: Sufficient
How can we split 14 so that the biggest Number of the List is > 3x smallest and <4x smallest. Only 3 and 11 fit. Range 11-3.

Answer D.

avg of 5 integers is 7 means total is 7x5=35
if integers 6,7,8 are in list means sum of remaining 2 nos is 35-21=14

option A: 3 is also in the list means other no is 11. so the range is 8.

Sufficient

Option B: take smallest term as 3 gives other term as 10 & 11.
other term could only be 14 because sum of these 2 terms should be 14.

sufficient

Ans is - D

Jimmy

Explaination :
Let the two missing integers are x and Y.
So list is x,y,6,7,8 and average is 7

Henc X + Y + 6+7+8 = 35
X + Y = 14.

If X = 1 and Y = 13 - range = 12
If X = 10 and Y = 4 Range = 6

Basically to find the range we need to find X and Y.

Statement 1 :
The integer 3 is in list A.
So either X or Y is 3, hence Either X or Y is 11. Consider X = 3 and Y = 11
Range = 8 - definite answer

Hence statement 1 is sufficient.

Statement 2 :
The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term.
Consider extremes,
consider Largest Term (Y) is equal to 3 time smallest term (x)
X + y = 14
X + 3X = 14
4X = 14
X = 3.5 So one of the term has to less than 3.5

consider Largest Term (Y) is equal to 4 time smallest term (x)
X + y = 14
X + 4X = 14
5X = 14
X = 2.8 so one of the term has to be more than 2.8

Only integer 3 satisfies these extremities, hence one of the missing integer is 3 and another is 11.

Range = 8

Hence statement 2 is sufficient.

Hence Option D, each of the statement is sufficient.

Solution -

Average of the list is (6+7+8+x+y) = 7*5 = 35 -> x+y=14.

Stmt 1 -
(6+7+8+3+y) = 35 - > Missing number from the list is y=11 ---> Range = 11-3 = 8. Sufficient.

Stmt 2 - The largest term y in list is greater than 3 times and less than 4 times the size of the smallest term x.

3x<y<4x -> After substituting y=14-x in the inequality 3x<14-x<4x -> 4x<14<5x. The only integer x satisfy the given inequality is x=3 (9<14<15).

Then Y=11. Range is 11-3=8. Sufficient.

ANS - D

Thanks,

Please give me kudos.

Given : Five Positive integers have 7 as average. If the integers 6, 7, and 8 are in list A, let us assume that other two integers are x and y.

So, \((x + y + 6 + 7 + 8) / 5 = 7\)
=> \(x + y + 21 = 35\) => \(x+y=14\)
So (x,y) can be (1,13), (2,12), (3,11), (4,10), (5,9), (6,8), (7,7), (8,6), (9,5), (10,4), (11,3), (12,2) or (13,1).
Now,
1.) The integer 3 is in list A.
So lets say, x = 3, that means y will be 11 (14-3).

So the list A is (3,6,7,8,11) And Range of A is 8 (11-3).

SUFFICIENT.

(2) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term.
\(3*smallest term < largest term < 4*smallest term.\)
Now we will check this condition for (x,y) as (1,13), (2,12), (3,11), (4,10), (5,9), (6,8) or (7,7).
As we can see, out of the values of (x,y) specified above only for (3,11) or (11,3) the list A satisfies this condition.
So the list A is (3,6,7,8,11) And Range of A is 8 (11-3).
Hence again SUFFICIENT.

So Answer is D.

5 Pos int in list A
Avg = 7
6,7,8 are in list A

5x7 is 35, total in list A must be 35
6+7+8 = 21

1) 3 is in the list

3 + 6 + 7 + 8 = 24, must get to 35, so last number is 11

Range 3 - 11

S.

2) Largest number is greater than 3 times, less than 4 times smallest numbers

Test it!

6+7+8 = 21

1*3 = 3
1*4 = 4
not possible

2*3 = 6
2*4 = 8
not possible

3*3 = 9
3*4 = 12
possible

4*3 = 12
4*4 = 16
not possible

S.

D.

MANHATTAN GMAT OFFICIAL SOLUTION:

The average of all five integers in the set is 7. Three of the integers in the set are given (6, 7, and 8) and they all have an average of 7. Therefore, the floating terms in this problem must also have an average of 7. If we assign x and y to represent these terms, we have:

(x + y)/2 = 7;
x + y = 14.

Our rephrased question is thus, “Given that x + y = 14, what is either x or y?” Once we know one of the values, we can solve for the other, and thereby determine the range of the set.

(1) SUFFICIENT: If 3 is one of the unknown integers, the other must be 11. The range is thus 11 – 3 = 8.

(2) SUFFICIENT: This statement might seem a little too vague to be sufficient, but by visually listing the possible pairs that add up to 14, we can rule out pairs that don't fit the constraint from this statement:


Notice that the pairings represent the constraint x + y = 14. Visually, this means that x and y are balanced around 7.

Among these pairs:
8 is 1.33 times the size of 6 (the ratio is too low).
9 is 1.8 times the size of 5 (the ratio is too low).
10 is 2.5 times the size of 4 (the ratio is too low).
11 is 3.66 times the size of 3 (an acceptable ratio).
12 is 6 times the size of 2 (the ratio is too high).

Only one pair of integers results in a ratio strictly between 3 and 4. The unknown terms must therefore be 3 and 11, and the range is 11 – 3 = 8.

The correct answer is D.

In this problem we saw the constraint x + y = 14: a fixed sum. Another common constraint is a fixed difference, such as a – b = 2. A fixed difference can be represented visually as a fixed distance between a and b on the number line, with a to the right because it is larger. That distance could move left or right:

(1)
\(x+y=14\)
\(3+y=14\)
\(y=11\)

Sufficient.

(2)
\(x+y=14\)
\(x+3x<14<x+4x\)
\(4x<14<5x\)
\(\frac{14}{5}<x<\frac{14}{4}\)
\(2.8<x<3.5\)

Since we are restricted to integers,

\(x=3\)

Sufficient.

---------------------------------------------------------------------------------------------

Can anyone help explain how to algebraically get from:
\(4x<14<5x\)

to

\(\frac{14}{5}<x<\frac{14}{4}\)

Or is it strictly intuitive?

(6+7+8+x+y)/5 = 7

Since the average of 6+7+8 is 7, the average of x+y must be 7 as well.

(x+y)/2 = 7 ---> x+y = 14

1.) The integer 3 is in the list A
3+y = 14 ---> y =11 so range = 3-11 SUFF

2.) The largest term in list A is greater than 3 times and less than 4 times the size of the smallest term
In order for x+y = 14, x and y must be the smallest and largest numbers in the set.
3x = y ---> 4x = 14 x = 3.5
4x = y ---> 5x = 14 x = 2.8
If x is an integer then x = 3 and y = 11 SUFF

---> Answer is D

I think there is a lot of complicated answers. Hence here is my simplified solution.

Given A.M = 35
Sum/5 = 7 => Sum = 35

Set : {x, y, 6,7,8} => x+y+6+7+8 => x+y = 14

Stmt 1 :
x=3 y=11 (Sufficient)

Stmt 2 :
Largest Term > 3*Smallest Term {Let a denote the smallest term}
Also,
Largest Term < 4*a

a=2
6<Largest Term<8, i.e, 7
Does not fit with x+y = 14

a=3
9<Largest Term<12, i.e, 10 & 11

we have already seen in stmt 1 that 11 satisfies.

Hence Sufficient.

Answer :: D

Lets say that 2 unknown integers in the set are X and Y

We know that X + Y + 6+ 7+ 8 = 7 x 5 = 35
X + Y = 35 – 21 = 14
X + Y = 14

Statement 1:
One integer is 3
If X = 3 then Y = 11
So your set is 3, 6, 7, 8, 11 ---- hence your conclusive range is 11-3 = 8
SUFFICIENT

Statement 2:
Assume that X is smallest and Y is largest term
3X < Y < 4x
But we know that X+Y = 14
Lets check values
X = 1 then y < 4 -----wont work
X = 2 then Y < 8 --- wont work
X = 3 then Y < 12 e.g. Y = 11 ----- this work
If X = 4 then y > 12 and y < 16 ---- but this will go beyond X+Y = 14
Only value that works is X = 3 and Y = 11
Set = (3, 6, 7, 8, 11)----------- range = 11-3 = 8
SUFFICIENT

Answer = D

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