A set of nonnegative integers consists of {x, x + 7, 2x, y, y + 5}.

Question

A set of nonnegative integers consists of {x, x + 7, 2x, y, y + 5}. The numbers of this set have four distinct values. What is its average (arithmetic mean)?

(1) x ≠ 5
(2) 4y + 12 = 6(y + 2)

Bunuel · Accepted Answer

A set of nonnegative integers consists of {x, x + 7, 2x, y, y + 5}. The numbers of this set have four distinct values. What is its average (arithmetic mean)?

Notice couple of things:
1. We are told that all numbers in the set are  integers ;
2. We are told that all those integers are  non-negative ;
3. We are told that the set contains four distinct values out of five (so two integers out of 5 are alike and other three are distinct).

(1) x ≠ 5. Clearly insufficient.

(2) 4y + 12 = 6(y + 2) --> 4y+12=6y+12 --> 2y=0 --> y=0. So, we have that our set is {x, x + 7, 2x, 0, 5}. Since we know that two integers out of 5 are alike then:

1. x, x+7 or 2x is 0. 
If x=0, then 2x=0 too, so in this case we'll have three alike integers x, 2x, and y. Thus this case is out.
If x+7=0, then x=-7 and we know that all integers must be non-negative so this case is out too.

2. x, x+7 or 2x is 5
 If x=5, then x+7=12 and 2x=10. This scenario is OK. The set in this case would be: {5, 12, 10, 0, 5} 
If x+7=5, then x=-2 and we know that the integers in the set must be non-negative. Thus this case is out.
If 2x=5, then x=5/2 and we know that the numbers in the set must be integers. Thus this case is out.

3. Two out of x, x+7 and 2x are alike.
x=2x is not possible, since in this case x=2x=0 and in this case we'll have three alike integers x, 2x, and y.
x=x+7 has no solution.
 2x=x+7 --> x=7. This scenario is OK. The set in this case would be: {7, 14, 14, 0, 5}.

Two cases are possible. Not sufficient.

(1)+(2) Since from (1) x ≠ 5, then from (2) x=7. So, the set is {7, 14, 14, 0, 5}. Sufficient.

Answer: C.

Hope it's clear.

siddharthismu87 · Answer

Hi, It goes like this: Statement 1: X not equal to 5, Which is INSUFFICIENT as we need to know the value of Y. Statement 2 says: 4y+12=6y+12 Solving STATEMENT 2, We get Y=0, Hence the series become, X, X+7, 2X, 0, 5 Now, we know that 2 numbers are identical, Which arises following possibilities: 1: x=2x Which means X=0, (But this is not possible as Three numbers in the series will be equal, which refuted the given condition) 2: X+7=2x, which means, X=7 3: X=5 4: X+7=5, (This is also not possible as it gives negative values of X) So Statement 2 says either X=7 or X=5, So Statement 2 is also Insufficient, Combining bo S 1 and 2, X=7, Hence C..

EvaJager · Answer

(1) If x = 0, the numbers are 0, 7, 0, y, y + 5.
If we take y different from 0, 7 and 2, for sure we get four distinct values. Therefore, we cannot know the average of the numbers.
For example y = 1: 0, 7, 0, 1, 6, for y = 10 - 0, 7, 0, 10, 15.
Not sufficient.

(2) From the given equality, y = 0.
We have the numbers x, x + 7, 2x, 0, 5.
x cannot be 0 - 0, 7, 0, 0, 5 - only 3 distinct values
x can be 5 - 5, 12, 10, 0, 5 - exactly 4 distinct values
x can be 7 - 7, 14, 14, 0, 5 - exactly 4 distinct values
Not sufficient.

(1) and (2) together:
If x is not 5, and x cannot be 0, then x, 2x, 0, and 5 are distinct numbers. It means that x + 7 must be one of them.
Since x + 7 > x, x + 7 > 0, x + 7 > 5 (x is positive), the only possibility left is x + 7 = 2x, from which x = 7.
The numbers are 7, 14, 14, 0, 5.
Sufficient.

Answer C.

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an    alternative    solution!

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mau5 · Answer

From F.S 1, all we know is that x ≠ 5. Now as there are 4 distinct values, any 2 of the elements have to be exactly same. Thus, we can have :

x = 2x --> x= 0 OR x+7 = 2x--> x = 7. 
In both cases we could assume the value of y = 1 and have the set as (0,7,0,1,6)--> Average = 14/5 = 2.8 OR the set would be (7,14,14,1,6) --> Average = 42/5 = 8.4.

Thus,even after adhering to the condition given in the F.S; as we are getting two different values for average-->Insufficient.

From F.S 2, we know that y=0. Thus, the set reads as (x,x+7,2x,0,5).

To make the elements assume 4 distinct values, we can set x = 5, and have the set as (5,12,10,0,5)
We could also assume x = 7 just as above and get the set as (7,14,14,0,5). Again, Insufficient.

Taking both the fact statements together, we now know that x = 7. Sufficient.

C.

ziko · Answer

An interesting question.

1statement) It tells us tha x does not equal to 5, lets take x=1 and y=1 (I am taking these numbers because in the question stem i see that there are four distinct numbers but we have five numbers, this means that two out of five numbers should be identical, so it is easier to take x and y as the same numbers. But to make this question even clearer i would add the word EXACTLY or ONLY four distinct numbers) then we have 1; 8; 2; 1; 6. If we take x=2 and y=2 then we have 2; 9; 4; 2; 7. Two sets which satisfies the question conditions of having four distinct numbers. Not sufficient.

2 statement) if we solve the equation it tells us that y=0. In this case we have last two numbers: 0 and 5. Lets see what are possible other numbers. We have x; x+7 and 2x. According to the question conditions there should be four distinct numbers, in order that to happen x should be equal to 5 or to 7, because in other cases we would have more than 4 distinct numbers or less than 4 distinct numbers. Two possible options - not sufficient.

Combining both statements we see that x does not equall to 5 then it means it is equall to 7. Both statements together are sufficient - choice C.

MahBir · Answer

I miss-read the second stmt, but I think this can be an alternate question where stmt 2 is alone sufficient
 (1) x ≠ 5
 (2) 4x + 12 = 6(y + 2)
Came up with answer B – please let me know if my approach is right here.

Set: {x, x + 7, 2x, y, y + 5}
Stmt 2 gives: 4x=6y i.e. 2x=3y hence x ≠ y or y ≠ 2x
Given that all are integers and are non-negative, hence x ≠ 2x, so from above set only 2 possible scenarios
1. y=x+7 (when combined with 2x=3y) leads to x = -21 -negative -not possible
2. x=y+5 (when combined with 2x=3y) leads to y = 10
so the set becomes {15, 22, 30, 10, 15}
hence answer is B

Divyadisha · Answer

We are given 5 non-negative integers and told that 4 are distinct numbers. That means there are two integers that are same.

(1) x ≠ 5. We don't know about y. not sufficient.

(2) 4y + 12 = 6(y + 2)
Solving this equation we get the value of y=5

But we don't know the value of x. Hence not sufficient.

Combining both statements:-
y=5
y+5=10

one value from x, x + 7, 2x is 5 or 10 
We know x is not equal to 5
which means 2x is not equal to 10
x+7=10 
x= 3

And now we can calculate the average. C is the answer

nahid78 · Answer

A tough one.. Kudos to bunuel for such nice explanation.

newyork2012 · Answer

(1) x ≠ 5 not sufficient
(2) 4y + 12 = 6(y + 2) => y=0 still not sufficient

1 and 2 together {x, x + 7, 2x, 0, 5}. only four distinct values ,at the same time x ≠ 5 ===> x≠0 ,but we need to have 2 same numbers ,so 2x = x+7 ====> x=7

so {7, 7 + 7, 2*7, 0, 5} ..easy to get the mean

c

CrackverbalGMAT · Answer

Q.stem says: The set has non negative integers(so 0 included) & two are identical(since 4 of 5 elements are distinct).
To find mean,we need to know the elements in the set.

#St1: x not equal to 5 gives us no information on y.
Insufficient. Eliminate A,D

#St2: 4y + 12 = 6(y+2)
So y=0 and hence two of the elements are 0 and 5(y+5).

Can we have another 0 as x?
In that case x,2x,y would all be 0.3 identical elements. Not possible.

Can we have another 5 as x?
In that case the set {x,x+7,2x,y,y+5} would be {5,12,10,0,5}

Two elements identical. Possible.
Can we have the terms x+7 and 2x identical?

So x+7 = 2x ? ( Two elements identical)
Then the set is {7,14,14,0,5}.Possible.

Thus no definite answer from st2.Eliminate B.
Combine 1 & 2,

You have x as 7 as it cannot be 5(st 1)
So {7,14,14,0,5}

(Sufficient)
 (option c) 
  D.S
GMAT SME

Sushen · Answer

C Q: What is the mean? 
 --{x,x+7,2x,y,y+5}
 --All NON-Negative integers
 --Two of those integers are alike and the other 3 all different.

(1) x≠5 Simply Not Sufficient. 
 
(2) 4y+12=6y+12 y=0
 {x,x+7,2x,0,5}
Limitations: x+7≠0 (all non-negative. X would be -7)
 x≠0, 2x≠0 (Only 2 values are the same)
 x≠x+7

Possibilities: 
 If x=7, we would have 4 values and could calculate a mean.
 If x = 1, we would have 4 values and we could calculate a mean
 BUT, these means would be DIFFERENT. I CANNOT determine the mean of this set.
 Not Sufficient

TOGETHER: 
x≠0 or 5 x≠2x (that would mean x=0, see above. Can^' t be true.
SO, x+7=2x and x=7 and we CAN calculate the mean.

Sufficient-----> C is the answer!

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A set of nonnegative integers consists of {x, x + 7, 2x, y, y + 5}.

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