Is the range of the integers 6, 3, y, 4, 5, and x greater

SOLUTION

Is the range of the integers 6, 3, y, 4, 5, and x greater than 9?

Given integers are: {3, 4, 5, 6, x, y}

(1) y > 3x. If \(x=1\) and \(y=4\) then the range=6-1=5<9 but if \(x=100\) then the range>9. Not sufficient.

(2) y > x > 3. If \(x=4\) and \(y=5\) then the range=6-3=3<9 but if \(x=100\) then the range>9. Not sufficient.

(1)+(2) From \(x > 3\) we have that the least value of \(x\) is 4, and from \(y > 3x=12\) we have that the least value of \(y\) is 13, hence the least value of the range is 13-3=10>9. Sufficient.

Answer: C.
_________________

Hi,

Range = Largest value - smallest value.

6, 3, y, 4, 5, and x, where x & y are integers

Using (1),
y>3x, if x=1, then y = 4, 5,..100....
in each case range can be 5, 6,....So, range is greater than 5. Insufficient.

Using (2),
y>x>3.
Minimum value of x = 4,
y=5,6,7... We can't say whether range is greater than 9.

Combining both statements;
\(x_{min} = 4\)
& since, y > 3x, \(y_{min}=13,\)
thus, 3, 4, 4, 5, 6, & 13 has range (13-3)=10, which is greater than 9
and on increasing x, range will also increase.

Thus, answer is (C),

Regards,

Here is a visual that should help.

pretty easy for a 700 level question...

1. x can be 1, and y can be 4 - so the answer is NO
x can be 4, and y can be 13 - so the answer is YES.
1 alone is insufficient. A and D are out.

2. y>x>3.
x can be 4, y =5 - answer is no
x can be 4, y can be 20 - answer is YES.
2 alone is not sufficient. B is out.

1+2.
x>3
y>3x
minimum value for x is 4.
minimum value for y is 13
yes, the range is greater than 9.

sufficient.
answer is C.

Statement One Alone:

y > 3x

Statement one is not sufficient to answer the question. For example, if x = 0, y could be 1 and the range is 6 - 0 = 6, which is less than 9. However, if x = 10, y could be 31 and the range is 31 - 3 = 28, which is greater than 9.

Statement Two Alone:

y > x > 3

Statement two is not sufficient to answer the question. For example, if x = 4, y could be 5 and the range is 6 - 0 = 6, which is less than 9. However, if x = 14, y could be 15 and the range is 15 - 3 = 12, which is greater than 9.

Statements One and Two Together:

From the two statements, we know that x > 3 and y > 3x. So the smallest integer x can be is 4 and the smallest integer y can be is 3(4) + 1 = 13. Thus the smallest range of the integers is 13 - 3 = 10, which is greater 9. Since 10 is the smallest range, any other range of the integers will be greater than 10 and hence greater than 9.

Answer: C
_________________

In the solution of these questions it is not asked to take minimum value and x and y can take any numbers so why we have solved this by taking minimum values

Hi arpitalewe,

Welcome to the GMAT Club!

So the logic behind trying to take the minimum permissible values for x and y is: we want to see if the minimum permissible value of y is such that range of the given set is greater than 9. If for minimum value this condition holds then for all other values it will hold as well.

Hope this solves your doubt.

Let me know if you need further clarification or if the above solutions make sense with the above logic.

Regards,
Gladi
_________________

Statement 1)

y > 3x

if x = 1 then y must be at least 4 but we cannot tell any further. 6-1 = 5 > 9 no.

If x = 5 y must be at least 16. 16 - 3 = 13 > 9 yes

Insufficient.

Statement 2)

y > x > 3 we can try the x = 4 and y = 5 we get an answer 6-3 = 3 > 9 no.

Try x = 4 y = 16, 16-3 = 13 > 9 yes.

Insufficient.

Now combine (1 and 2)

If x = 4 then y at least 13 (given that we took the minimum possible value for x)

13 - 3 = 10 > 9

Try x = 10 then y at least 31

31 - 3 = 28 > 9

Sufficient. Answer choice C

arpitalewe check the above solution it might help.

Also as Gladiator59 explained that the logic is to test if the minimum value satisfies the condition (range > 9) then the maximum will satisfy it too since y > 3x

Posted from my mobile device

Hello Bunuel,

In no explaination i see anyone considering x or y to be negative. In quesion it is mentioned they are integers, so will be right to consider negative values? Am i missing somethinghere?
Also i assumed all the integers would be distinct, but unless stated , we can consider that integers are repeating, something i missed.

Thank you in advance for answering

(2) says that y > x > 3, so neither of them can be negative.
_________________

I have one small doubt:

Doubt statement: How do we know that x and y are integers since there is no mention of their nature. As per GMAT, unless not mentioned or derived from the available information, all numbers are real numbers. So, does it means that in a range, all values in the list must be the same type?

Is the range of the integers 6, 3, y, 4, 5, and x greater than 9?

3,4,5,6,x,y

Suppose y is the largest, then the question is asking whether y - 3 > 9 or y > 12?

(1) y > 3x
Let's say x = 3 ---> y > 9 so y = 10,11,12,13 are all valid
Insufficient
(2) y > x > 3
y = 5 --> R = 3
y = 10000000 --> R > 9
Insufficient

C: x = 4 at minimum so y > 3(4) ---> y > 12 at minimum

C

anonymous19 It states explicitly that x and y are integers. One should never assume indeed and your suspicion that they are all of the same type is fair. I don't recall ever seeing too many questions on here that mix types (unless otherwise stated)
_________________

Statement 1: y > 3x
Let us say, x = -5000, then y> -15000; Say, y = -1000
Range: 6 - (-5000) > 9

However, if x is 0, then y> 3(0) or y>0; Say, y=1;
Range: 6-0 = 6<9;
So, (1) is unsufficient

Statement 2: 3<x<y;
Say, x = 4, y = 5;
Range = 6-3 = 3 <9
But if x = 1000; y = 1001;
Range = 1001- 3 > 9

Therefore, (2) is insufficient

Combining (1) and (2): 3<x<y
And y>3x

Say, x=4 (as low within the constraints as possible)
Therefore, y>12
Say, y=13 (as low within the constraints as possible)

Range = 13-3 = 10 > 9

Any integer value of x > 4 would yield larger y and therefore, a range even larger than 9

Therefore, (C)

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________