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Since the elements are {5, 8 , x, y} and x<=4 , y>=10, the range is simply y-x

I. 3x+5y=56
y-x can't be determined with this equation. We need one more
Not Sufficient

II. -7x+7y=56
7(y-x)=56 => y-x = 8
Sufficient

Answer is B
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Here is my solution to this one =>

Since x≤4
and y≥10
The range of the set will be y-x


Statement 1-->

Clearly Not sufficient

Statement 2-->
-7x+7y=5
Hence y-x=8

Hence Range =8

Hence Sufficient

Hence B
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The range is the difference between the largest and the smallest numbers in a set.
In the given set, the maximum value that "x" can take is "4" and the minimum value that "Y" can take is 10, hence it indicates Y is the largest data point and "x" is the smallest data point.
Thus Range = Y-X

This question does not require us to find different values of X and Y, we just need to find "Y-X" for our answer.
First equation alone is not sufficient as there is no way we can find single value of Y-X. This equation to be true shall have multiple value to X and Y

Second equation if solved intelligently yields Y-X=8 ==> Answer - B
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chetan2u
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If x ≤ 4 and y ≥ 10, what is the range of the numbers in the set {5, 8, x, y}?

(1) 3x + 5y = 72
(2) –7x + 7y = 56

Hi,
a good Q...
as can be seen from the restrictions in the upper limit of x as 4 and lower limit of y as 10, the range would depend on the values of x and y only......

lets see the statements...
(1) 3x + 5y = 72
since y is the bigger number, lets substitue y as 10, then 11, 12,13... and find if there is any integer value fitting in for x..
let y=10.. 3x+5*10=72... 3x=22.... x cannot be integer..
let y=11... 3x+5*11=72...3x=17...x again cannot be integer
let y=12.. 3x+5*12=72... 3x=12.... x =4.. so x,y can be 4,12and we get one set of answer..range=8
let y=13... 3x+5*13=72...3x=7...x again cannot be integer
let y=14... 3x+5*14=72...3x=2...x again cannot be integer
let y=15.. 3x+5*15=72... 3x=-3.... x =-1.. so x,y can be -1,15 and we get another set of answer..range=16
no need to look forward.. we have more than one range so insuff.

(2) –7x + 7y = 56
7(y-x)=56..
y-x=8..
suff as it gives us the range as 8 irrespective of values of x and y
x,y can be 2,10...3,11...4,12 and n

ans B
Why is it important for x and y to only be integers? I am confused since its not mentioned in the question and Range can be of non-integer values also right?
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chetan2u
Bunuel
If x ≤ 4 and y ≥ 10, what is the range of the numbers in the set {5, 8, x, y}?

(1) 3x + 5y = 72
(2) –7x + 7y = 56

Hi,
a good Q...
as can be seen from the restrictions in the upper limit of x as 4 and lower limit of y as 10, the range would depend on the values of x and y only......

lets see the statements...
(1) 3x + 5y = 72
since y is the bigger number, lets substitue y as 10, then 11, 12,13... and find if there is any integer value fitting in for x..
let y=10.. 3x+5*10=72... 3x=22.... x cannot be integer..
let y=11... 3x+5*11=72...3x=17...x again cannot be integer
let y=12.. 3x+5*12=72... 3x=12.... x =4.. so x,y can be 4,12and we get one set of answer..range=8
let y=13... 3x+5*13=72...3x=7...x again cannot be integer
let y=14... 3x+5*14=72...3x=2...x again cannot be integer
let y=15.. 3x+5*15=72... 3x=-3.... x =-1.. so x,y can be -1,15 and we get another set of answer..range=16
no need to look forward.. we have more than one range so insuff.

(2) –7x + 7y = 56
7(y-x)=56..
y-x=8..
suff as it gives us the range as 8 irrespective of values of x and y
x,y can be 2,10...3,11...4,12 and n

ans B
Why is it important for x and y to only be integers? I am confused since its not mentioned in the question and Range can be of non-integer values also right?

x and y don’t need to be integers. Given that x ≤ 4 and y ≥ 10, x is the smallest value and y is the largest value in the set {5, 8, x, y}, making the range equal to y - x.

For (1), there are multiple values of x and y, whether integer or non-integer, that satisfy 3x + 5y = 72, resulting in different values for y - x. Therefore, (1) is not sufficient.

For (2), dividing –7x + 7y = 56 by 7 gives y - x = 8, which directly answers the question. Hence, (2) is sufficient.
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