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Pick numbers 2, 4, 6 so that:
R = 12
X = 2
Y = 2

Statement 1)
12 - 2(2) - 2 / 3
=6/3
=2 -> Smallest #

Statement 5)
12 + 2 + 2(2) / 3
=18/3
=6 -> Largest #
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gmatt1476
The real number R is expressed as the sum of three numbers so that the median of the three numbers is x greater than the least of the three numbers and y less than the greatest of the three numbers, where x and y are positive real numbers. Select for Least an expression equivalent to the least of the three numbers and select the Greatest an expression equivalent to the greatest of the three numbers. Make only two selectionA, one in each column.
The three numbers are a, b and c.

Given : b=a+x and c=a+x+y

R= a+a+x+a+x+y = 3a+2x+y

\(a=\frac{R-2x-y}{3}\)...a is the least.
Greatest = a+x+y = \( \frac{R-2x-y}{3}+x+y= \frac{R-2x-y+3x+3y}{3}=\frac{R+x+2y}{3}\)...
­
Can you please elaborate a bit? Its really difficult for me to translate the question stem to actual equations and i get lost in between especially in this question its really confusing
­
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Jake7Wimmer Those numbers you picked work for multiple options since 2x and 2y both give 4.
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Jake7Wimmer
Pick numbers 2, 4, 6 so that:
R = 12
X = 2
Y = 2

Statement 1)
12 - 2(2) - 2 / 3
=6/3
=2 -> Smallest #

Statement 5)
12 + 2 + 2(2) / 3
=18/3
=6 -> Largest #
­This method does not work since we get the same value for both options 5 and 6 (as the greatest value). Also, unequal spacing; say 2,5,6 or 1,2,5 will also give 2 seperate answers here since the answer (if we assume numbers to plug in the question) will always be dependent on the numbers picked by us (while calculating the bigger value). 

Could someone suggest an alternate method here?
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harshheda2002
Jake7Wimmer
Pick numbers 2, 4, 6 so that:
R = 12
X = 2
Y = 2

Statement 1)
12 - 2(2) - 2 / 3
=6/3
=2 -> Smallest #

Statement 5)
12 + 2 + 2(2) / 3
=18/3
=6 -> Largest #
­This method does not work since we get the same value for both options 5 and 6 (as the greatest value). Also, unequal spacing; say 2,5,6 or 1,2,5 will also give 2 seperate answers here since the answer (if we assume numbers to plug in the question) will always be dependent on the numbers picked by us (while calculating the bigger value).

Could someone suggest an alternate method here?
It actually works, pick x=1 and y=3 {1,2,5}
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I picked numbers 3; 5; and 8. (So x=2 and y=3)

For the least I got (R-x-2y)/3 not (R-2x-y)/3

Where did I go wrong?
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Hey, those same numbers also work for the second choice (R-x-2y)/3

Am I missing something?
Jake7Wimmer
Pick numbers 2, 4, 6 so that:
R = 12
X = 2
Y = 2

Statement 1)
12 - 2(2) - 2 / 3
=6/3
=2 -> Smallest #

Statement 5)
12 + 2 + 2(2) / 3
=18/3
=6 -> Largest #
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In your case \(\frac{R - x - 2y}{3} = \frac{16 - 2 - 2 * 3}{3} = \frac{8}{3}\), which isn't the least of the chosen numbers (3,5,8). However, the other equation actually works \(\frac{R - 2x - y}{3} = \frac{16 - 2 * 2 - 3}{3} = \frac{9}{3} = 3\), which is the least of 3,5,8.

I hope that helped you!
MyNameisFritz
I picked numbers 3; 5; and 8. (So x=2 and y=3)

For the least I got (R-x-2y)/3 not (R-2x-y)/3

Where did I go wrong?
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No, you aren't missing anything, you instead found a problem with this type of solution. Just picking numbers and the substituting the values in the equations will give you the correct answers. However, it can be unreliable since other answer options, for that choice of numbers, might give the same results.
MyNameisFritz
Hey, those same numbers also work for the second choice (R-x-2y)/3

Am I missing something?
Jake7Wimmer
Pick numbers 2, 4, 6 so that:
R = 12
X = 2
Y = 2

Statement 1)
12 - 2(2) - 2 / 3
=6/3
=2 -> Smallest #

Statement 5)
12 + 2 + 2(2) / 3
=18/3
=6 -> Largest #
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Thank you, I understood where I went wrong. The question asked to chose the least and greatest of the three numbers. I simply chose the smallest and largest number out of the 6 equations. Got to be a lot more careful in DI problems.
Bismuth83
In your case \(\frac{R - x - 2y}{3} = \frac{16 - 2 - 2 * 3}{3} = \frac{8}{3}\), which isn't the least of the chosen numbers (3,5,8). However, the other equation actually works \(\frac{R - 2x - y}{3} = \frac{16 - 2 * 2 - 3}{3} = \frac{9}{3} = 3\), which is the least of 3,5,8.

I hope that helped you!
MyNameisFritz
I picked numbers 3; 5; and 8. (So x=2 and y=3)

For the least I got (R-x-2y)/3 not (R-2x-y)/3

Where did I go wrong?
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KarishmaB why is my way wrong
R = L + M + G
Where
L is least number
G is greatest number
M is median

M = x+L
M =G-y

x+L = G -y
G-L = x+y
Difference of the two numbers is X + Y
So the options that worked are
Greatest : (R+2x+y)/3
Lowest : (R-x-2y)/3


Difference = X+Y
gmatt1476
The real number R is expressed as the sum of three numbers so that the median of the three numbers is x greater than the least of the three numbers and y less than the greatest of the three numbers, where x and y are positive real numbers. Select for Least an expression equivalent to the least of the three numbers and select the Greatest an expression equivalent to the greatest of the three numbers. Make only two selectionA, one in each column.
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Your logic is correct and you showed that the difference must be x+y. However, you didn't show that if the difference between two answer choices is x+y, then they must be the answer. In other words, you could have two numbers with that difference but are impossible to fit all of the conditions.

Hopefully that helped!
Kavicogsci
KarishmaB why is my way wrong
R = L + M + G
Where
L is least number
G is greatest number
M is median

M = x+L
M =G-y

x+L = G -y
G-L = x+y
Difference of the two numbers is X + Y
So the options that worked are
Greatest : (R+2x+y)/3
Lowest : (R-x-2y)/3


Difference = X+Y
gmatt1476
The real number R is expressed as the sum of three numbers so that the median of the three numbers is x greater than the least of the three numbers and y less than the greatest of the three numbers, where x and y are positive real numbers. Select for Least an expression equivalent to the least of the three numbers and select the Greatest an expression equivalent to the greatest of the three numbers. Make only two selectionA, one in each column.
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