The real number R is expressed as the sum of three numbers so that the

Question

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 selection A, one in each column.

https://gmatclub.com/forum/download/file.php?id=74497

chetan2u · Accepted Answer

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} ...

Jake7Wimmer · Answer

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 #

shwetasood · Answer

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

Yes2GMAT · Answer

Jake7Wimmer  Those numbers you picked work for multiple options since 2x and 2y both give 4.

harshheda2002 · Answer

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?

Bismuth83 · Answer

1. Let the 3 numbers be a , b , and c such that a \leq b \leq c . The question asks us to find a ( the least number ) and c ( the greatest number ). 2. The numbers can be drawn on a number line . https://gmatclub.com/forum/download/file.php?mode=view&id=78781 3. Now we can write the distances between each other. https://gmatclub.com/forum/download/file.php?mode=view&id=78782 4. Since R = a + b + c , it equals the sum of the distances to 0. https://gmatclub.com/forum/download/file.php?mode=view&id=78783 5. So, R = ( a ) + ( a + x ) + ( a + x + y ) = 3 a + 2 x + y \rightarrow a = \frac{R - 2x - y}{3} . In addition, R = ( c - y - x ) + ( c - y ) + ( c ) = 3 c - 2 y - x \rightarrow c = \frac{R + 2y + x}{3} . NumberlineR1.jpg NumberlineR3.jpg NumberlineR2.jpg

Juangui17 · Answer

It actually works, pick x=1 and y=3 {1,2,5}

MyNameisFritz · Answer

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?

MyNameisFritz · Answer

Hey, those same numbers also work for the second choice  (R-x-2y)/3

Am I missing something?

Bismuth83 · Answer

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!

Bismuth83 · Answer

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 · Answer

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.

Kavicogsci · Answer

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

Bismuth83 · Answer

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!

egmat · Answer

Your setup is correct, but there's a subtle  trap  here.

What you did right: 
- R = L + M + G
- M = L + x and M = G - y
- G - L = x + y

Where it breaks down:

You found that G - L = x + y and then picked options where the difference equals x + y. Your choice was:
- Greatest:  (R+2x+y)/3 
- Least:  (R-x-2y)/3

The difference is indeed x + y.  But  having the correct difference is  necessary but not sufficient !

The missing check:  Do these three numbers actually sum to R?

Let's test your pair:
- L =  (R-x-2y)/3 
- M = L + x =  (R+2x-2y)/3 
- G =  (R+2x+y)/3

Sum =  (R-x-2y)/3 + (R+2x-2y)/3 + (R+2x+y)/3 
 =  (3R + 3x - 3y)/3 
 =  R + x - y

This does  NOT  equal R!

The correct approach:

Set up equations and solve for each variable directly:

From R = L + M + G, substitute M = L + x and G = M + y = L + x + y:
R = L + (L + x) + (L + x + y) =  3L + 2x + y

Therefore:  L =  (R - 2x - y)/3

Similarly, G = L + x + y =  (R - 2x - y)/3 + x + y = (R + x + 2y)/3

Verification: 
- L =  (R-2x-y)/3 
- M = L + x =  (R+x-y)/3 
- G =  (R+x+2y)/3 
- Sum =  (R-2x-y + R+x-y + R+x+2y)/3 = 3R/3 = R

Answer: Least = (R-2x-y)/3, Greatest = (R+x+2y)/3

Key Takeaway:  When you have multiple constraints (difference AND sum), you must verify  ALL  constraints are satisfied, not just one.

The real number R is expressed as the sum of three numbers so that the

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Least	Greatest
		(R-2x-y)/3
		(R-x-2y)/3
		(R-x-y)/3
		(R+x+y)/3
		(R+x+2y)/3
		(R+2x+y)/3

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