Statistics - Range

linfongyu wrote:If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?
1) The average of the set containing the numbers x, y, z, and 8 is 12.5
2) The mean and the median of the set containing the numbers x, y, and z are equal.

[spoiler]OA: C. I disagree and think that it should be A.[/spoiler]

Range is the distance between the smallest and largest numbers in the set.

For the sake of simplicity, let's say that x is the smallest number and z is the biggest.

From the original, then, we know that z - x = 8. The question is "what's the value of x?".

1) (x + y + z + 8)/4 = 12.5

We now have 2 equations and 3 unknowns. There's no way to combine them to solve for x: insufficient.

2) with an odd number of terms, the median is the middle term. By our previous definition, that's y. So:

y = (x + y + z)/3

We now have 2 equations and 3 unknowns. There's no way to combine them to solve for x: insufficient.

Combined:

We now have 3 equations and 3 unknowns - ding ding - sufficient, choose (C).

The "n linear equations" rule is the most powerful data sufficiency tool known to GMATkind; the better you understand the rule (including all of its subtleties), the fewer calculations you'll need to make on Test Day.

x <=y <= z and z-x = 8


1. x+y+z = 42. Insufficient, since x value varies when ya vary y.
2. (x+y+z)/3 = y or x+z = 2y. with z-x =8, x = y -4. Insuff

Combined sufficient.

I must be missing something. The question asks for the value of the smallest number in the set. Doesn't that mean just that, the value of the smallest number in the set?

According to statement 1, x+y+z+8 = 50, x+y+z = 42, (x+y+z)/3 = 14. The range of {x,y,z} is 8. The only set of number I can think of, with a range of 8, that satisfies this condition is {10, 14, 18}, and therefore the value of the smallest number in the set is 10. What am I missing?

linfongyu wrote:I must be missing something. The question asks for the value of the smallest number in the set. Doesn't that mean just that, the value of the smallest number in the set?

According to statement 1, x+y+z+8 = 50, x+y+z = 42, (x+y+z)/3 = 14. The range of {x,y,z} is 8. The only set of number I can think of, with a range of 8, that satisfies this condition is {10, 14, 18}, and therefore the value of the smallest number in the set is 10. What am I missing?

How about {9, 16, 17} or {11, 12, 19}?

Also remember, x, y and z don't even have to be integers. Once you consider non-integers, the number of possible sets is infinite.

linfongyu wrote:I must be missing something. The question asks for the value of the smallest number in the set. Doesn't that mean just that, the value of the smallest number in the set?

According to statement 1, x+y+z+8 = 50, x+y+z = 42, (x+y+z)/3 = 14. The range of {x,y,z} is 8. The only set of number I can think of, with a range of 8, that satisfies this condition is {10, 14, 18}, and therefore the value of the smallest number in the set is 10. What am I missing?

How abt {11, 12, 19}, whose average is 14 n whose range is 8.

the smallest number <= average
the largest >= average

14 -a is the smallest number
22 -a is the largest number

14 -a, 6 +2a, 22- a are the numbers.

14 -a <= 6+2a
3a >= 8

3a <= 16

8 <= 3a <= 16
8/3 <= a <= 16/3
a = 3 or 4 or 5

We got two lists.

a = 3, {11, 12, 19}
a = 4 {10, 14, 18}
a =5 {9, 16, 17}

Insufficient

Thanks guys. I see my mental block... I got to (x+y+z)/3 = 14, found the set of 3 numbers 10, 14, and 18, with a range of 8 and fell in love with them. BIG mistake!

palvarez wrote:
8 <= 3a <= 16
8/3 <= a <= 16/3
a = 3 or 4 or 5

We got two lists.

a = 3, {11, 12, 19}
a = 4 {10, 14, 18}
a =5 {9, 16, 17}

Insufficient

You're correct, it's insufficient - however, you've made a very dangerous assumption.

3, 4 and 5 aren't the only numbers between 8/3 and 16/3.

Even if your inequality had reduced to:

8/3 <= a <= 10/3,

the statement would have been insufficient; even though there's only one integer in that range, there's an infinite number of non-integers, and nowhere in the question does it say that x, y and z must be integers.

I solved the DS this way, Is the procedure correct.

Statement 1: x+y+z+8/4=12.5
x+y+z=48

There will be more than 1 combination of x,y,z to add up to 48 and in the range 8.
Eliminate A,D
Statement 2: Given x,y,z are consecutive integers.
Mean=Median.

let the three values be
n,n+1,n+2
But clearly Statement 2 by itself is not sufficient.

From 1 and 2 how ever
We get all the numbers
3n+3=48
3n=45
n=15,
The three numbers are 15,16,17

Hence the smallest number is 15 C:

So , the Answer is E.

One more query. If it is mentioned that x, y & z are intergers , answer would be C.

list1 a = 3, {11, 12, 19}
list2 a = 4 {10, 14, 18}
list3 a =5 {9, 16, 17}

list1 & list3 will be go by statement.