The average of a certain set of numbers is 20 and its

Question

The average of a certain set of numbers is 20 and its range is 20. By adding one more number to the set, we obtain set S. Is the range of set S greater than 20?

(1) The number we have added is 32.
(2) The average of set S is 24.

EMPOWERgmatRichC · Accepted Answer

Hi All,

This is an interesting statistics-based question. To solve it in an efficient way (notice that I did not say "fast"), you'll likely need a mix of TESTing VALUES, Number Property knowledge and pattern-matching skills.

We're told a number of facts about a set of numbers:
1) The average of the set = 20
2) The range of the set = 20

3) We are NOT told how many terms there are in this set
4) We are NOT told if the terms are integers

Next, we're told that we're going to add one additional number to this set. We're asked if the range of this NEW set is > 20. This is a YES/NO question.

Fact 1: The number added is 32

Let's TEST VALUES...

IF....
The starting set is {10, 30}
The average is 20 and the range is 20
Adding a 32 makes the range 32 - 10 = 22 and the answer to the question is YES.

IF....
The starting set is {12, 16, 32}
The average is 20 and the range is 20
Adding a 32 makes the range 32 - 12 = 20 and the answer to the question is NO.
Fact 1 is SUFFICIENT

Fact 2: The average of the new set is 24.

This is an interesting piece of information - it means that the one new value that is added is enough to "pull up" each of the existing values by 4 (from an average of 20 to 24). As such, this new number has to be big enough to offset ALL of the existing values in the set. Here's where 'pattern-matching' comes in handy.

IF....
The starting set is {10, 30}, the missing number has to "pull up" each of those 2 numbers by 4. The new number = 24+4+4 = 32.
32 - 10 = 22 and the answer to the question is YES.

IF....
We have MORE terms, then the missing number has to "pull up" MORE numbers by 4....
The starting set is {12, 16, 32}, then the new number = 24+4+4+4 = 36.
36 - 12 = 24 and the answer to the question is YES.

If you choose a set with a "lowest number" that gets closer to 20, the largest number gets closer to 40 (because the range = 20). However, since the AVERAGE = 20, you end up needing MORE numbers to pull the average down....

Consider if you had {nineteen 19s and one 39}, here we have 20 terms, each of which would need to be "pulled up" by 4. The new number would be 24 + 20(4) = 104
This range is considerably bigger....104 - 19 = 85 and the answer to the question is still YES.

This pattern proves that Fact 2 is SUFFICIENT. It's not something you can likely deduce by staring at the screen, so you have to be ready do enough work to pattern-match this deduction.

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

smyarga · Answer

I really don't like this problem. May be somebody will find better solution?

(1)  Insufficient 
Example 1: 11, 18, 31 (range 20, average 20) and new set S: 11, 18, 31, 32. The range of S is greater than 20
Example 2: 13, 14, 33 (range 20, average 20) and new set S: 13, 14, 32, 33. The range of S is 20.

(2)  Sufficient 
Let first set be T and the number of elements in T be  n . 
Since the average of T is 20, the sum of elements is  20n .
If we add one number, the new sum will be  24(n+1)=24n+24 . So we add  24n+24-20n=4n+24 . 
Let us show that  4n+24  is greater than the greatest number in T.

Suppose that the first element in T is  x , so the last one is  x+20 . To make  x+20  the greatest possible we need to make others the least possible, because our sum is fixed as  20n . So suppose we have  n-1  numbers  x  and the last one  x+20 . 
 x\cdot (n-1)+x+20=20n 
 xn=20n-20 
 x=20-20/n 
So, the greatest number possible in the set is  x+20=40-20/n

If  4n+24<40-20/n , then  4n+20/n-16<0  or  n+5/n-4<0  or  n^2-4n+5<0 , that is impossible since the discriminant is negative.

So, the new element in S is always greater than the greatest in T.

The correct answer is B.

manpreetsingh86 · Answer

let total no. of terms in the set be x; then sum of all of the x terms will be 20x (because average of these x numbers =20). Also the range of the set =20

st. 1 number added =32

let set consists of two numbers.
if set has 10,30 in it, then by adding 32 in the set range of this set will increase. and it will be 22

now suppose set consists of 3 numbers 13, 14, 33. by adding 32 in the set. range of this set will not increase.

hence clearly statement 1 is not sufficient

st.2
let the number added in the set be n
such that (20x+n)/x+1 =24
or 4x=n-24

here x must be integer, because no. of terms in the set can't be fraction.

thus minimum value of n=32, for which x=2. now, as per the question range of the set =20. thus two terms of the set will be 10 and 30 and clearly 32 is greater than 30. thus range of the set will be more than 20

now consider n=36
x=3

let's see if we can form the set of three numbers in which maximum term is greater than or equal to 36. let's try 36. if 36 is in the set then 16 is also in the set ( range=20). 
sum of 16 and 36 =52, thus third term =8. which means the set =8,16,36 which is not possible as it violates the range condition. thus range of the set will be more than 20

similar results can be obtained for other sets containing different terms.

thus range of the new set will be more than 20.

smyarga · Answer

Why you decide that " because no. of terms in the set can't be fraction"?

manpreetsingh86 · Answer

because you can't have a set in which no. of terms will be in fraction. for example we can't form a set containing 3.5 terms
because set will have either 3 or 4 terms (1,2,3) or (1,2,3,4).

KarishmaB · Answer

Avg - 20
Number of numbers in set - Not known. 
Range - 20

So the set could look something like this:
10, 20, 20, ... 20, 30
If the range has to stay the same, it should lie within 10 to 30 (in this case only)

(1) The number we have added is 32.
If we add 32 to the set above, we know that its range will change.
Can the original set already have 32 such that its range will not change?
12, 16, 20, 20, ... 20, 32
Here range will be 20, average will be 20. If we add 32 to it the range will not change. 
So this statement alone is not enough.

(2) The average of set S is 24.
The new number increases the average by 4. So it is average + 4 extra for each number in the set. We don't know how many numbers are there in the set.
The added number could be much larger than the greatest number such that it will increase the range. 
In case the number of numbers in the set is small, is it possible that the range does not change?
The minimum number of elements in the set will be 1. The two elements would be equidistant from 20: 10 and 30
To increase the average by 4, the new number should be 20 + 3*4 = 32. This will change the range.
Let's look at another case in which the set has 3 elements. To increase the average by 4, the new number should be 20 + 4*4 = 36.
Is it possible that 36 is already a part of the original set? 12, 12, 36
Here the range is more than 20. So not possible. 
Hence, in any case, if the average goes up by 4, the added number will increase the range.
We find a pattern. As we keep going up, the added number will be larger and larger and will increase the range.
This statement alone is sufficient.

Answer (B)

dishadesai · Answer

bb  is there a quicker way to solve this? The above solutions seems a bit tricky to think of within the stipulated time

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