Each of the five divisions of a certain company sent representatives

Question

Each of the five divisions of a certain company sent representatives to a conference. If the numbers of representatives sent by four of the divisions were 3, 4, 5, and 5, was the range of the numbers of representatives sent by the five divisions greater than 2 ?

(1) The median of the numbers of representatives sent by the five divisions was greater than the average (arithmetic mean) of these numbers.
(2) The median of the numbers of representatives sent by the five divisions was 4.

DS06110.01

shridhar786 · Accepted Answer

If the numbers of representatives sent by four of the divisions were 3, 4, 5, and 5 was the range of the numbers of representatives sent by the five divisions greater than 2?

STATEMENT (1) The median of the numbers of representatives sent by the five divisions was greater than the average (arithmetic mean) of these numbers.

let that 5th division send = 2 representatives
-- 2,3,4,5, and 5 (average =  \frac{19}{5} =3.8 median = 4 and median>average)
range of the number of representatives = 5-2 = 3- was the range of the numbers of representatives sent by the five divisions greater than 2?--YES

let that 5th division send = 5 representatives
-- 3,4,5,5, and 5 (average =  \frac{22}{5} =4.4 median = 5 and median>average)
range of the number of representatives = 5-3 = 2- was the range of the numbers of representatives sent by the five divisions greater than 2?--NO

we cant get the definite answer
INSUFFICIENT

STATEMENT (2) The median of the numbers of representatives sent by the five divisions was 4.
--median of five divisions = 4
--representatives sent by other divisions = 3,4,5, and 5
so the representatives sent by the fifth division --will be less than equal to 4
let 5th division sent = 4 representatives
--3,4,4,5, and 5
range = 5-3 = 2 -was the range of the numbers of representatives sent by the five divisions greater than 2?--NO

let 5th division sent = 2 representatives
--2,3,4,5, and 5
range = 5-2 = 3 -was the range of the numbers of representatives sent by the five divisions greater than 2?--YES
INSUFFICIENT

combining both statements together
median = 4 and median > average of the representatives
so, from here we know that 5th division sent less than equal to 4 represantatives
4 is not possible --3,4,4,5, and 5 (average =  \frac{21}{5} =4.2 median = 4 and median<average)
3 is not possible --3,3,4,5, and 5 (average =  \frac{20}{5} =4 median = 4 and median = average)

--1 and 2 are the possible number of representatives 5th division can send
in both cases median = 4 and median > average
range--1,3,4,5, and 5 = 5-1 =4
range--2,3,4,5, and 5 = 5-2 =3
was the range of the numbers of representatives sent by the five divisions greater than 2?--YES
SUFFICIENT

C is the answer

KarishmaB · Answer

List: 3, 4, 5, 5 Add another number x, which could be any integer from 1 onwards i.e. 1/2/3/4/5/6/7/8.... Statement 1: Median > Mean Median of the 5 number list can be 4 or 5 only If median = 4, x is 1 or 2 to give mean < median. If x is 1 or 2, range is greater than 2. Here, x cannot be 3 or more than 3 because then mean is equal or greater than median. If median = 5, x is 5 or 6 or 7. If x = 5, mean < median and range is 2. If x = 6 or 7, mean < median and range > 2. Not sufficient . Statement 2: Median = 4 i.e. 4 is the 3rd value. So x can be 1/2/3/4. If x is 1 or 2, range is greater than 2. If x is 3 or 4, range is 2. Not sufficient. Using both together, we know that median is 4 so x is 1 or 2 only. Hence range is certainly more than 2. Sufficient. Answer (C)

ZoltanBP · Answer

Let integer x be the 5th number. The original question: Is R>2 ? The range of the four known numbers is 5-3=2. The rephrased question: Is 3\leq x\leq 5 ? 1) We know that Me>A and can evaluate three possible cases. Case 1: If x\leq 4 , then Me=4 and A=(17+x)/5. 4>(17+x)/5 x<3 The solution is x<3, which would give us a No answer to the rephrased question. Case 2: If 4(17+x)/5 4x>17 x>4.25 The solution is 4.255, then Me=5. We don't need to evaluate this case since we already have two different answers. Thus, we can't get a definite answer to the rephrased question. \implies Insufficient 2) We know that Me=4, so x\leq 4 . If x is 4, then the answer to the rephrased question is Yes. However, if x is 1, then the answer to the rephrased question is No. Thus, we can't get a definite answer to the rephrased question. \implies Insufficient 1&2) Case 1 is the only valid case from statement 1). Thus, the answer to the rephrased question is a definite No. \implies Sufficient Answer: C

reynaldreni · Answer

Given set of values {3,4,5,5} & Let F be the fifth value Range for the given set of four values = 2 (1) The median of the numbers of representatives sent by the five divisions was greater than the average (arithmetic mean) of these numbers. case#1: F=2 New Set ={2,3,4,5,5} Median = 4 Mean = 19/5 = 3.8 < Median Range = 5-2 = 3 New Range > 2 --> YES case#2: F=5 New Set = {3,4,5,5,5} Median = 5 Mean = 22/5 = 4.4 < Median Range = 5-3 = 2 New Range > 2 --> NO Hence statement (1) is not sufficient. (2) The median of the numbers of representatives sent by the five divisions was 4. case#3: F=1 New Set = {1,3,4,5,5} Median = 4 Range = 5-1 = 4 Mean = 18/5 = 3.6 < Median New Range > 2 --> YES case#4: F=4 New Set ={3,4,4,5,5} Median = 4 Range = 5-3 = 2 Mean = 21/5 = 4.2 > Median New Range > 2 --> NO Hence statement (2) is not sufficient. Combining (1) & (2) Only case #1 & #3 are valid. Sufficient! In general, if the distribution of data is skewed to the left, the mean is less than the median. Based on this fact, number picking can be done to build cases appropriately.

suminha · Answer

Each of the five divisions of a certain company sent representatives to a conference. If the numbers of representatives sent by four of the divisions were 3, 4, 5, and 5, was the range of the numbers of representatives sent by the five divisions greater than 2 ?

-> If we find the number of representatives sent by fifth division, we can answer yes or no.
-> Let the number of representatives sent by fifth division is \(x\).
-> \(0<x<n\), \(x=integer\)

(1) The median of the numbers of representatives sent by the five divisions was greater than the average (arithmetic mean) of these numbers.

-> When \(x=1\), this statement stands true. Also when \(x=5\), this statement stands true.
-> However, then the rage is greater than 2 when \(x=1\) and less than 2 when \(x=5\).
-> Insufficient

(2) The median of the numbers of representatives sent by the five divisions was 4.

-> \(x\) can be 1, 2, 3, and 4. Thus the rage differs too.
-> Insufficient

(1)&(2) The average (arithmetic mean) of these numbers is less than 4, which is the median of the numbers of representatives sent by the five divisions.

-> \(avg<4\)
-> \(\frac{x+3+4+5+5}{5}<4\)
-> \(x<3\)
-> Therefore the range is greater than 2. -> Sufficient

Posted from my mobile device

Will2020 · Answer

Hi KarishmaB ! I did not understand your analysis of statement 1. How do you know that the possible median values for the list are only 4 or 5? Thank you!

KarishmaB · Answer

The list with the 4 numbers is 3, 4, 5, 5. 
If I add one more integer to it, the median will be the 3rd number in the list when all numbers are arranged in ascending order.

x  can be such that the current 2nd element becomes 3rd element. e.g.  1 , 3,  4,  5, 5 (median 4)
 x  can be such that the current 4th element becomes 3rd element. e.g. 3, 4,  5 , 5,  6  (median 5)
 x  can be such that it itself is the 3rd elements i.e. 3, 4,  x , 5, 5. Now since x is an integer, and there is no integer between 4 and 5, x must be either 4 or 5. In either case, median will be 4 or 5.

Hence, median must be 4 or 5.

So do this simply, just visualise them on a number line. Where can I place x and how will the rest of the number shift? Also, any number can shift by only one rank so 2nd can become 3rd or 4th can become 3rd at the most. 1st number cannot become the 3rd number.

Engineer1 · Answer

KarishmaB - I just have one question. Why in statement 2 the fifth number (x) <=4? It is an a sequence with odd number of terms. Should x not be qual to 4? I am sorry, not sure what I am missing.

KarishmaB · Answer

The given numbers are 3, 4, 5, 5
One more number (say x) is added to this list and their median becomes 4. This means that the third number in the list in increasing order is certainly 4. We already have a 4 in the list so that could be the median. So the new number can be 4 or less than 4 too. 
The value of x could be 1 or 2 or 3 or 4.

1, 3, 4, 5, 5 - Median = 4, x = 1
2, 3, 4, 5, 5 - Median = 4, x = 2
3, 3, 4, 5, 5 - Median = 4, x = 3
3, 4, 4, 5, 5 - Median = 4, x = 4

scrantonstrangler · Answer

Let the last division send 'a' representatives.
Now the set becomes (3,4,5,5,a)

Note: We cant assume 'a' is the largest or smallest or in the middle for now, but what we can infer from the stem is this.
The median for the set can only be 4 or 5.
This is because the set has only integers. Divisons cant exactly send people in decimals. (Dark jokes aside).

So, why do I say we can only take 4 or 5 as medians?
Here is why.
The set can be as follows
(a 3 4 5 5) Median = 4 (a would have to be less than or equal to 3)
(3 a 4 5 5) Median = 4 (a would have to less than equal to 4 but greater than or equal to 3)
(3 4 a 5 5) Median = (a can be 4 or 5 only) a has to be less than or equal to 5 but greater than equal to 4)
(3 4 5 a 5) Median = 5 (a has to be 5)
(3 4 5 5 a) Median = 5 (a has to be at least 5 or more)

This is how you can be sure median is 4 or 5.

Now with S1.
We see that median for this set is greater than the mean.
Mean of the set = (17+a)/5
Because median has two possible values
We will consider both:
(17+a)/5<4
..a<3

or

(17+a)/5<5
..a<8.

For a<3, we can easily say that the range of the set is definitely greater than 2,
but for a<8, we can't say range is surely greater than 2. if a = 7, Range = 7-3 =4: YES
 for a = 3, Range = 5-3 = 2, then NO.

Hence S1 is insufficient.

Now for S2)
Median is 4.
Set can now be as follows:
(a 3 4 5 5) Median = 4 (a would have to be less than or equal to 3) (Range is greater than or equal to two)
(3 a 4 5 5) Median = 4 (a would have to less than equal to 4 but greater than or equal to 3) (Range = 2)
(3 4 a 5 5) Median = a (can be 4 or 5 only) a has to be less than or equal to 5 but greater than equal to 4) Range =2.

Hence even with S2, we cant say for sure.

Combining S1 and S2, we can say median is 4 and median is greater than mean, so (17+ a)/5 is less than 4, implying a < 3. Hence 'a' can be 1 or 2, range can then be 4 or 3 only, which is greater than 2.

Hence C is the option.

KarishmaB  request your tips on coming up with ways to reduce the time taken to solve this.

Gemmie · Answer

Given: 3, 4, 5, 5
x: the unknown (x > 0)
Question: Is the range > 2?

Statement (1): The median of the numbers of representatives sent by the five divisions was greater than the average (arithmetic mean) of these numbers.

(i) Calculate the average of the known four numbers:

Average = \(\frac{3+4+5+5}{4} = \frac{17}{4} = 4.25\)

(ii) Determine the possible median values when adding the fifth representative number.

+) If x < 3 => x, 3, 4, 5, 5
=> median = 4
=> Avg < \(\frac{17 + 3}{5} = 4\) = Median
=> x can be 1 or 2

+) If x = 3 => 3, 3, 4, 5, 5
=> median = 4
=> Avg = \(\frac{17 + 3}{5} = 4\)

+) If x = 4 => 3, 4, 4, 5, 5
=> median = 4
=> Avg = \(\frac{17 + 4}{5} > 4\)

+) If x >= 5 => 3, 4, 5, 5, 5
=> median = 5
For avg < median => 17 + x < 25
=> x < 8
=> x can be 5, 6 or 7

=> Statement (1) is NOT sufficient.

Statement (2): The median of the numbers of representatives sent by the five divisions was 4.

It could be either

3, 4, 4, 5, 5 => Range = 2

or

x, 3, 4, 5, 5 (x < 3) => Range > 2

=> Statement (2) is NOT sufficient.

Combined:
There is only possibility: x, 3, 4, 5, 5 (x < 3)
=> Range > 2

=> Sufficient

==> Answer is C

Kavicogsci · Answer

Took a lot of time unnecessarily - finally arrived at a solution that can be done in <2 min

Each of the five divisions of a certain company sent representatives to a conference. If the numbers of representatives sent by four of the divisions were 3, 4, 5, and 5, x
We know x is an integer as it is a representative

So 3,4,5,5,x
What could median be?
For x>=5 median is 5.......(1)
For x <=4 Median is 4........(2)

Now,
1. Median is greater than Avg
Implications - the set is not symmetrical around median (symmetrical means the distance of each number from median in left side is equal to exact same distance of each number from median in right side).....(3) So we cannot have 3,3,4,5,5

But if x = 2, (Median is 4 from (2) and Mean by deviations method will be less than median) range is 3 Yes answer is NO to the Q that range is more than 2
2,3,4,5,5

if x =5, (Median is 5 from (1) and Mean by deviations method will be less than median) range is 2 answer is NO to the Q that range is more than 2
3,4,5,5,5

Insufficient

2. Median is 4
From (2) we know x>=4 so x =0,1,2,3,4
In each case range is different leading to both yes and no

Both
From A we know the set can't be symmetrical around the median and from B we know x<=4
So
3,3,4,5,5 is not allowed as median = mean
3,4,4,5,5 is not allowed as mean > median [Don't waste time in calculations! Solve by deviations method - to the right of 4 we have +1,+1 to the left of 4 we have 0,-1, hence mean will be greater than 4]

X = 0/1/2 in each cases all conditions are held
Median>mean
And x>=4
Hence range>2

gmatt1476

Each of the five divisions of a certain company sent representatives to a conference. If the numbers of representatives sent by four of the divisions were 3, 4, 5, and 5, was the range of the numbers of representatives sent by the five divisions greater than 2 ?

(1) The median of the numbers of representatives sent by the five divisions was greater than the average (arithmetic mean) of these numbers.
(2) The median of the numbers of representatives sent by the five divisions was 4.

DS06110.01

bumpbot · Answer

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Each of the five divisions of a certain company sent representatives

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