Each of the five divisions of a certain company sent representatives

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

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 \(4<x\leq 5\), then Me=x.

x>(17+x)/5
4x>17
x>4.25

The solution is \(4.25<x\leq 5\), which would give us a Yes answer to the rephrased question.

Case 3: If x>5, 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

Rephrase Q: Is 3 <= x <= 5?

1) Median > Mean

Case 1: x <=4

4 > (17+x)/5
3 > x => No

Case 2: 4<x<5

Not possible since we have an integer constraint (i.e. # of Human).

Case 3: x >= 5

5 > (17 + x)/5
8 > x => Not Sufficient

If x = 5, then yes.
If 5 < x < 8, then no.

Not Sufficient

2) Median = 4

Don't know x. Not sufficient
If 3<= x <=4, then yes.
If x < 3, then no.

1+2

Case 1: x <=4

4 > (17+x)/5
3 > x => No

Sufficient

ANSWER: C

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.

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

Hard one!

Let's rephrase:
question asks if x = 1/2 or >5.

(1) median > average.
our median shall be 4 or 5 in any case.
if x=2 (yes), our median (4) is larger than our average (3.something).
if x=5 (no), our median (5) is larger than our average (4.something).
IN.

(2) median = 4.
x can be 4 (no), 3 (no), 2 (yes) and 1 (yes).
IN

(1+2)

If our median is 4, than x cannot be 4 (because then the average > median)
and x cannot be 3 (because then average = median)

so x has to be 1 or 2.

S.

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 ?

Give the numbers 3, 4, 5, and 5, the ONLY way for the range of the numbers to be less than 2 is if \(x < 3\) or \(x > 5\)

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

Scenario 1: x =1
\(\frac{1 + 3 + 4 + 5 + 5 }{ 5} = \frac{18}{5} = 3.6\)
\(median = 4\), \(average = 3.6\)
range > 2; YES

Scenario 2: x = 5
\(\frac{3 + 4 + 5 + 5 + 5 }{ 5} = \frac{22 }{ 5 }= 4.4\)
median = 5, average = 4.4
range < 2; NO

Insufficient.

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

Scenario 1: 3, 4, 4, 5, 5
\(range < 2\)

Scenario 2: 2, 3, 4, 5, 5
\(range > 2\)

Insufficient.

(1&2) median = 4, median > mean

\(\frac{1 + 3 + 4 + 5 + 5 }{ 5} = \frac{18}{5} = 3.6\)
median = 4, average = 3.6
range > 2; YES

Answer is C.
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ii'm not sure i understand how you got Is \(3\leq x\leq 5\). Why are we capping the range at 5. couldn't the fifth division send 6,7,8,9 + people?

Hi KarishmaB and GMATNinja! Can you help me with this question? Thank you!

Since the Range asked has to be greater than 2. We need to see whether the 5th team sends more than 5 members or less than 3.
Case1: Median>Mean
Lets see if 6, makes the relation valid.
Median=5; Mean=4.6
Range=3; True
Lets consider 5
Median=5; Mean= 4.4
Range=2; False
NOT Sufficient
Case 2: Median=4
Lets consider 2
Median= 4; Mean=3.8; Range=3; True
Lets consider 3
Median=4; Mean=4; Range=2; False
NOT Sufficient

Case1+Case2
No of team members has to be less than 3.
Sufficient.

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)
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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!

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.
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