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# Each of the five divisions of a certain company sent representatives

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Each of the five divisions of a certain company sent representatives [#permalink]
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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.

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.
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Each of the five divisions of a certain company sent representatives [#permalink]
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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

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Re: Each of the five divisions of a certain company sent representatives [#permalink]
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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

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.

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Re: Each of the five divisions of a certain company sent representatives [#permalink]
KarishmaB
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

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.

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!
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Re: Each of the five divisions of a certain company sent representatives [#permalink]
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KarishmaB
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

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.

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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Re: Each of the five divisions of a certain company sent representatives [#permalink]
KarishmaB
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

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.

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.
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Re: Each of the five divisions of a certain company sent representatives [#permalink]
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KarishmaB
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

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.

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.

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
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Re: Each of the five divisions of a certain company sent representatives [#permalink]
­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.

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Each of the five divisions of a certain company sent representatives [#permalink]
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