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I think there is a typo. it should be 10 as median is 5 NOT 3.

Edited the post...

Thanks!!
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Abhishek009
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Set A consists of 2 positive integers, and set B consists of 3 positive integers. If the median of set A is 5, and the average (arithmetic mean) of set B is 3, then which of the following must be true?

I. The median of the combined sets is less than 5
II. The mean of the combined sets is less than 4
III. The median of set B is less than 5

A) II only
B) III only
C) I and II
D) II and III
E) I, II and III

A = ( a , b )
B = ( c , d , e )

Now, ( a + b ) = 5 { Median of 2 numbers is Mean }
And ( c + d + e ) = 9

So, ( a + b + c + d + e ) = 15

I. If c = 15 then other numbers must be -ve , hence incorrect

II. Mean of Combined set is 15/4 = 3.75 , True, Less than 4

III. If d < 5 then - d = 4 , 3 , 2 , 1

If d = 4

c + e = 5


If d = 3

c + e = 6


If d = 2

c + e = 7


If d = 1

c + e = 8

Hence, both II and III are possible....

Answer must be (D)...

Hey Abhishek009
please let me know if I am mistaken anything here
you said median is mean of two and directly taken sum ... i think sum of elements of set will 10 and combined will be 19
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Abhishek009
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Set A consists of 2 positive integers, and set B consists of 3 positive integers. If the median of set A is 5, and the average (arithmetic mean) of set B is 3, then which of the following must be true?

I. The median of the combined sets is less than 5
II. The mean of the combined sets is less than 4
III. The median of set B is less than 5

A) II only
B) III only
C) I and II
D) II and III
E) I, II and III

A = ( a , b )
B = ( c , d , e )

Now,( a + b ) = 5 { Median of 2 numbers is Mean }
And ( c + d + e ) = 9

So, ( a + b + c + d + e ) = 15

I. If c = 15 then other numbers must be -ve , hence incorrect

II. Mean of Combined set is 15/4 = 3.75 , True, Less than 4

III. If d < 5 then - d = 4 , 3 , 2 , 1

If d = 4

c + e = 5


If d = 3

c + e = 6


If d = 2

c + e = 7


If d = 1

c + e = 8

Hence, both II and III are possible....

Answer must be (D)...

Abhishek009, few corrections (please refer to the highlighted part)
a+b = 5*2 = 10
( a + b + c + d + e ) = 10 + 9 = 19
II. Mean of Combined set is 19/5 = 3.8

:)
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I feel that the stem should mention the word ONLY. A consists of only 2 positive integers and set B consists of only 3 positive integers.
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i assumed the following-

set A ={1,9}
mean=median= 5

set B ={1,2,6}
mean = 3
median=2

combined set = {1,2,6,9}
mean=18/4=4.5
median=4

statement 1- true
statement 2- false
statement 3- true

since 1 and 3 were not in option, i picked 3 only, i.e B

can anyone tell me where did i go wrong?

thanks in advance.
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(1st) every integer is a positive integer. Also, we have no constraint against repeated integers (we can have the same integers within a set

Set A (X , Y)

Set B (P , Q , R)


(2nd)
Median of Set A - for 2 positive integers, the Median = Mean = Mid-point between those 2 integers

Mean = 5 = (X + Y)/2


If 5 is the mid-point between the 2 integers and we assume X is the smaller value, then Y must be just as far from 5 going in the other direction on the number line

Possible values in Set A:

(1 , 9) —— (2 , 8)——(3 , 7)——(4 , 6)——(5 , 5


(3rd) the arithmetic mean of set B is given by:

(P + Q + R)/3 = 3

P + Q + R = 9

—In which P, Q, and R must be a positive integer —

WOTF must be true?

-I- the median of combined sets is less than 5

Let X = Y = 5

And

P = 5
Q = 2
R = 2
such that P + Q + R = 5 + 2 + 2 = 9

The combined set would be:

[2 , 2 , 5, 5, 5]

in which case the median = 5, a value that is NOT less than 5

-I- does NOT have to be true

-II- the Mean of the combined sets is less than 4

We know that:
X + Y = 10
And
P + Q + R = 9

Mean of Combined sets = (Sum of integers) / 5 = (X + Y + P + Q + R) / 5 = (10 + 9) /5

19/5 is indeed < 4

II must be true


-III- the median of set B is less than 5

Since P + Q + R = 9

And we are told that each variable must be a positive integer

And the median of any set with an ODD number of terms will always be the middle number when values are arranged in ascending order

It is impossible to come up with 3 positive integers that SUM to 9 and have a median of 5 or greater

If we make them all equal
3 - 3 - 3

Whenever we try to add (+1) to an integer we must take away (-1) from another integer in order to keep the SUM and average constant

III must be true.


II and III must be true only

I can be false

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Set A consists of 2 integers and has a media of 5, since there are only 2 integers the mean will also be 5 and the sum of both the integers will be 10.

Set B consists of 3 integers with an average of 3, which means the sum of the integers is 9.

The 1st statement says that median of the combined sets is less than 5
This statement will not always hold true because the integers in Set A can both be 5 and Set B can be 2, 2, 5
Now the combined set will look like this
{2,2,5,5,5}
The median here is not less than 5 but equal to 5, hence statement 1 is false.

The 2nd statement says that the mean of the combined sets is less than 4
The sum of both the sets is 10+9=19 and the total number of integers is 5 making the mean just under 4 which makes the statement true.

The 3rd statement says that the median of set B is less than 5
The sum of Set B is 9, with 3 integers no matter how you arrange the numbers the median will always be less than 5.
Hence the 3rd statement is also true.

The answer is D
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Set A (a1,a2)
Set B (a3,a4,a5,a6)

1) Median of Set A is 5
2) Mean of Set B is 3

To satisfy condition 1, Set A (1,9 ) (2,8) (3,7), (4,6) (5,5) (6,4) etc.
To satisfy condition 2, Set B make the total sum be 9 so that mean is 3, (1,3,5) (1,4,4) (1,1,7)

I. The median of the combined sets is less than 5 - Must not be true
take (5,5,9,1,1,7) rearranging it as (1,1,5,7,9) median is 5

II. The mean of the combined sets is less than 4 - Must be true
take any sets from above but ultimately it will be less than 4 as (10+9)/5 = 19/5 <4

Also please note, If there are only two numbers in a data set, then their median will be the same as their mean

III. The median of set B is less than 5 -Must be true

Now for set B, median will always be less than 5 as there is no other case which meets our mean criteria to be 3

take (1,5,5) =11/3 >3
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