X is a set containing 7 different numbers. Y is a set containing 6 dif

Question

X is a set containing 7 different numbers. Y is the set containing 6 different positive numbers all of which are members of set X. Is the mean of X equal to the mean of Y?

(1) Range of X is greater than range of Y
(2) Sum of all the numbers in set X< Sum of all the numbers in set Y

chetan2u · Accepted Answer

Hi..
 If all 7 numbers are different and 6 of them have a certain MEAN, M..
This would be EQUAL to mean of 7 numbers ONLY when 7th number itself is EQUAL to M ...

Let's see the statements...

a) Range of X is greater than range of Y
 This means the 7th number is either greater than or less than the 6 numbers in set Y..
So the 7 number cannot be EQUAL to mean of 6 numbers..
Ans is NO 
Sufficient

b) Sum of all the numbers in set X< Sum of all the numbers in set Y
 Clearly a lower number /7 will be less than greater number/6..
Clearly mean of Y>mean of X..
Ans is again NO 
Sufficient

D

Flower_Child · Answer

This is an interesting question which took a lot of my time .

First of all the main thing to notice is that X is a set containing 7 different numbers.
And, Y is a set containing 6 different positive numbers.This tells us that the one extra number in set X can be 0 or negative also.

Now Y is a subset of X.
So lets say that the extra number in set X is k, for sake of ease .
We can note that if sum of numbers of set Y (smaller set) is S then that of set X is S+k.

Now lets jump to answer choices:

a) Range of X is greater than range of Y

Ok. So that means the extra number in set X is either largest or smallest. That's why the range is different .
Example - Consider a set 4,6,8,10 . If I want to change range of this set I have to add a number that is either less than 4 or greater than 10. I can add 5,7,and 9 but that will not change the range .

Also, the question is asking -Is the mean of X equal to the mean of Y?

Think about this small set again 4,6,8,10. the mean is middle value 7 . Now IF we want to add another number in this set AND want to keep the mean same , then what that number would be? It might be easier to understand this via number line:
_________4__6_7_8__10____

If I add a number on the left side of 7 , the mean will shift to the left .Then the mean will be less than 7.
Similarly , if I add a number to the right of 7 , the mean will shift to the right . The mean will then be greater than 7 .

Thus to keep the mean same in an existing set we have to add the number that is equal to mean.
In our question the mean will be somewhere in middle of 6 different numbers but the one extra number in set X is outside this range of 6 number. Hence the mean of X will not be same as that of Y .
SUFFICIENT.

b) Sum of all the numbers in set X< Sum of all the numbers in set Y

The set x contains 7 numbers .
Set Y contains 6 numbers .

Let, Sum of Set Y = S => Mean = \(S/6\) ---(1)
Sum of Set X = S+k => Mean =\( (S + k)/7\) ----(2)

Now (2) can be equal to (1) only if S+k is greater than S.
Eg: 4 = 24/6 = 28/7

Since its given Sum of all the numbers in set X< Sum of all the numbers in set Y , mean of two sets will not be same.
SUFFICIENT

JeffTargetTestPrep · Answer

We know that n_X = 7 (different numbers) and n_Y = 6 (different positive numbers). We also know that Y is a subset of X.

Is mean_X = mean_Y ?

Statement One Alone:

=> Range of X is greater than range of Y

We see that the extra element in set X must be less than the smallest element in set Y or greater than the greatest element in set Y.

Adding an extra element to a set won’t change the mean of the set if and only if the extra element is equal to the mean of the original set.

The mean of set Y is between the smallest and greatest elements in set Y, because the mean is always between the smallest and greatest elements if the elements are not all equal.

Therefore, with the extra element, set X has a mean that is different from the mean of set Y.

We see that we have a definite No answer to the question. Statement one is sufficient. Eliminate answer choices B, C, and E.

Statement Two Alone:

=> Sum of all the numbers in set X < Sum of all the numbers in set Y

Since set Y contains only positive numbers, the sum of all the numbers in set Y must be positive.

Therefore, the extra number in set X must be negative in order for the sum of all the numbers in set X to be less than the sum of all the numbers in set Y.

We again see that the extra number in set X cannot be equal to the mean of set Y. So, with the extra element, set X has a mean that is different from the mean of set Y. Again, we have a definite No answer to the question. Statement two is sufficient.

Answer: D

Ahi · Answer

X is a set containing 7 different numbers. Y is the set containing 6 different positive numbers all of which are members of set X. Is the mean of X equal to Y?

a) The range of X is greater than the range of Y
b) Sum of all the numbers in set X< Sum of all the numbers in set Y

for 1 option 
Range of x> Range of Y So we can take an example like X set {-4,2,3,4,5,6,7} and set Y={2,3,4,5,6,7}
The range is (Highest-Lowest) So for set X, the range is 6, and for set Y, the range is 5. So find the mean of set X and Set Y are not equal 
sufficient try with other numbers too

for 2 option 
The sum of set X is smaller than the sum of set Y 
mean is not equal again 
sufficient 
hence D is the answer

arunbhati · Answer

Hi Sir,

if we use number as X=(1,2,3,4,5,6,7) and Y as (1,2,3,4,5,7) then in this case range of X and Y is same how statement 1 is sufficient

sum of X is 28 and sum of Y is 22. Statement 2 is not valid here

can you explain me where i am wrong

Bunuel · Answer

What are you trying to do there? Your example is invalid because it does not satisfy either statement.

• Statement 1 requires range of X > range of Y, but your example has equal ranges.
• Statement 2 requires sum of X < sum of Y, but your example has 28 > 22.

X is a set containing 7 different numbers. Y is a set containing 6 dif

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