Number Properties

Aman verma wrote:Q: The average of a set of whole numbers is 27.2. When 20% of the elements(i.e Numbers) are eliminated from the set of numbers then the average becomes 34. The number of elements in the new set of Numbers can be :

a) 27

b) 35

c) 52

d) 63

e) 71

PS: Please solve this algebraically. This problem can be easily solved by plugging in the options, but solving this algebraically will clarify the underlying logic and the mechanics of the problem.

Let original number of elements be n.
After removing 20% elements, remaining elements = (100-20)%n =0.8n
New total = 0.8n*34 = 27.2n

As all elements are whole numbers, we need new total = integer.
This is only possible if 27.2 is multiplied by a multiple of 5. Let n=5m.
Therefore 0.8n = 0.8*5m = 4m. So we need to find the option which is a multiple of 4.
Only 52 fits the bill. Hence C is correct.

New total = 0.8n*34 = 27.2n

How?
IF S = {1,2,3,4,5}, average of five elements of set S is 3
If 20% elements are to be removed (= 1 element is removed) then let's assume "1" is removed from set S.
If 1 is removed: New set S1 ={2,3,4,5} and new average of S1= 14/4 = 3.5
Then using your equation above for New Total: 0.8(5)(3.5) = 14 ╪ (3)(5)
!?

kartikshah wrote:

New total = 0.8n*34 = 27.2n

New Total: 0.8(5)(3.5) = 14 ╪ (3)(5)
!?

New Total: 0.8(5)(3.5) = 14. Correct.
Why does the new total have to equal 3*5?

In your equation why have you equated the new total with the old total?
I did not understand the basis of your equation 0.8*34*n = 27.2n

kartikshah wrote:In your equation why have you equated the new total with the old total?
I did not understand the basis of your equation 0.8*34*n = 27.2n

Oh I understand where you're coming from.

I haven't equated the old total with the new one.

0.8*34 equals 27.2.

It just happens that in this question that the old total equals the new one. (Which also tells us that the elements removed must have been a bunch of zeroes). It doesn't have to.
Is is clear now?

Aman verma wrote:Q: The average of a set of whole numbers is 27.2. When 20% of the elements(i.e Numbers) are eliminated from the set of numbers then the average becomes 34. The number of elements in the new set of Numbers can be :

a) 27

b) 35

c) 52

d) 63

e) 71

PS: Please solve this algebraically. This problem can be easily solved by plugging in the options, but solving this algebraically will clarify the underlying logic and the mechanics of the problem.

Let N = the number of elements and A = the average of elements.
Sum = N*A.

When old N decreases by 80%, new N = (4/5)N.
When old A increases from 27.2 to 34 -- an increase of 25% -- new A = (5/4)A.
Thus, new sum = (4/5)N * (5/4)A = N*A.
Thus, the new sum is equal to the old sum -- REGARDLESS of the number of elements.

Since the number of elements has no effect upon the sum, the only requirement is that the number of elements be a WHOLE NUMBER.
Since new N = (4/5)N, we know that N must be a multiple of 5 and that (4/5)N -- the new number of elements -- must be a multiple of 4.
To illustrate:
If old N=5, then new N = (4/5)N = 4.
If old N=10, then new N = (4/5)10 = 8.
If old N=15, then new N = (4/5)15 = 12.
In each case, new N is a multiple of 4.

Among the 5 answer choices, only 52 is a multiple of 4.

The correct answer is C.

@ EAGLEEYE: Yep! It's clearer now. Thanks for the added explanation on how zeroes could have been removed to leave the average unaffected! I hadn't thought about that.