Average again.

:mrgreen: I love math :mrgreen: !

Let a and b be the number of employees in town A and town B respectively
Average wages of employees in town A = X
Average wages of employees in town B = Y

Hence, new average salary = (aX + bY)/(a + b)

Question posed :

Is new average salary smaller than (X + Y)/2?
Implies, Is (aX + bY)/(a + b) < (X + Y)/2

=> 2(aX + bY)< (X + Y)(a + b)

=> 2aX + 2bY < aX + bX + aY + bY

=> aX + bY < bX + aY

=> a(X-Y)-b(X-Y) < 0

=> (a-b)*(X-Y) < 0

So the word question is reduced to Is (a-b)*(X-Y) < 0 ?, which is possible only in two cases

i) (a-b)<0 and (X-Y)>0
ii)(a-b)>0 and (X-Y)<0

(1) There are more employees in Town A than B

Implies a > b => (a-b) > 0 but we are not sure of the value of x-y hence insufficient

(2) Y-X = 4200

Implies X-Y = -4200 i.e X-Y < 0 but we are not sure of the value of a-b hence insufficient

[spoiler]Using both the conditions you get X-Y < 0 and (a-b) > 0, resulting in (a-b)*(X-Y) < 0 , Hence Option C[/spoiler]

lenagmat wrote:In a certain company, the average wages of employees in Town A is X, the average wages of employees in Town B is Y. If both types of employees are added together, is the new average salary smaller than (X + Y)/2?

(1) There are more employees in Town A than B
(2) Y-X = 4200

Average age of employees in Town A & B can be (X + Y)/2 , if the population of both the town were same...

The New average salary smaller than (X + Y)/2 - This suggests that the Population size is not the same...

So either Population in town A > Population in Town B

OR

Population in Town B > Population in Town A...


We can not arrive at a specific solution , so option 1 can not be possible.

Again , from the given statement we can not say Y-X = 4200 , because there is no mention of any figure..

So neither of the given answer choices follows !!!

lenagmat wrote:In a certain company, the average wages of employees in Town A is X, the average wages of employees in Town B is Y. If both types of employees are added together, is the new average salary smaller than (X + Y)/2?

(1) There are more employees in Town A than B
(2) Y-X = 4200

Let's assume no of Employees from town A to be a, and no of employees from town B to be b.

Hence,

Total of wages for Town A = aX (No of Employees X Avg. wages)

Total of wages for Town B = bY

Total wages for town A and B = aX + bY

Total no of Employees = a+b

New Average = ( Total Wages of Employees in Town A and B ) / (Total no of Employees in Town A and B)

= (aX+bY) / (a+b)

The stem asks us to check whether,

= (aX+bY) / (a+b) <. (X+Y)/2 - Cross Multiply and apply FOIL

= 2aX + 2bY < aX + aY + bX + bY. - Taking like terms on each side

= 2bY-bY-bX < aY + aX - 2aX

= bY - bX < aY- aX

= b(Y-X) < a(Y-X). - So, if this is true than the New Average is less than ( X + Y ) / 2 )

Statement 1: - There are more Employees i n Town A than in Town B, Hence a > b

If Y- X is positive then b(Y-X) < a(Y-X) holds true.

But if Y-X is negative then b(Y-X) > a ( Y-X) note that the sign flips hence the original inequality we are testing does not hold true

As there are two possible outcomes : Insufficient - Eliminate AD

Statement 2 :- (Y-X) = 4200

This does not tell us about a and b hence : Insufficent : Eliminate B

Both, a > b and Y-X = 4200 and hence positive we can conclude that,

b(Y-X) < a (Y-X)

Correct Answer C [/b]