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 again.
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Say, number of employees in town A and B are a and b respectively.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
Hence, new average salary = (aX + bY)/(a + b)
Statement 1: a > b
Hence, the new average salary is more weighted towards X.
Therefore, if X > Y, then the new average salary will be greater than (X + Y)/2 and vice versa. But we don't know whether X is greater than Y or not.
NOT sufficient
Statement 2: Y = X + 4200
As we don't know the relation between a and b we cannot answer the original question.
NOT sufficient
1 & 2 Together: As a > b and X < Y, the new average salary must be smaller than (X + Y)/2.
SUFFICIENT
The correct answer is C.
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- neelgandham
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: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]
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]
Anil Gandham
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- Abhishek009
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Average age of employees in Town A & B can be (X + Y)/2 , if the population of both the town were same...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
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 !!!
Abhishek
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Let's assume no of Employees from town A to be a, and no of employees from town B to be b.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
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]