In a company of 140 employees, there are three departments,

Question

In a company of 140 employees, there are three departments, A, B and C. The average number of safety infractions committed by each employee in department A in 2005 was 10, and the average number of safety infractions committed by each employee in department B was 12.25. Was the average number of safety infractions committed by each employee in the company greater than 11?

(1) The 45 employees in department C each committed an average of 13 safety infractions in 2005.
(2) There are 44 employees in department B.

amorica · Answer

C...

though i have this sneaky feeling that i fell for a trap answer...

casalucas · Answer

According to me, the right answer is A.

Starting from the average asked about (11), let's check the global quantity of accidents it gives for the company : 11 * 140 = 1540.

From Statement (1), you know the average (13) for department C, as well as the quantity of employees (45). Therefore, we know the "total" for this department is 13 * 45 = 585.

The department B has a bigger average, but the average is a fraction, so let's see what's the least quantity of employees in the department, as well as the total it could be. The average is 12.25, so logically the smallest quantity of employees in this department is 5. It gives us a "total" of 61.

The department A has an average of 10. With the 45 employees in C and the 5 in B, it leaves 90 employees maximum in A. It means a total of 900.

From there, we have a total of 61+585+900 = 1546. This is already bigger than the global quantity of accidents we need to be sure that the average is going to be larger than 11. If we changed the distribution of employees between department A and B, this global total would just get higher, since there are more accidents on average in department A versus B. Therefore, Statement (1) is SUFFICIENT.

From Statement (2), we don't learn enough information.

ANSWER : A.

Regards,

Pierre-Luc
Montreal, Canada

gauravgoyal_g · Answer

I would go with C too.
We would need the "weight" of each department to know the company average.

kevincan · Answer

Wow! A fellow Canadian! (That's where kevinCAN comes from). A is the right answer, but the minimum number of employees in B is 4, not 5. Also, how could we answer this question without so much multiplication?

gauravgoyal_g · Answer

Cool question!!! I should treat each of your questions as a trick question

Googlore · Answer

Priyah · Answer

According to me, A is the answer bcoz 
the avg of the 3 depts=(10+12.25+13)/3 = 11.75 >11
though I'm not sure if this is the right approach.

cicerone · Answer

Priya, this is not true in all cases.
U can do this with average calculation provided the strengths of A,B&C are equal........

Priyah · Answer

Yes, you are absolutely right....I am waiting for the correct way to solve this which doesnt involve all the math(as per kevincan).

Zagros · Answer

I would go with C.

I suspected a trap also. However, I could not figure out anyway that either one on it's own is sufficient.

paddyboy · Answer

Wow! Are all Canadians knee deep in math?

kevincan · Answer

(1) We know that there are 45 employees in C, a positive multiple of 4 employees in B and thus at most 91 in A. Suppose that the average number of infractions commited by each employee was at most 11. Thus the number of infractions would be at most 11*140

We know, however, that the number of infractions is
45(11+2)+(4k)(11+1.25)+(95-4k)(11-1)=140*11 +90-95+9k=11*140+9k-5 > 11*140 since k is positive. Thus our original supposition was wrong. i.e the average is more than 11. SUFF.

(2) Clearly INSUFF

OA=A

In a company of 140 employees, there are three departments,

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