At a certain company, the average (arithmetic mean) salary of 10 of

Question

At a certain company, the average (arithmetic mean) salary of 10 of the employees is $30,000, the average salary of 30 other employees is $40,000, and the average salary of the remaining 20 employees is $60,000. What is the average salary of the 60 employees at the company?

(A) $40,000
(B) $43,000
(C) $45,000
(D) $50,000
(E) $55,000

(A) $40,000
(B) $43,000
(C) $45,000
(D) $50,000
(E) $55,000

sashiim20 · Answer

Average salary of  10  employees  = $30,000

Total Salary of  10  employees  = 30,000 * 10 = 300,000

Average salary of  30  employees  = $40,000

Total Salary of  30  employees  = 40,000 * 30 = 1,200,000

Average salary of  20  employees  = $60,000

Total Salary of  20  employees  = 60,000 * 20 = 1,200,000

Total number of employees  = 60

Total Salary of  60  employees  = 1,200,000 + 1,200,000 + 300,000 = 2,700,000

Average salary of  60  employees  = \frac{2,700,000}{60} = $45,000

Answer (C)...

Abhishek009 · Answer

\frac{(10*30000) + (30*40000) + (20*60000)}{10 + 30 + 20}

=  \frac{300000 + 1200000 + 1200000}{60}

=  \frac{2700000}{60}

=  45000

Thus, the answer must be (C)

phoenix128 · Answer

10*30 + 30*40 + 20*60 = 2700. The only option that has a perfect divisor of 2700 is (C) (since 45 * 60 = 2700). Eliminate others. Answer is Option (C).

Heseraj · Answer

Since every number is very round, therefore, we can easily reduce to case to one digit number for employees and two digit numbers for income and just multiply to a 1000 when the final average is gotten.

$\frac{1*30+3*40+2*60}{6} = \frac{270}{6}=45 * 1000$

Therefore, option C

\frac{1*30+3*40+2*60}{6} = \frac{270}{6}=45 * 1000

Therefore, option C

ScottTargetTestPrep · Answer

We can create the following equation for a weighted average:
 
(10 x 30,000 + 30 x 40,000 + 20 x 60,000)/60
 
(300,000 + 1,200,000 + 1,200,000)/60
 
2,700,000/60
 
45,000
 
Answer: C

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At a certain company, the average (arithmetic mean) salary of 10 of

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At a certain company, the average (arithmetic mean) salary of 10 of

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