Question

A company employs three analysts, two programmers and one salesperson. - The analysts are paid a combined total of $264K. - The third analyst makes 50% less than the first two analysts combined. - The first analyst makes 20% more than the second analyst. - The first programmer makes $30K more than the second. - The salesperson makes $20K less than the second programmer. - The company spends a total of $505K on salaries. Determine the salary (in $K) of each employee.

Answer 1

Answer:

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Answer 2

Percentage

Given:

3 analysts, 2 programmers, one salesperson are given in question.

The analysts are paid a combined total of $264K.

The third analyst makes 50% less than the first two analysts combined.

The first analyst makes 20% more than the second analyst.

The first programmer makes $30K more than the second.

The salesperson makes $20K less than the second programmer.

The company spends a total of $505K on salaries.

To find:

Salary of each employee.

Explanation:

Let the employees be A1, A2, A3, P1, P2, S1

Let the salary of second analyst A2 be x.

So, the salary of 1st analyst A1 will be 120% of x$=1.2x$

Hence, the salary of 3rd analyst A3 will be 50% less than (x + 120% of x)$=1.1x$

According to question,

$x + 1.2x + 1.1x =\ 264K$

$3.3x=\ 264K\\\\\implies x=\frac{\ 264K}{3.3} \\\\\implies x =\ 80K$

Let the salary of 2nd programmer P2 be y

So, the salary of 1st programmer P1 be $(y+\ 30)K$.

The salary of salesman S1 be $(y-\ 20)K$.

According to condition given in question,

$x+1.2x+1.1x+y+(y+\ 30)+(y-\ 20)=\ 505K$

$3.3x+3y=\ 495K\\\\\implies 1.1x+y=\ 165\\\\substituting\ the \ value \ of \ x \ in \ this\ equation\\\\1.1(80)+y=\ 165\\\\y=\ 77K$

So, salary of :

A1 = $1.2x=\ 96K$

A2 = $x= \ 80K$

A3 = $1.1x=\ 88K$

P1 = $y+\ 30K=77+30=\ 107K$

P2 = $y= \ 77K$

S1 = $y-\ 20K=77-20=\ 57K$