Two men X and Y started working for a certain company at similar jobs

We are totaling 10 years of salary for each person, since X has an increase every year, we can group the monthly salaries into annual ones to have:

$3600 base salary, increased by $360 each year.

His salary for the last year would be \(3600 + 360*9\), the mean and median of each year's salaries would be \(\frac{3600 + 3600 + 360*9 }{ 2} = 3600 + 180*9 = 3600 + 1620 = 5220\). Finally multiply by 10 since that is the average for 10 years, to get \(52,200\).

For Y we repeat the same steps but we can consider 20 periods of 6 months, hence base salary 1200 with increments of 90.

At the end of 10 years his salary would be \(1200 + 90*19\). The mean and median of these 20 salaries would be \(\frac{1200 + 1200 + 90*19}{2}\). Multiply by 20 to get \(24000 + 900*19 = 24000 + 17100 = 41,100\).

Finally, add both salaries to get 93,300.

Ans: A

I did not read it well...

Salary of X = 300x 12 + 330 x 12 .... upto 10 terms
= 12 (300+330+360... upto 10 terms)

Sum of salaries of x for 10 years = 12 x (2(First term) x (n-1)difference) x n/2 = 12 x(2(300) + (10-1)30)x 10/2
= 12 x 870 x 10 /2 = 60 x 870 = 52200

Salary of Y = 200 x 6 + 215 x 6 --- upto 20 terms
= 6 ( 200 +215 + ... upto 20 terms)

Sum of Salaries of Y for 10 years = 6 x (2(First term) x (n-1)difference) x n/2 = 6 x(2(200) + (20-1)15)x 20/2
= 6 x 685 x 20 /2 = 60 x 685 = 41100

Total salary of X & Y = 52200 + 41100 = 93300

Why are we multiplying by 12 for X and 6 for Y. I do not see anywhere in the question 300$/month or 200$/month

Or am I missing something?

annual hike for X
where as bi annual hike for Y