# Difficult Percentages Solved QuestionAptitude Discussion

 Q. Two persons Raj and Ramu started working for a company in similar jobs on January 1, 1991. Raj's initial monthly salary was Rs 400, which increases by Rs 40 after every year. Ramu's initial monthly salary was Rs 500 which increases by Rs 20 after every six months. If these arrangements continue till December 31, 200. Find the total salary they received during that period.
 ✖ A. Rs 1,08,000 ✖ B. Rs 1,44,000 ✖ C. Rs 1,32,000 ✔ D. Rs 1,52,400

Solution:
Option(D) is correct

Raj's salary as on 1 jan 1991 is Rs 400 per month

His increment in his month salary is Rs 40 per annum

His total salary from 1 jan 1991 to 31st dec 2000

i.e. in ten years

$=12[2(400)+(10-1)40]\times \dfrac{10}{2}$

$=\text{Rs }69,600$

Ramu's salary as on Jan 1st 1991 is Rs 550 and his half yearly increment in his month salary is Rs 20.
His total salary from 1 jan 1991 to dec 31, 2000

$=6[2(500)+(20-1)20]\times \dfrac{20}{2}$

$=\text{Rs }82,000$

Total salary of Raj and Ramu in the ten year period:

= Rs. $69600+$ Rs. $82800$

⇒ Rs 1,52,400

## (5) Comment(s)

PRATYUSH ANAND
()

Hi 2 all,

Here they used Arithmetic Progression(A.P) formula,

Sum of $n$ terms ,where $a$ is initial term ,which is incremented by $d$

$=\dfrac{n}{2}(2a+(n-1)d)$

here,

$a=$ Rs 400,Rs 500 for Raj and Ramu salary respectively.

$n=10, 20$ for Raj and Ramu respectively.

$d=40, 20$ for Raj and Ramu salary increments respectively.

12 is multiplied for Raj's 10yrs calculation as after a year(12 months),his salary is incremented.

Similarly ,6 is multiplied for Ramu's 10yrs calculation as after 1/2 year(6 months),his salary is incremented.

Deepak
()

we are having sum of series I .e yearly salary of raj 12(400)+ 12(400+40)+ 12(400+80) and so on for 10 years and formula for sum of number series with difference is n/2(2a+ (n-1)d) here n =10 and a = 400 and d = 40. common is factor 12

Rakesh
()

Plz explain it again....

how have you calculated raj's money for 10 yr

Balvinder Singh
()

by using ap formula for sum=n/2(2a+(n-1)d)

Rohit
()

I haven't understood the solution ....