Quantitative Aptitude - Arithmetic Ability Questions

Given year is divided by 4, and the quotient gives the number of leap years.

Here, 100%4 = 25.

But, as 100 is not a leap year => 25 - 1= 24 leap years.

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.

R=100*60100*6 =10% p.a

Now, P = Rs. 12000. T = 3 years and R = 10% p.a.

C.I=Rs.12000*1+101003-1
       =Rs.3972

Let , S -  sample space        E - event of selecting 1 girl and 2 boys. 

Then, n(S) = Number ways of selecting 3 students out of 25 

                = 25C3 

                = 2300.

n(E) = 10C1×15C2 = 1050. 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

Required numbers are 10,15,20,25,...,95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)*5=95,then n=18.
Required sum=n/2(a+l)=18/2(10+95)=945.

 gain in 2 years =Rs.5000*254*2100-5000*4*2100  

= Rs. (625 - 400) 

= Rs. 225 

 gain in 1 year =Rs.2252=112.50