Quantitative Aptitude - Arithmetic Ability Questions

Let each instalment be Rs.x .

1st year =  [x + (x * 12 * 2)/100] 

2nd year = [ x + (x *12 * 1)/100] 

3rd year = x 

Then, [x + (x * 12 * 2)/100] + [ x + (x *12 * 1)/100] + x =1092 

3x + ( 24x/100 ) + ( 12x/100 )  = 1092 

336x =109200

Therefore, x = 325 

Each instalment = Rs. 325

 H. C. F of two prime numbers is 1.  Product of numbers = 1 x 161 = 161.

 Let the numbers be a and b . Then , ab= 161.

 Now, co-primes with  product 161 are (1, 161) and (7, 23).

 Since x and y are prime numbers and x >y , we have x=23 and y=7.

 Therefore,  3y-x = (3 x 7)-23 = -2

Area to be plastered = 2l+b×h+l×b 

=225+12×6+25×12=744sq.m 

Cost of plastering = 744×75100=Rs.558                       

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

Herein the given sequence, 4, 2, 1, 0.5, 0.25, 0.125,...

Common Ratio r = 2/4 = 1/2 = 0.5/1 = 0.25/0.5 = 0.125/0.25 == 0.5.