Quantitative Aptitude - Arithmetic Ability Questions

S.I. for 1 ½ years = Rs (1164 - 1008) = Rs 156 . 

S.I. for 2 years = Rs (156 x 23 x 2)= Rs 208. 

Therefore, Principal = Rs (1008 - 208) = Rs 800. 

Now, P = 800, T= 2 and S.I. = 208.

Therefore, Rate = (100 x S.I.) / (P x T) = [ (100 x 208)/(800 x 2)]% = 13%

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

The smallest value will be obtained by giving x the smallest possible value while y gets the largest possible value. The smallest value for x = 7. This means that y will have to be less than or equal to 3 (so that x + y <11), but we want the largest value of y which is 3.

The difference, x - y will be 4.

 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