Quantitative Aptitude - Arithmetic Ability Questions

Let the required length be x meters

 More men, More length built     (Direct proportion) 

Less days, Less length built      (Direct Proportion)

Men20:35Days6 : 3⋮⋮ 56 :x 

=> (20 x 6 x X)=(35 x 3 x 56) 

=> x = 49 

Hence, the required length is 49 m.

Excess of B's height over A's = [(40/(100 - 40)] x 100% =  66.66%

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 given number series is 14, 27, 105, 525, 4993

14

14 x 2 - 1 = 27

27 x 4 - 3 = 105

105 x 6 - 5 = 625 (not 525)

625 x 8 - 7 = 4993

Hence, the odd man or number in the given number series is 525.