Quantitative Aptitude - Arithmetic Ability Questions

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
13.5*10000m2=135000m2 
Let altitude = x metres  and  base = 3x metres.
Then, 12*3x*x=135000⇒x2=90000⇒x=300 

Base = 900 m and Altitude = 300 m.

Let sum = S. Then, amount = 7S/6

S.I. = 7S/6 - S = S/6; Time = 3 years.

Rate = (100 x S) / (S x 6 x 3) = 5 5/9 = 50/9 %.

Let the article costs 'X' to A
Cost price of B = 1.2X
Cost price of C = 0.75(1.2X) = 0.9X
Cost price of D = 1.4(0.9X) = 1.26X = 252
Amount paid by A for the article = Rs. 200.

Circumference = No.of revolutions * Distance covered

Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.
Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.
Number of revolutions per min. =(1100/4.4) = 250.

Total number of elementary events = 50C5
Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 = 29C2 ways and remaining two in 20C2 ways.

Therfore,favourable number of events = 29C2*20C2

Hence,required probability = 29C2*20C2/50C5 =551 / 15134

(A+B)'s two days work = 140+150=9200 

Evidently, the work done by A and B duing 22 pairs of days 

 i.e in 44 days = 22×9200=198200 

Remaining work = 1-198200= 1/100

 Now on 45th day A will have the turn to do 1/100 of the work and this work A will do in 40×1100=25

 Therefore,  Total time taken = 4425daya