Quantitative Aptitude - Arithmetic Ability Questions

Converting each of the given fractions into decimal form, we get

 5/9 = 0.55, 7/11 = 0.63,  8/15 = 0.533, 11/17 = 0.647

Clearly, 0.647>0.63>0.55>0.533

so , 11/17 > 7/11 > 5/9 > 8/15

100C10 + 90C20 + 70C30 + 40C40 = 100!10!x20!x30!x40!

Let us assume Amount be 100 Rs and we calculate in CI

 First year 60% of 100 = 60 amount (100+60) is 160

 Second year 60% of 160 = 96 amount (160+96) is 256

 Third year 60% of 256 =153.6 amount (256+153.6) is 409.6

 Here the Amount of 100 Rs is quadrapled in 3 years.

 One year contains 2 half years

 Three year has 6 half years.

 Therefore,  It takes 6 half years.

To cover a distance of 800 kms using a 800 cc engine, the amount of diesel required = 80 litres.

=> the amount of diesel required to cover a distance of 1000 kms = 1000 x 80/800 = 100 litres

However, the vehicle uses a 1200 cc engine and it is given that the amount of diesel required varies directly as the engine capacity.

i.e., for instance, if the capacity of engine doubles, the diesel requirement will double too.

Therefore, with a 1200 cc engine, quantity of diesel required = 1200/800 x 100 = 150 litres.

Maneela lent Rs. 8000 in two parts,

1 Oct, 1994 = (1993 years + Period from 1.1.1993 to 1.10.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

93 years = (70 ordinary years + 23 leap year) = (70 x 1 + 23 x 2)= 4 odd days

Jan  Feb  March  April  May  June  July  Aug  sept  Oct

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 1 ) = 274 days

274 days = (39 weeks + 1 day) =1 odd day.

Total number of odd days = (0 + 1 + 4 + 1) = 6 odd days.

Given day is Saturday.

   I. gives, Rate = 5% p.a.

 II. gives, S.I. for 1 year = Rs. 600.

III. gives, sum = 10 x (S.I. for 2 years).