Quantitative Aptitude - Arithmetic Ability Questions

Area of a square SxS = 1024 => S = 32 cm
Length of the rectangle = 32x2 cm = 64 cm
Breadth of the rectangle = 32 - 12 = 20 cm
Required ratio = 64 : 20 = 16 : 5

Let the total number of valid votes be 'V'.

70% of V = 70/100 x V = 7V/10

Number of votes secured by the other candidate = V - 7V/100 = 3V/10

Given, 7V/10 - 3V/10 = 172 => 4V/10 = 172

=> 4V = 1720 => V = 430.

Income from Rs. 100 stock = Rs. 8.

Income from Rs. 2500 stock = Rs8100*2500 = Rs. 200.

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.

Let the numbers be x and y. Then, x + y = 20 and x - y = 8.

 x^2 - y^2 = (x + y) (x - y) = 20 x 8 = 160.

let breadth = x then length = x+20  

perimeter = 5300/26.50 = 200m 

2(x+20+x) =200      => 4x+40 =200  

x = 40 and length = 40+20 = 60m

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.