Quantitative Aptitude - Arithmetic Ability Questions

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.

Let the length of the rectangle = L cm

Then the breadth of the rectangle = (L - 10) cm

Perimeter of a rectangle = 2(L + B) cm

=> 44 = 2(L + L - 10)

=> 44 = 4L - 20

=> 4L = 64

=> L = 16 cm

=> B = L - 10 = 6 cm

Diagonal = L2+B2 = 162+62 = 17.08 cm

Given that train K crosses a pole in 30 seconds and train L crosses the same pole in one minute and 15 seconds.

Let the length of train K be Lk and that of train L be Ll

given that Lk = 3/4 Ll

As the train K and L crosses the pole in 30 seconds and 80 seconds respectively,
=> Speed of train K = sk = Lk/30

Speed of train L = sl = Ll/80

Lk = 3/4 Ll
=> sk = 3/4 Ll/(30) = Ll/40

Ratio of their speeds = sk : sl
= Ll/40 : Ll/80

=> 1/40 : 1/80  =  2 : 1

Let the side of the square be 's' cm

length of rectangle = (s+5) cm

breadth of rectangle = (s-3)cm

(s+5) (s-3) = 

s2 - 5s - 3s - 15 = s2 

2s = 15

Perimeter of rectangle = 2(L+B) = 2(s+5 + s–3) = 2(2s + 2)

= 2(15 + 2) = 34 cm 

Given total number of students in the class = 21

So each student will have 20 greeting cards to be send or receive (21 - 1(himself))

Therefore, the total number of greeting cards exchanged by the students = 20 x 21 = 420.

Let the number of correct answers be X.

Number of incorrect answers = (60 – X).

Therefore, 4x – (60 – x) = 130 

=> 5x = 190  => x = 38

-10+10-10*10100 % = -1%

 Decrease in revenue = 1%

circumference of A = 2πr = 8 => so r = 4π
volume = π×4π2×10 = 160π
circumference of B = 2πr   = 10 => so r = 5π
volume = π×5π2×8 =  200π
so ratio of capacities = 160/200 = 0.8
so capacity of A will be 80% of the capacity of B.