Quantitative Aptitude - Arithmetic Ability Questions

Total number of elementary events = 20C2 ways =190

There are 7 white balls out of which one white can be drawn in 7C1 ways and one ball from remaining 13 balls can be drawn in 13C1 ways.

So,favourable events = 7C1*13C1

Therfore,required probability = (7C1*13C1)/20C2=91/90

Rate = 4%

Time = 8 years

Simple Interest = PTR/100

= 32P/100

Given, P –32P/100 = 34068

P = 34000

P = 500

We know, the perimeter of a triangle = sum of all the sides of a triangle = a + b + c

Here given that sides a = 38, b= 36 and c = 18

Now, perimeter of the triangle = a + b + c = 38 + 36 + 18 = 92 units.

In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E

Total = 13 letters

But last letter must be N

Hence, available places = 12

In that odd places = 1, 3, 5, 7, 9, 11

Owvels = 4

This can be done in 6P4 ways 

Remaining 7 letters can be arranged in 7!/3! x 2! ways

Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.

Total weight of 50 boys = 45 X 50 = 2250

Average weight of 49 boys = 44.9

Total weight of 49 boys = 44.9 X 49 = 2200.1

Weight of boy who left the class=2250–2200.1=49.9

Let the smaller number be 'm'

Then, from the given data, the larger number is '3m'

Given that m + 3m = 44

=> 4m = 44

m = 44/4 = 11

=> m = 11

=> 3m = 3 x 11 = 33

Hence, the two numbers are 11 and 33.