Quantitative Aptitude - Arithmetic Ability Questions

We know the rule that,

At particular time for all object , ratio of height and shadow are same.

Let the height of the pole be 'H'

Then

 188=H48

=> H = 108 m.

29 x 2 = 58
20 x 3 = 60
         -----------
               2 years

Therefore, the age of child is 2 years.

Total number of events=6 x 6 x 6=216
Let A be the event of getting a total of atleast 6 and B denoted event of getting a total of less than 6 i.e.,3,4,5.

So,B={(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)}

Favourable number of cases=10

Therefore,P(B)=10/216

=> P(A) = 1 - P(B)

             = 1 - (10/216) = 103/108

Let breadth = x metres.
Then, length = (x + 20) metres.

Perimeter =5300/23.50

2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.

x = 90% of y = 

=> y = 1.1% of x.

Total numbers in a die=6

P(mutliple of 3) = 2/6 = 1/3

P(multiple of 4) = 1/6

P(multiple of 3 or 4) = 1/3 + 1/6 = 1/2

Next train comes at 17:00 hrs.
So, last train will be = 17:00hrs - 2:30hrs
= 14:30hrs
Announcement made after 37 min of the last train.
S0, 14:30hrs + 00:37 min
= 15:07 hrs.

Ratio C1:C2 = 3:5 and C2:C3 = 7:11
So C1:C2:C3 = 21 : 35 : 55
Let the strength of three classes are 21x, 35x and 55x respectively, then
Given that 21x + 35x + 55x = 333
=> 111x = 333 or x=3
So strength of the class with highest number of students =55x = 55x3 = 165.