Quantitative Aptitude - Arithmetic Ability Questions

Time = (100 x 81)/(450 x 4.5) years

= 4 years.

Let the required number of days be x.

Less cows, More days (Indirect Proportion)

Less bags, Less days (Direct Proportion)

$$\begin{gathered} cows 1:40 :40:x \\ bags 40:1 :40:x \\ \end{gathered}$$

1 x 40 x x = 40 x 1 x 40

=>  x = 40.

Let the slower pipe alone fill the tank in x minutes. 

Then,faster pipe will fill it in x/3 minutes. 

=>1x+3x=136 =>4x=136 =>x=144mins

Numbers of students in section A = x
∴ Numbers of students in section B and C = (100 – x)
∴ x 70 + (100 – x) 87.5 = 84 100
=> 70x + 87.5 100 – 87.5x = 8400
=> 8750 – 17.5x = 8400
=> 17.5x = 8750 – 8400 => x = 20.

Let the distance be d.

 d8-2-d8+2 = 4

=> 2d = 120 

=> d = 60 kms.

Let the first number be K and second number be L
Then, 2K + 3L = 100.....(1)
3K + 2L = 120.......(2)
solving (1)&(2), we get
K = 32 and L = 12

Hence the largest number is 32.

In a circle of n different persons, the total number of arrangements possible = (n - 1)!
Total number of arrangements = n(S) = (15 – 1)! = 14 !
Taking three persons as a unit, total persons = 13 (in 4 units)
Therefore no. of ways for these 13 persons to around the circular table = (13 - 1)! = 12!
In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =
n(E) = 12! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!
= 12!x3x2 / 14x13x12! = 6/14x13 = 3/91