Quantitative Aptitude - Arithmetic Ability Questions

Given Expression = 18800470÷20 = 40÷20 = 2

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

Let the distance be d.

 d8-2-d8+2 = 4

=> 2d = 120 

=> d = 60 kms.

    Quantity  X  Rate =  Price

         1  x  1  =  1

       0.7 x 1.25 = 0.875

      Decrease in price = (1-0.875) x 100 = 12.5%

Let the middle digit be x.

Then, 2x = 10 or x = 5. So, the number is either 253 or 352

Since the number increases on reversing the digits, so the hundred's digit is smaller than the unit's digit.

Hence, required number = 253.

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

= 4 years.

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.