Quantitative Aptitude - Arithmetic Ability Questions

9321 = 3 x 13 x 239 = 31 x 13 1 x 239 1 ; add one to all the powers ie.,

(1+1) x (1+1) x (1+1)= 8 ;

we get eight factors.

Days in 4 years => 

Let the first year is Normal year i.e, its not Leap year. A Leap Years occurs once for every 4 years.

4 years => 365 + 365 + 365 + 366(Leap year)

4 years => 730 + 731 = 1461

Therefore, Number of Days in 4 Years = 1461 Days.

Rs.1440 - 1200 = Rs.240 is the interest on Rs.1200 for one year.

Rate of interest = (100 x 240) / (1200) = 20% p.a

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 x 2)/(15 x 14) = 1/35

Probability that both are yellow = ²C₂/¹⁵C₂ = (2 x 1)/(15 x 14) = 1/105

Probability that one blue and other is yellow = (³C₁ x ²C₁)/¹⁵C₂ = (2 x 3 x 2)/(15 x 14) = 2/35

Required probability = 1/35 + 1/105 + 2/35 = 3/35 + 1/105 = 1/35(3 + 1/3) = 10/(3 x 35) = 2/21

Let x litre pepsi is required.

(10-9)=1           :                3=(9-6)

Therefore  in 1:3 one part = 5 

=>  x= 5 litres

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 numbers be 3x, 4x, 5x.

Then, their L.C.M = 60x.

So, 60x=3600 or x=60.

Therefore,  The numbers are (3 x 60), (4 x 60), (5 x 60).

Hence,required H.C.F=60

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking. 

(i) 1 lady out of 4 and 4 gentlemen out of 6 

(ii) 2 ladies out of 4 and 3 gentlemen out of 6 

(iii) 3 ladies out of 4 and 2 gentlemen out of 6 

(iv) 4 ladies out of 4 and 1 gentlemen out of 6 

In case I the number of ways = C14×C46 = 4 x 15 = 60 

In case II the number of ways = C24×C36 = 6 x 20 = 120 

In case III the number of ways = C34×C26 = 4 x 15 = 60

In case IV the number of ways = C44×C16 = 1 x 6 = 6 

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246