Quantitative Aptitude - Arithmetic Ability Questions

Each of the numbers except 28, is a multiple of 3.

Let A, B, C be the respective events of solving the problem and A , B, C be the respective events of not solving the problem. Then A, B, C are independent event

∴A, B, C are independent events

Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

 PA=12, PB=23, PC= 34

∴ P( none  solves the problem) = P(not A) and (not B) and (not C)  

                  = PA∩B∩C 

                  = PAPBPC         ∵ A, B, C are Independent                       

                  =  12×23×34  

                  = 14  

Hence, P(the problem will be solved) = 1 - P(none solves the problem) 

                = 1-14= 3/4

Here the given series is 396, 462, 572, 427, 671, 264.

In all the terms, the middle digit is the sum of first and third digit except 427.

So the Odd number in the given series is 427.

Let the third number be x.

Then, first number = 120% of x =120x/100 = 6x/5  

Second number =150% of x = 150x/100 = 3x/2

Ratio of first two numbers = 6x/5 : 3x/2 = 12x : 15x = 4 : 5

Given 92613 x 62742241 = ? + 89

Using unit digit method,

On LHS,

1 x 1 x 1 = 1

sqrt of last digit 1 => it must end with either 9 or 1

so now

1 x 1 = 1

1 x 9 = 9

On RHS,

? + 89

=> ? = ...1 - 89 = last digit is 11- 9 = 2

or

=> ? = ...9 - 89 = last digit is 9 - 9 = 0

Here in the options last digit 2 is option C.

Each number is twice the preceding one with 1 added or subtracted alternatively.

So, the next number is (2 x 261 + 1) = 523.

Given 373+353+283-3x37x35x28372+352+282-37x35-35x28-37x28

It is in the form of

 a3+b3+c3-3abca2+b2+c2-ab-bc-ca= a + b + c

Here a = 37, b = 35 & c = 28 

=> a + b + c = 37 + 35 + 28 = 100

Therefore,   = 100

20% of a = b 

=> (20/100) * a = b 

 b% of 20 =(b/100) x 20 = [(20a/100)  / 100] x 20= 4a/100 = 4% of a.