Aptitude and Reasoning Questions

As we know we have 10 letter and 10 different address and one more information given that exactly 9 letter will at the correct address....so the remaining one letter automatically reach to their correct address
P(E) = favorable outcomes /total outcomes
Here favorable outcomes are '0'.
So probability is '0'.

From figures (i) and (ii) we conclude that the number 1, 2, 3 and 4 appear adjacent to 6. Thus, the number 5 will appear opposite 6. Therefore, when six is at the bottom, then 5 will be at the top.

ab×bc=45×1516=34=16-916+9

=725

Let the Present age of the son be P, which means he was born P years ago.At the time the son was born, the age of the man was : (48 - P). So, according to the statement made by the man, his age when the son was born should be equal to 2/3 of P.

Therefore, 

Solving, we get P = 28.8 years; which is the present age of the son.

By selling Rs. 100 stock , cash realised = Rs.100-4-14=Rs.3834  

By selling Rs. 2400 stock, cash realised =Rs.3834*1100*2400 = Rs 2298.

When we divide 1234 by 8, remainder is 2

When we divide 1235 by 8, remainder is 3

When we divide 1237 by 8, remainder is 5

---> 2 x 3 x 5 = 30 

As 30 will not be the remainder because it is greater than 8,

when 30 divided by 8, remainder = 6.

16200 = 23×34×52

A perfect cube has a property of having the indices of all its prime factors divisible by 3.

Required number = 32×5 = 9x5 = 45.

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4
= 24-1 ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
=23-1........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + … + 5C5
= 25..............(C)

From (A), (B) and (C), required number of ways
=  2524-123-1