Quantitative Aptitude - Arithmetic Ability Questions

When at least 2 women are included.

The committee may consist of 3 women, 2 men : It can be done in  4C3*6C2  ways

or, 4 women, 1 man : It can be done in  4C4*6C1ways

or, 2 women, 3 men : It can be done in 4C2*6C3 ways.

Total number of ways of forming the committees

= 4C2*6C3+4C3*6C2+4C4*6C1

= 6 x 20 + 4 x 15 + 1x 6

= 120 + 60 + 6 =186

Given a x a = 12

asked to find a x a x a x a =  (a x a) x (a x a)

a⁴ = a² x a² 

= 12 x 12

= 144.

More :: Mathematical operations questions and answers.

Ratio of times taken by A and B = 1 : 2.

The time difference is (2 - 1) 1 day while B take 2 days and A takes 1 day.

If difference of time is 1 day, B takes 2 days.

If difference of time is 30 days, B takes 2 x 30 = 60 days.

So, A takes 30 days to do the work.

A's 1 day's work = 1/30

B's 1 day's work = 1/60

(A + B)'s 1 day's work = 1/30 + 1/60 = 1/20

A and B together can do the work in 20 days.

The given number series 100, 101, 97, 105, 90

100

100 + 1^2 = 101

101 - 2^2 = 97

97 + 3^2 = 106 ( not 105)

106 - 4^2 = 90

Hence, the odd man in the given series is 105.

Let the total number of children be x. 

Then, x * (20% of x) = 405  

=> x * 20x/100 = 405 

15x2 = 405 

=> x=45 

Number of sweets received by each child = 20% of 45 = 9.