Quantitative Aptitude - Arithmetic Ability Questions

Assume both trains meet after 'p' hours after 7 a.m.
Distance covered by train starting from A in 'p' hours = 20p km
Distance covered by train starting from B in (p-1) hours = 25(p-1)
Total distance = 200
=> 20x + 25(x-1) = 200
=> 45x = 225
=> p= 5
Means, they meet after 5 hours after 7 am, ie, they meet at 12 p.m.

55 min. spaces are covered in 60 min.
60 min. spaces are covered in ((60/55) x 60) min = 720/11 min.
Loss in 65 min. = (720/11 - 65 ) = 5/11
Loss in 24 hrs = (5/11 x 1/65 x 24 x 60)min = 1440/143 min.

Here, S={HH,HT,TH,TT}
Let E be the event of getting one head
E={TT,HT,TH}
P(E )= n(E)/n(S) =3/4

The number of exhaustive events = 100 C₁ = 100.

We have 25 primes from 1 to 100.

Number of favourable cases are 75.

Required probability = 75/50 = 3/2.

In this type cases we always have loss only.

Loss% = Common Gain or Loss102= 2102= 1/25 = 0.04 %

Let their ages one year  ago be 4x and 3x years

 Sum of their present ages =(4x +1+3x+1) = 16 years.

Let C.P =Rs. x

Then as given, (753 - x ) = (x - 455)

=>2x = 1208 => x = 604

 There fore S.P= 150% of 604 => Rs. 150100×604= Rs.906

Since order does not matter it is a combination. 

The word AND means multiply. 

Given 4 basketball, 3 volleyball, 2 soccer. 

We want 2 basketball games and 1 other event. There are 5 choices left. 

C(n,r) 

C(How many do you have, How many do you want) 

C(have 4 basketball, want 2 basketball) x C(have 5 choices left, want 1) 

C(4,2) x C(5,1) = (6)(5) = 30

Therefore there are 30 different ways in which you can go to two basketball games and one of the other events.