Quantitative Aptitude - Arithmetic Ability Questions

Required numbers are 10,15,20,25,...,95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)*5=95,then n=18.
Required sum=n/2(a+l)=18/2(10+95)=945.

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

= 10P4

= 5040.

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

P(E) =n(E)/n(S)=2/52=1/26.

A : B : C = 500 : 300 : 200 = 5 : 3 : 2.

Let their capitals be 5x, 3x and 2x respectively.

Then, 5x = 10000  

=>  x = 2000.

C's capital = 2x = Rs. 4000.

i) Let the ages of the five members at present be a, b, c, d & e years. 

And the age of the new member be f years. 

ii) So the new average of five members' age = (a + b + c + d + f)/5 ------- (1) 

iii) Their corresponding ages 3 years ago = (a-3), (b-3), (c-3), (d-3) & (e-3) years 

So their average age 3 years ago = (a + b + c + d + e - 15)/5 = x ----- (2) 

==> a + b + c + d + e = 5x + 15 

==> a + b + c + d = 5x + 15 - e ------ (3) 

iv) Substituting this value of a + b + c + d = 5x + 15 - e in (1) above, 

The new average is: (5x + 15 - e + f)/5 

Equating this to the average age of x years, 3 yrs, ago as in (2) above, 

(5x + 15 - e + f)/5 = x 

==> (5x + 15 - e + f) = 5x 

Solving e - f = 15 years. 

Thus the difference of ages between replaced and new member = 15 years.