Quantitative Aptitude - Arithmetic Ability Questions

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.

Let the ten's digit be x. Then, unit's digit = 2x + 1.

[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1

<=> (12x + 1) - (9x + 9) = 1 <=> 3x = 9, x =  3.

So, ten's digit = 3 and unit's digit = 7. Hence, original number = 37.

'LOGARITHMS' contains 10 different letters.

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

= 10P4

= 5040.

Let the present ages of son and father be x and (60 -x) years respectively.

Then, (60 - x) - 5= 4(x - 5)

55 - x = 4x - 20

5x = 75  => x = 15

j=mi

let length = 2x and breadth = x then
(2x-5) (x+5) = (2x * x)+75
5x-25 = 75 => x=20
length of the rectangle = 40 cm