Quantitative Aptitude - Arithmetic Ability Questions

  4             3             4             7               15        ?
×1-1      ×1+1       ×2-1       ×2+1           ×3-1

15x3 - 1 = 45 - 1 = 44

The given number series follows a pattern that

196, 169, 144, 121, 100, 81, ?

    -27  -25   -23   -21  -19  -17

=> 81 - 17 = 64

Therefore, the series is 196, 169, 144, 121, 100, 81, 64.

Given number series is 5  19  49  101  181  295 ?

Here the series follows pattern that

5   19   49   101   181   295  ?

  14   30   52    80     114   (difference)

      16  22   28     34 (difference)

         6     6     6  (Common difference)

Therefore, the missing number in the series is ? = 295 + 115 + 34 + 6 = 449.

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

At 5 o'clock, the hands are 25 min. spaces apart.
To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.

55 min. spaces are gained in 60 min

40 min. spaces are gained in 6055×40 min  

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.

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.

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4. 

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.