Quantitative Aptitude - Arithmetic Ability Questions

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 1897, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1897 + 11) = 1908, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1897 + 6) = 1903

Therfore, The calendar of the year 1897 can be used again in the year 1903

Let the Number be Y.

Then  Y = 296 q + 75

= (37 x 8)q +( 37 x 2) + 1

= 37 (8q + 2) + 1

Thus, when the number is divided by 37, the remainder is 1

perimeter = total cost / cost per m = 10080 /20 = 504m
side of the square = 504/4 = 126m
breadth of the pavement = 3m
side of inner square = 126 - 6 = 120m
area of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq m
cost of pavement = 246*6*50 = Rs. 73800

a+b+c2=a2+b2+c2+2ab+bc+ca

$$\begin{gathered} 2ab+bc+ca=a+b+c2-a2+b2+c2 \\ =169-69=100 \\ =ab+bc+ca=50 \end{gathered}$$

Here a = 3 and r = 6/3 = 2. Let the number of terms be n.
Then, t = 384 => a * r^(n-1) = 384
=> 3 * 2^(n-1) = 384 => 2^(n-1) = 128 = 2^(7)

=> n-1 = 7 => n = 8.

Let Rajeev's present age be  x years.

Rajeev's age after 15 years = (x + 15) years.

Rajeev's age 5 years back = (x - 5) years

Then ATQ, 

x + 15 = 5 (x - 5)

x + 15 = 5x - 25

=> x = 10

Hence, Rajeev's present age = x = 10 years.