Mathematical Operations Questions

Let the least value of the prize = Rs. x

Then the next value of the prize is x+30 , x+60, x+90, ....x+240.

Given total amount of cash prizes = Rs.1890

--> x + (x+30) + (x+60) + (x+90) + ....+ (x+240) = 1890

--> 9x + (30 + 60 + 90 + 120 + 150 + 180 + 210 + 240) = 1890

--> 9x + 30(1 + 2 + 3 + 4....+ 8) = 1890

--> 9x + 30(36) = 1890

--> 9x = 810 --> x=90

Hence the least value of the prize x=90

Unit digit of this expression is always 1 as the base ends with 1. 

For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit.

i.e, tenth place digit in 2151 is 5 and

tenth place digit in power 415 is 1

And the units digit in the product of 5 x 1 = 5

Therefore, last two digits of 2151415 is 51.

By the given data, we have  the expression :

(3 × 15 + 19) ÷ 8 - 6 = (45 + 19) ÷ 8 - 6 = 64 ÷ 8 - 6 = 8 - 6 = 2

Let the number of buffaloes be x

 and number of ducks be y.

Now, number of heads = x+y

Number of legs = 4x+2y  .......(1) 

But given that legs = 2(x+y)+24 ......(2)

solving (1) & (2),we get

4x+2y=2x+2y+24 

--> x=12.

Therefore number of buffaloes =12