The quadratic equation whose roots are reciprocal of can be obtained by replacing x by 1/x.
Hence, 2(1/x)(1/x)+ 5(1/x) + 3 = 0
=>
-4-(-10) = -4+10 = 6
-10-(-4) = -10+4= -6
Therefore, 6-(-6) = 6+6 = 12
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
? = (3 + 4 - 2 - 1) + ( 1/6 + 1/2 - 2/3 - 11/12)
= 4 + [(2+6-8-11)/12]
= 4 - (11/12 )= 31/12.
Given (49.001)2 = ?
=> =~ 49 x 49
=~ 2401
=~ 2400
Using Trial and error method,
From the options u = 1, v = 3/2 satisfies both the equations.
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 is 51.
This can be done in a method called Approximation.
Now,
Using BODMAS law,
3 x 3 + 3 - 3 + 3 =
3 x 3 = 12
= 12 + 3 - 3 + 3
= 9 + 3
= 12
Hence, 3 x 3 + 3 - 3 + 3 = 12.
? = 5068 x 4/37
? = 548
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.