Draw a table with Student, Subject and Sport.
P studies in Class II.
Q likes black.
R does not study in Class VI.
S likes pink and studies in Class I.
M likes blue.
Q likes black and does not study in Class IV or V.
The one who studies in Class IV does not like green.
M likes blue and does not study in Class IV.
The one who likes yellow studies in Class VI.
R does not study in Class VI.
Based on the above information, the classification of the group is as follow in the table.
Using Trial and error method,
From the options u = 1, v = 3/2 satisfies both the equations.
Given (49.001)2 = ?
=> =~ 49 x 49
=~ 2401
=~ 2400
? = (3 + 4 - 2 - 1) + ( 1/6 + 1/2 - 2/3 - 11/12)
= 4 + [(2+6-8-11)/12]
= 4 - (11/12 )= 31/12.
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
-4-(-10) = -4+10 = 6
-10-(-4) = -10+4= -6
Therefore, 6-(-6) = 6+6 = 12
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
=>
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.
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