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.
Let the runs scored by P, Q, R, S and T be a, b, c, d and e
From given data,
a + b + c + d + e = 36 x 5 = 180
b + c = 107
Let a = x
e = x-8
d = x-3
now 3x - 11 + 107 = 180
3x = 84
x = 28
e's score = 28-8 = 20
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
-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,
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