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
Using Correct Symbols , We have :
Given expression = 30 / 2 + 3 x 6 - 5 = 15 + 18 - 5 = 28
Here 12 x 2.5 = 30
In the same way 14 x 2.5 = 35
By the given data, we have the expression :
(3 × 15 + 19) ÷ 8 - 6 = (45 + 19) ÷ 8 - 6 = 64 ÷ 8 - 6 = 8 - 6 = 2
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
=>
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