The solution of 3(2u + v) = 7 uv and 3(u + 3v) = 11 uv is _____

 

? = (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 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

-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

Here 12 x 2.5 = 30

 In the same way 14 x 2.5 = 35

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

Given,

A + B = 2C ----(i)

C + D = 2A----(ii)

Adding (i) and (ii) we get : A + B + C + D = 2C + 2A  

=>  B + D = A + C

Clearly 5 - 0 + 3 X 5 = 20