Question

Pls solve from q 1 to q11

Answer 1

1) 3 $\frac{1}{3}$

in improper fraction = 3x3+1 = $\frac{10}{3}$

multiplicative inverse = $\frac{3}{10}$

2) as sum of exterior angle is always 360 , measure of each exterior angle is $\frac{360}{15}$

= 24 degree

3) sorry i do not know what is unbiased  but the prime numbers in a normal dice are 2,3,5  so probability in normal dice = $\frac{3}{6}$  

simpified = $\frac{1}{2}$

4) no it wont be a perfect square because in square roots the number of zeroes should always be even to make it perfect

5) 11 cube = 11 x 11 x 11 = 1331

6) price = 170 rupees

SECTION B

7) $\frac{5}{12}$  + $\frac{-3}{12}$   = $\frac{2}{12}$

simplified = $\frac{1}{6}$

$\frac{7}{5}$  x  $\frac{1}{6}$  = $\frac{7}{30}$

8) let the integers be x , x+1, x+2

so

x + x+1 + x+2 = 60

3x  + 3 = 60

3x = 60-3

3x = 57

x =  $\frac{57}{3}$  = 19

x = 19

so integers are :

x = 19

x+1 = 19 + 1 =20

x+2 = 19 + 2 =21

19,20,21

9) the question is incomplete we need the measure of angle q to find the answer

10) a )  $\frac{9}{10}$

     b)   $\frac{5}{10}$

11) triplets = 2m , $m^{2}$ + 1 , $m^{2}$ - 1

let 2m = 8

m = 8 divided by 2 = 4

$m^{2}$ + 1 = $4^{2}$ + 1 = 16 + 1 = 17

$m^{2}$ - 1 = $4^{2}$ - 1 = 16 - 1 = 15

triplets = 8,17,16

in which 8 is the smallest

hope it helps