Question

I'll give brainly to correct answer.7 and 8

Answer 1

Ans. 7)  We are given total no bangles:
                                 Pamela and sheila have : 50
                                       Sheila & Mala have : 40
                            and, Pamela and Mala have : 30


       Let sheila have x no of bangles, Pamela have y no of Bangles               and Mala have z no of bangles
 
    Then we can write, x+y = 50 .................................(1)
                                     x+z= 40 ......................................(2) 
                             and, y+z= 30 .........................................(3)

          By adding Eqn (1), (2) & (3)
 
              (x+y) + (x+z) = (y+z) = 50 + 40 + 30
              2x + 2y + 2z = 120 
              2(x+y+z)=120
             so, x + y + z = 60.......................................(4)
           
       By Putting y+z = 30 in (4)
  
             x = 60-30 = 30
     
          By putting  x+z = 40 in (4)
                              y = 60-40
                              y= 20
       and then z = 10
   
 so, No of Bangles:
      
         Sheila have = x = 30
         Pamela have = y = 20
  and,Mala have = z = 10


Here is the answer
        
            sorry bro! I can't solve ans 8)