math-solution-class-8

144 beads were distributed among some children| If there were 3 children fewer each child would have got 16 beads | How many children shared the beads?

Question: 144 beads were distributed among some children. If there were 3 children fewer each child would have got 16 beads . How many children shared the beads?Solution: Let the number of children be x.Therefore if there would be 3 children less then each would have get 16 beads. That means number of children now=x-3

math-solution-class-8

1/(1+x^(b-a)+x^(c-a) )+1/(1+x^(a-b)+x^(c-b) )+1/(1+x^(b-c)+x^(a-c) )=1

The text form of this sum (1/ (1+x^(b-a)+x^(c-a) )+(1/(1+x^(a-b)+x^(c-b) )+(1/(1+x^(b-c)+x^(a-c) )=1 is given below. So that students may not get confused with the text form of the sum. Watch video or see this solution:First term: (1/ (1+x^(b-a)+x^(c-a) )=(1/ (1+x^b*x^a)+x^c*x^a))=(x^a/ (x^a+x^b+x^c))Second term: (1/ (1+x^(a-b)+x^(c-b) )=(1/ (1+x^a*x^b)+x^c*x^b))=(x^b/ (x^a+x^b+x^c))Third term: (1/ (1+x^(b-c)+x^(a-c) )=(1/ (1+x^b*x^c)+x^a*x^c))=(x^c/ (x^a+x^b+x^c))ThereforeFirst term+Second term+third term=(x^a/ (x^a+x^b+x^c))+(x^b/ (x^a+x^b+x^c))+(x^c/ (x^a+x^b+x^c))=((x^a+x^b+x^c)/