Answers

2014-07-18T00:50:59+05:30
Let, no. of chocolate = x  ,no. of banana = y,& no. of coconut =z
therefore x+y+z=100----------------eqn(1)
1 banana=1 rupees  , 1chocolate=5/100=1/20 rupees  , 1 coconut=5 rupees
a/q,       1/20x+y+5z=100
  or,         x+20y+100z=2000-------------eqn(2)
  or,      (100-y-z)+20y+100z=2000  ----------------------from eqn(1)
  or,       99z+19y=1900
  or,       80z+19z+19y=1900
  or,       80z+19(y+z)=1900
  or,       80z+19(100-x)=1900
  or,       80z-19x=0
                  x/z=80/19
so, when x=80,then z=19
it means when x is less than 80 then z will be in fraction which is not possible ,because x,y,z are numbers of things .or if x is greater than 80 then z will be in fraction.
therefore x =80 ,z=19,y=1 


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2014-07-18T22:15:21+05:30
Let, no. of chocolate = x
no. of banana = y
no. of coconut =z
therefore x+y+z=100----------------(1)
1 banana=1 rupees
1chocolate=5/100=1/20 rupees
1 coconut=5 rupees
         1/20x+y+5z=100
⇒         x+20y+100z=2000-------------eqn(2)
⇒    (100-y-z)+20y+100z=2000  ----------------------from eqn(1)
⇒       99z+19y=1900
⇒      80z+19z+19y=1900
⇒       80z+19(y+z)=1900
⇒      80z+19(100-x)=1900
⇒       80z-19x=0
                  x/z=80/19
therefore when x=80,then z=19
⇒ when x is less than 80 then z will be in fraction which is not possible
⇒x≠80
since because x,y,z can take only numeric values
therefore x =80 ,z=19,y=1 
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