haresh went to buy fruits.she bought 1 apple , 3 bananas , 5 oranges and paid 21 rupees.another person bought 7 apple ,5 bananas , 3 oranges and paid 51 rupees.if i want to buy 1 apple, 1 banana , 1 orange then how much i have to pay?
a) 8
b) 9
c) 10
d) data insufficient

• There can be multiple possible rates of apple, banana and orange to satisfy given conditions, But cost of 1 apple, 1 banana , 1 orange will be Rs 9.
  1x+3y+5z = 21........(1)
  7x+5y+3z = 51........(2)
  
  Adding (1) and (2), we get
  8x+8y+8z = 72
  8*(x+y+z)=72
  x+y+z=9
  
  so option b)..9
• 12 years agoHelpfull: Yes(6) No(2)
• b)9
  
  Let, x = apple; y = banana; z = orange;
  
  1x+3y+5z = 21--------->(i)
  7x+5y+3z = 51--------->(ii)
  x+y+z = ??---------->(iii)
  
  By solving equation(i)&(ii), we get
  x=5; y=2; z=2
  
  Substitute x,y and z value in equation(iii)
  x+y+z = ??
  5+2+2 = 9
  
  So, answer is 9.
  
  
• 12 years agoHelpfull: Yes(3) No(5)
• data insufficient.
• 12 years agoHelpfull: Yes(2) No(2)
• b) 9
  
  Cost of 1 apple = 5rs.
  Cost of 1 orange= 2rs.
  Cost of 1 banana= 2rs.
  So, total cost of each 1 fruits = 5+2+2 = 9rs.
  
  Haresh bought,
  1apple,3bananas,5oranges = (1 *5) + (3 *2) + (5 *2)
  = 5+6+10
  = 21rs
  Another person bought,
  7apples,5bananas,3oranges = (7 *5) + (5 *2) + (3 *2)
  = 35+10+6
  = 51rs
  
  So, for
  1apple,1banana,1orange = (1 *5) + (1 *2) + (1 *2)
  =5+2+2
  =9rs.
• 12 years agoHelpfull: Yes(2) No(4)
• Data is insufficient.
  
  But by Hit and trial, u can get
  A-Rs 4
  B--Rs 4
  orange--Rs 1
  
  or as 5,2,2
  to satisfy both eqns.
  
  sum in both cases = Rs 9.
  
  Vignesh may be having some logical solution with given data.
• 12 years agoHelpfull: Yes(2) No(2)
• Any cost of banana and orange satisfying condition
  
  B+ 2 O = 6
  
  will give result.
  like B= O =2
  B=4, O =1
  B=1, O =2.5
• 12 years agoHelpfull: Yes(2) No(2)