In an army of a country, there are (3x^2-5x-2) fighter planes & (x^2-x-2) cargo planes. Both types of planes are put in different group in such a way that every group consists of equal number of planes. And an official is appointed to take care of each and every group. Find out the least number of official to be appointed so that there is atleast one official for every group.

a. 4x+1
b. 4x+2
c. x+1
d. x-2

• x-2 is the answer........
• 10 years agoHelpfull: Yes(12) No(0)
• factorize above two equation
  (3x+1)(x-2),(x+1)(x-2) hcf=(x-2)
  for least group =(((3x+1)(x-2))/hcf) +(((x+1)(x-2))/hcf))=3x+1+x+1
  ans: 4x+2
• 10 years agoHelpfull: Yes(10) No(3)
• Add the equations,u will get 4x^2-6x-4.Factorize it and u will get (x-2)*(4x+2).
  as least number is asked.therfore x-2 is the answer.
• 10 years agoHelpfull: Yes(7) No(1)
• 4x+2
  On factorizing two equations we get (3x+1)(x-2) and (x+1)(x-2)
  this means we can divide both types of planes into 3x+1+x+1 no of groups containing x-2 no of planes in each group..hence 4x+2 is the ans...
• 10 years agoHelpfull: Yes(3) No(0)
• x-2, since it is common in both equations !!!
• 10 years agoHelpfull: Yes(2) No(0)
• plz provide solution in detail...
• 10 years agoHelpfull: Yes(1) No(0)
• assume x=3.so there are 10+4 PLANES.Means 14 groups are made where each group consists of 1 plane and hence 14 officials are to be appointed that is=4x+2.
  
• 10 years agoHelpfull: Yes(0) No(0)