If the sum of two numbers is 55 and the H.C.F. and L.C.M of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

Option 1 : 55/601 Option 2 : 601/55 Option 3 : 11/120 Option 4 : 120/11

• given a+b=55,hcm and lcm is 5 and 120
  axb=lcmx hcf
  600=120x5
  1/a+1/b=a+b/ab=>55/600=>11/120
  
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• is there any criteria for direct interiew in hcl for 2015 passouts?
  
  
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• option c) 1/a+1/b=(b+a)/ab=55/600=11/120
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• Ans: option 3
  
  Suppose sum of of num a+b=55
  given dat H.C.F and L.C.M of these num are 5 and 120 dat is equal to product of two num means
  ab=5*120=600
  So sum of reciprocals of the numbers i.e1/a+1/b =a+b/ab
  =55/600
  =11/120
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• 15 40
  5*3 2*2*2*5
  LCM
  5*2^3*3=120
  
  1/15+1/40
  =11/120
  
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• given: (x+y)=55 .lcm=120 , hcf=5
  let the two no: be x & y.
  the reciprocal of the no: is 1/x & 1/y.
  we know that
  num1* num2=lcm*hcf
  x*y=120*5=600
  ques is sum of reciprocal of given no
  therefore we have to find 1/x+1/y
  (x+y)/xy=55/600=11/120
  
  
  
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• a+b=55
  hcf=5
  lcm=120
  product(ab)= lcm*hcf=120*5=600
  1/a + 1/b =(a+b)/ab =55/600 = 11/120
• 10 years agoHelpfull: Yes(0) No(1)