of the 4 nos,whose average is 60,the first one is 1/4th of the sum of the sum of the last 3.the first no is what?

• Let four number be p q r and s, then p+q+r+s = 240 and It is also given that p=1/4(q+r+s). So on combining equation numbers 1 and 2. We get p=48
• 7 years agoHelpfull: Yes(6) No(0)
• (a+b+c+d)/4=60
  a+b+c+d=240;
  a=1/4(b+c+d)
  b+c+d=4a;
  substitute value of b+c+d in equation
  a+4a=240;
  5a=240;
  a=48
• 7 years agoHelpfull: Yes(3) No(0)
• the sum of the last three is 48 so 1/4th is 12
• 7 years agoHelpfull: Yes(1) No(2)
• let a,b,c,d be numbers
  their avg is 60 i.e (a+b+c+d)/4=60 -----eq 1
  also a=1/4*(b+c+d) -----eq 2
  i.e b+c+d =4a
  substituting in eq 1
  a+4a=4*60
  5a=240
  a=48
  thus first number is 48
• 7 years agoHelpfull: Yes(1) No(0)
• (q+r+s)/4 +q+r+s=240
  5(q+r+s)=960
  q+r+s=192
  so, (q+r+s)/4=48
• 7 years agoHelpfull: Yes(0) No(0)
• Let First No=x
  Let sum of rest 3 no's = y
  (x+y)/4=60
  x+y=240.....(A)
  also it is given that 4x=y
  therefore,
  4x+x=240
  5x=240
  x=48
• 7 years agoHelpfull: Yes(0) No(0)
• a=0.25(240+a)
  => a=58
• 7 years agoHelpfull: Yes(0) No(0)
• A + B + C + D = 240.
  
  NOW A = B + C + D/4
  B + C + D + 4B + 4C + 4D =960
  5 (B + C + D ) = 960
  B + C + D = 192
  SINCE A IS 1/4 (B + C + D) SO WE GET 48 (192/4).
• 7 years agoHelpfull: Yes(0) No(0)
• given (a+b+c+d)/4=60
  therefore, a+b+c+d=240
  now given
  a=1/4of(b+c+d)
  therfore,we can also write a+b+c+d=240 as,
  1/4(b+c+d)+(b+c+d)=240
  {(b+c+d)+4(b+c+d)}/4=240
  5(b+c+d)=240*4
  b+c+d=240*4/5
  b+c+d=192
  now a= 1/4 of (b+c+d)
  a=1/4*192
  ie, 48 ans.
• 7 years agoHelpfull: Yes(0) No(0)