if x-y + z = 19 , y + z =20 , x-z=3 , find d value of x+4y-5z
a)22 b)38 c)17 d) none

• ans is option D) none
  
  x-y+z=19---------(1)
  y+z=20
  thus y=20-z---------(2)
  x-z=3
  thus x =3+z---------(3)
  substitute x and y in equation (1)
  you will get,
  (3+z) - (20-z) + z = 19
  3z = 36
  z = 12
  substitute z in (2) and (3)
  thus x = 15
  and y = 8
  substitute value of x,y,z in x+4y-5z
  answer is - 13
• 10 years agoHelpfull: Yes(10) No(0)
• option d
  
  x-z=3 (given)
  x=3+z..(1)
  sub in x in this
  in x-y+z=19..(given)
  z+3-y+z=19
  -y+2z=16...(2)
  y+z=20..(given)
  z=12;
  then x=15,y=8
• 10 years agoHelpfull: Yes(1) No(0)
• d)
  y=8,x=15,z=12
• 10 years agoHelpfull: Yes(0) No(0)
• d) none
  Exp : By solving 3 equation we get
  x = 17, y = 6, z= 14.
  So x+4y-5z = -29
• 10 years agoHelpfull: Yes(0) No(0)
• 31 is the answer...
  
  let y be the total no. of loaves
  a be the remaining loaves after 1st thief
  b be the remaining loaves after 2nd thief
  .
  .
  e = 0 be the remaining loaves after 5th thief
  
  solution:
  a = y - 0.5y - 0.5
  = 0.5y - 0.5
  b = a - 0.5a - 0.5
  = 0.5a - 0.5
  c = 0.5b - 0.5
  d = 0.5c - 0.5
  e = 0.5d - 0.5 but e = 0,
  thus,
  d = 1
  c = 3
  b = 7
  a = 15
  y = 31 -> answer
• 10 years agoHelpfull: Yes(0) No(0)
• op:D
  y=8,x=15,z=12
• 10 years agoHelpfull: Yes(0) No(0)
• ans is option D is correct
  x-y+z=19......(1)
  
  y+z=20.........(2)
  
  x-z=3............(3)
  
  from eq (3)
  x=3+z.....(4)
  from eq(2)
  y=20-z....(5)
  
  put the value of eq(4) AND eq(5) in eq(1)
  we get z=12
  similarly we get x=15 and y=18
  
  we get in putting in eq x+4y-5z
  we get result -13
  
  S0,option D is correct....
• 9 years agoHelpfull: Yes(0) No(0)
• option (D) is correct ans .
  bcoz i found x=15,y=8,z=12
  and then put x+4y5z= 15+32-60= -13
• 9 years agoHelpfull: Yes(0) No(0)