A number when divided by D leaves a remainder of 8 and when divided by 3D leaves a remainder of 21.
What is the remainder left, when twice the number is divided by 3D?
13 b) cannot be determined c) 3 d) 42

• ans is 3, it is because
  let number is x,and quotient is A . D is divisor,and R is remainder= 8
  then x = A*D + 8 --------------(I)
  in second situation let quotient is B , 3D is divisor and R is remainder = 21
  then x = B*3D +21 -----------(II)
  =>
  AD +8 =3BD +21
  or AD - 3BD =21 - 8 = 13
  or D (A- 3B) =13*1
  or D=13 ,and A -3B = 1 , IF B = 1 ,then A = 4
  NUMBER X = 4*13 + 8 = 52+8 = 60
  If 60*2 is divided by 3D =3*13 =39,
  120/39 = 3 is quotient,and REMAINDER = 3
• 9 years agoHelpfull: Yes(22) No(3)
• X=aD+8.....(1)
  X=b3D+21.....(2)
  Multiplying (1) with 3
  3X=3aD+24.....(3)
  Subtracting 2 from 3...
  2X=3D(a-b)+3...
  So 2X/3D=(a-b)+3/3D...
  So remainder is 3.... :-)
• 9 years agoHelpfull: Yes(10) No(0)
• 42
  3Dk+21=y
  2(3Dk+21)=2y
  3D(2k)+42=2y
  remainder =42
• 9 years agoHelpfull: Yes(0) No(4)