A certain number when successively divided by 5 and 7 leaves the remainders 3 and 4 respectively. What is the remainder, if the same number is divided by 35?
(a)21
(b)23
(c)24
(d)29

• ANSWER==> b.23
  let's find the number
  Number= X(say)
  X/5= y+ 3 remainder
  Y/7= 1 + 4 remainder
  so Y=7*1 +4 =11
  X = 11*5 + 3 = 58
  
  Now 58/35 gives remainder = 23
• 10 years agoHelpfull: Yes(6) No(12)
• Let X is the number
  Given Points are
  X=5Y+3 . . . . . (1)
  Y=7Z+4 . . . . . (2)
  Putting Y value in 1st equation we will get
  X=35z+23
  Hence when X is divided by 35 it will leave 23 as remainder.
• 7 years agoHelpfull: Yes(5) No(1)
• Overall we can solve this using remainder understanding. Here we use successive division for this.
  In this case as number was successively divided by 5 and 7 leaving remainders 3 & 5 respectively.
  If a number leaves remainder 3 when divided by 5, then it can be expressed as (5p+3). Now 'p' is the quotient after first division.
  When ‘p' is divided by 7, it leaves remainder 4.
  So, ‘p' can be expressed as (7q+4)
  Thus the number should be: [5 x (7q+4) + 3] = (35k + 23)
  Now when the number is divided by 35, the remainder will be:
  R[(35k+23)/35]
  = R[(35k)/35] + R[23/35]
  = 0 + 23
  = 23 Which is the required answer! Thanks. Option B.
• 5 years agoHelpfull: Yes(0) No(0)
• b) refernce rs agarawal chapter 1 - numbers
• 4 years agoHelpfull: Yes(0) No(0)