how many numbers are there below 1000 which are neither perfect square nor perfect cube?

A) 961
B)958
c)955
D) not remb

• 31^2= 961 and 32^2= 1024... so square of 1 to 31 is less than 1000.
  
  Again 10^3= 1000.... so cube of 1-9 are below 1000.
  
  So, total no. of perfect square and perfect cube is (31+9)-1 [as 1 is common for both].
  
  So the ans is 1000-39= 961
• 9 years agoHelpfull: Yes(38) No(16)
• ans :962
  for numbers 1 to 31 there are squares below 1000
  for numbers 1 to 9 there are squares below 1000
  square and cube of 1 is 1 , 4*4*4=64 and 8*8=64,9*9*9=729 and 27*27=729 so we have to consider it thrice
  then 31+9=40-3=37
  ans 962
• 9 years agoHelpfull: Yes(33) No(5)
• 962 is the right answer
  because there are 31 digits having square below 1000
  and 9 digits having cubes below 1000
  and 1,64,729 are counted twice so u have to subtact these choices
  
• 9 years agoHelpfull: Yes(10) No(0)
• Total no of squares of a number below 1000 are 31
  Total no of cubes below 1000 are 9 in which 1 ,64, 729 already taken in squares.so required no = 999-31-6=962
• 9 years agoHelpfull: Yes(10) No(0)
• ans :961
  for numbers 1 to 31 there are squares below 1000
  for numbers 1 to 9 there are squares below 1000
  square and cube of 1 is 1 so we have to consider only once
  then 31+9=40-1=39
  ans 961
• 9 years agoHelpfull: Yes(5) No(6)
• cubes of no 1 to 9 less than 1000 = 9 nos
  squares of no 1 to 31 less than 1000= 31 nos
  the coinciding cubes and squares are ^ 6 therefore 1^6 ,2^6,3^6 =3 nos
  total nos =31+9-3=37
  nos below 1000 therefore from 999
  999-37=962 ans
• 9 years agoHelpfull: Yes(5) No(0)
• @gona 962 is given in options.
• 9 years agoHelpfull: Yes(1) No(0)
• @SUDHEER KUMAR 'MORON' ques is below 1000 so ans will be 962 at least read ques properly before solving
• 9 years agoHelpfull: Yes(1) No(0)
• We use the Inclusion-Exclusion Principle. There are 1000 integers from 1 to 1000;
  among these numbers, 31 are perfect squares (indeed, 312 = 961 ≤ 1000, but 322 > 1000), 10 are perfect cubes (this is because 103 = 1000), and 3 are both squares and cubes (these three numbers are 16 = 1, 26 = 64, and 36 = 729). Thus, by the Inclusion-exclusion principle, there are
  1000 − (31 + 10) + 3 = 962
  numbers that are neither perfect squares nor perfect cubes.
• 7 years agoHelpfull: Yes(1) No(1)
• 962 is not given in options surely..option d is either 956 or 949
• 9 years agoHelpfull: Yes(0) No(1)
• the ans is 958 bcoz we have 10 cubes below 1000 and square are 32 so from 1000 minus these number we get ens 1000-42=958
• 9 years agoHelpfull: Yes(0) No(2)
• 1,2,4,8,9,16,25,27,36,49,64,81,100,121,125,144,169,196,216,225,256,289,324,343,361,400,441,484,512,529,576,625,676,729,784,841,900,961=== total 38 numbers...so anwwer is 962
• 9 years agoHelpfull: Yes(0) No(0)
• 1 and 64 are common for both square and cube.
  so it should be 31(for square numbers)+9(for 1-9 cube numbers)-2=38
  Is it correct?
• 9 years agoHelpfull: Yes(0) No(3)