In school there some bicycles and 4 wheeler wagons. On Tuesday there are 58 wheels in the campus. How many bicycles are there ?
Option
a) 7
b) 8
c) 9
d) 6

• May be it's a dummy que as both 7 and 9 are suitable for the given condition.
  
• 10 years agoHelpfull: Yes(12) No(0)
• let no of bicycles(2wheels) be B
  and no of wagons(4wheels) be W
  total no of wheels = 2B+4W =58
  B+2W=29------->this shows B must be an odd number.
  therefore from the given options ans is 9
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• data insufficient
  
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• either 7 or 11(a/c).
  Bcz (58-(7*2))/4=11,
  similarly for 9 is 10 four wheelers.
  where as for 8 and 6 it is in fractions. so nither b nor c
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• 7/9 can be the ans
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• either 7 or 9.....
  
  data insufficient questions
  
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• ans can be both 7 or 9.
  let the no of bicycles be x and no of 4wheeler wagons be y.so 2x+4y=58. on substituting the values 7 and 9 in place of x we get, 2*7+4*11=58 and 2*9+4*10=58 hence both the options 7 and 9 are valid
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• as per options ans may be 7 or 9 both are possible.
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• 7 bicycles are there
  
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• 7
  because 7*2=14 and 58-14=44/4=11
  7 bycyles and 11 four wheel vagons
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• from ans we can find
  9*2=18
  58-18=40
  divisible by 4
  ans = 9
  
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• a)7. coz....7*2=14....
  58-14=44
  44/4=11(absolutely divisible).
  actually:4 wheelers shld be seperated.
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• it will be 48/4=12 four wheel and 58-48=10 10/2=5 bycyles
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• Ans. 9 Bicycles
  because 9*2=18 and 58-18=40/4=10
  9 bicyles and 10 four wheel Wagons
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• 7 or 9
  
  and data insufficient
  
  
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• let no of bicycles be x and wagons be y
  then 4y+2x = 58
  only 7 and 9 satisfies this
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• answer is d)6
  as bicycles are 2 wheelers
  and assume 4x for 4 wheeler wagons
  so
  equation will be
  (2wheelers)+4x=58
  now value for 2 wheeler i.e cycle select from option
  let us assume first a)7
  so,
  7+4x=58
  4x=58-7=51
  x=51/7 as it is not divisible by seven so it is incorrect option
  
  same is for option b and c
  now assume option (d)6
  (2wheeler)+4x=58
  6+4x=58
  4x=52
  x=13
  so it is divisible so answer is (d).
• 9 years agoHelpfull: Yes(0) No(2)