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A very big story.on Tuesday college parking place have only 4wheelers & bicycles,total no of wheels was 182,yhen what is the possible no of bicycles?
Read Solution (Total 3)
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- Both options 2 and 4 are possible.
Let 'f' be the fourwheelers and 'b' be the bicycles
Then,
4f+2b = 182 If b = 20, 4f=142 So, f is not an integer..hence this option is not valid If b = 19, 4f+2b = 182
If b = 20,
4f=142
So, f is not an integer..hence this option is not valid
If b = 19,
4f = 144
So, f = 36, an integer (VALID)
If b = 18,
4f = 146,
So, f is not an integer (NOT VALID)
If b = 17,
4f = 148
f is an intger (VALID) - 12 years agoHelpfull: Yes(5) No(0)
- 91 possible number of bicycles.
Consider, there is no 4wheelers in parking.
So, No. of wheels/2
=182/2
=91 - 12 years agoHelpfull: Yes(2) No(2)
- Soln. = 89 possibles of bicycles..
Expln.= Total Wheels, X=182
Four Wheelers must contain 4 wheels as Y
Two Wheelers must contain 2 wheels as Z
Therefore Max(Z)=???
=X - Min(Y)/2
=182-4/2
=178/2
=89..
$ Vignesh(India,TamilNadu) $ - 13 years agoHelpfull: Yes(1) No(6)
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