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I have a machine which has four cog wheels in constant mesh. The largest cog has 242 teeth and the others have 160, 64 and 22 respectively. How many revolutions must the largest cog make before each of the cogs is back in its starting position?

2006-11-26 14:05:32 · 3 answers · asked by freddy197120032003 3 in Science & Mathematics Mathematics

3 answers

242*160=38,720 cogs
38,720/64=605(not divisible
by 242,160,64 or 22) therefore,
same cogs cannot coincide
38,720/22=1760(not divisible by
242 or 64) therefore same cogs
cannot coinside

therefore,minimum revolutions required
to get back to original position
=160

i hope that this helps

2006-11-26 21:19:20 · answer #1 · answered by Anonymous · 0 0

u r write dupi

take lcm of160,64,22
and divide it by 242

ans 14.54
but if uwant integral answer then

take lcm of242,160,64,22
and divide it by 242
ans is160

2006-11-26 14:15:39 · answer #2 · answered by sidharth 2 · 0 0

take lcm of160,64,22
and divide it by 242

2006-11-26 14:09:11 · answer #3 · answered by Dupinder jeet kaur k 2 · 0 0

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