Artihmetic commercial maths

[sanju]

I have a question, will ask with example ;

Say : One Cycle is made up of
Handle - 1 NO ; Bell - 1 No and wheel : 2 NO

Question 1: If I have Handle - 10 ; Bells - 10 No and Wheel - 20 No
How many complete cycles can be made ?
Answer : 10 complete Cycles can be made


Question 2 : if I have Handle - 10 ; Bells - 10 NOs and wheel - 19 No
How many complete cylces can be made ?
Answer : 9 complete cycles can be made.

Question 3 : If I have Handle - 100; Bells ;110 and Wheel - 220 No
How may complete cycles can be made ?

What is the method you adopted to answer the question 3 ?

Is there any simple way like HCF,GCD,LCM by which no of cycles can be cmpleted for given handles, bells and wheels ?


Sanju

 

[Guest]

If I understand your question, define int(x) as the greatest integer less than or equal to x. For example, int(1.2) = 1, int(6.87) = 6, and int(4) = 4. The number of complete cycles is the least number out of these three:

int((number of handles) / 1)
int((number of bells) / 1)
int((number of wheels) / 2)

 

[sanju]

alexflury :lol:

That was wonderful solution.

Thank you very much

Sanju