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So the question is - Six bells commence tolling together at intervals of 2,4,6,8,10 and 12 seconds respectively In thirty minutes how many times do they toll together? (A) 16 (B)15 This question is real question from a recent competitive exam. Now half the students marked 15 and other half 16. But the answer key says correct answer is 16. Now the argumennt in favour of 15 is page 3 of https://docs.google.com/viewer?a=v&pid=explorer&srcid=0B2dG8EJJrqn7YXVPZ2ZxU09lbWs And the peoople say 16 they argue that we have to include start point as well as the end point. Please let me know your answer and the reasoning behind it

divakar
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3 Answers3

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It's a poor question, but wording is difficult. One of the challenges of mathematics is clearly stating our ideas and recognizing things that we have left open to interpretation. This is why we strive to be clear in our definitions. "Commence" and "times" are both undefined, this is why the students were not able to answer the question.

Rylan
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We can imagine a universe $U_1$ in which all humans agree that all intervals specified in plain English include only the start point, unless explicitly stated otherwise. In $U_1$, if a student answers 15 to this question, they are right; if they answer 16, they are wrong. In $U_1$, if the author of an exam asks this question, it is reasonable for them to offer both 15 and 16 as choices.

We can imagine a universe $U_2$ in which all humans agree that all intervals specified in plain English include both endpoints, unless explicitly stated otherwise. In $U_2$, if a student answers 16 to this question, they are right; if they answer 15, they are wrong. In $U_2$, if the author of an exam asks this question, it is reasonable for them to offer both 15 and 16 as choices.

Now, it is an empirical fact that we do not live in $U_1$ or in $U_2$.

On present-day Earth, this question should not be asked on a competitive multiple-choice exam. If it must be asked, the grader should mark both 15 and 16 as correct. Otherwise, the exam degenerates into a game of "guess which universe the author (incorrectly) thinks we live in".

Chris Culter
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You have to calculate the smallest common muliple of the 6 time intervals. If you factor out 120 it is: $2 \times 2 \times 2 \times 3\times 5=120$ Thus all the 6 intervals are included. And there is no smaller common multiple. 1800s/120=15

15 times the bell toll together, if they do not toll together at the beginning. Otherwise they toll 16 times together.

callculus42
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  • @calculas question says they toll in the beginning also. But if we include the beginning bell also do 16th bell fall in 30 minutes? – divakar Jul 31 '14 at 06:42
  • Yes. That´s correct. – callculus42 Jul 31 '14 at 06:44
  • @calculas then how you argue agaimst the explaination given on the page 3 of shared link in the question – divakar Jul 31 '14 at 06:53
  • @divakar Maybe you can argue with the following example: 2 ships are in a harbor. They leaving together the harbor. They return to the harbor at the intervalls of 2 and 3 days. How many times the ships are together in 12 days ? They are together today (midday) . Then they leave the harbor. Ship 1 return at day 2,4,6,8,10,12 and ship 2 return at day 3,6,9,12. Thus the ships are together today and at the days 6,12- Overall 3 times. – callculus42 Jul 31 '14 at 07:42