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Problematics | Cycling to college
Classes start at 9:30 and two students arrive separately, each a few minutes late. How far are their respective villages from the college?
There was a time when the bicycle was a widespread mode of commutation even in urban centres. I grew up in a city where many students cycled to school, although I was not one of them: I took the school bus. Today, of course, bicycles are rarely used by students in larger cities, especially metros, but there are still places where students ride them to school or college. You may have come across such cyclists when going on a road trip away from busy urban centres.
Here are a couple of such students.
#Puzzle 203.1
A district college lies at the end of a highway. Along the highway are several villages from where students cycle to the college every day. Classes start at 9:30 am, and students by practice know when to set off in order to reach college on time, give or take a few minutes.
Two of the villages are exactly 1 km apart. For ease of identification, let us call them Nearer Village and Farther Village (with respect to the college). One morning, a student sets off from Farther Village at 9 am, cycling at his usual speed. Another student sets off from Nearer Village at 9:09, also cycling at her usual speed. The girl rides faster than the boy, and given that she also lives 1 km closer to the college, she can afford to start a little later.
At 9:24, the girl catches up with her classmate, greets him, and continues ahead. She reaches the college at 9:33, 3 minutes after the bell rings. The teacher arrives at 9:35. The boy rides into the college at 9:36 and is allowed into the classroom by the teacher, who had arrived only a minute earlier.
What are the distances of the two villages from the college, and the cycling speeds of the two students?
#Puzzle 203.2
Father #1 gives his son ₹150. Father #2 gives his son ₹100. Together, the two sons have become richer by 150.
Explain how.
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 202.1
Dear Mr Kabir,
The basis for the winning strategy in the described game (Fibonacci nim) is to express the existing number of sticks as the sum of distinct, non-consecutive Fibonacci numbers. This representation is unique for every positive integer, and is called a Zeckendorf representation.
The game in the puzzle begins with 20 sticks. This can be expressed as 20 = 13 + 5 + 2. The strategy is to remove the smallest Fibonacci term in this expression, i.e. 2. After the crucial opening move of removing 2 sticks, each subsequent move by the first player depends on what the second player chooses.
The key rule is not a fixed one-line script after the opening move; it is a response strategy. After each move by the opponent, recalculate the remaining sticks in Zeckendorf form, and take the smallest Fibonacci number in that representation. Repeat this until winning.
— Shri Ram Aggarwal, Palam, New Delhi
(Thank you. The term “Zeckendorf representation” is new to me. That apart, the strategy described above is exactly what I had read in a published description of the game. — KF)
#Puzzle 202.2
Hi,
Let N be the total number of errors. Say Proofreader #1 spots a fraction p of all errors, and Proofreader #2 spots a fraction q of all errors. Since Proofreader #1 found 30 errors, pN=30. And since Proofreader #2 found 24 errors, qN=24.
Now 20 errors were found by both. Since each proofreader independently spots a fixed fraction of errors, the fraction spotted by both is pq. Therefore, pqN=20.
Substituting p=30/N and q=24/N in the above equation gives (30/N)(24/N)N=20, which simplifies to N = 36. So, there were 36 errors in total.
The number of errors found by at least one proofreader is 30 + 24 – 20 = 34. Therefore, the number missed by both is 36 – 34 = 2.
— Ajay Ashok, New Delhi
Solved both puzzles: Yadvendra Somra (Sonipat), Vinod Mahajan (Delhi)
Solved #Puzzle 202.1: Shri Ram Aggarwal (Delhi), Dr Sunita Gupta (Delhi), Professor Anshul Kumar (Delhi)
Solved #Puzzle 202.2: Ajay Ashok (Delhi)
