Problematics | The renovated guesthouse
Given two bills for paint, lights and fans without being told any of the individual prices, can you work out the amount of a refund?
The puzzle that launched Problematics in another newspaper decades ago (April 1993) involved a few unknowns and fewer variables. There are at least two kinds of such puzzles. In one kind, you look for usually two integer solutions from one equation, with various constraints ensuring that the solution is unique. We have had a few of those In these columns previously.
The other kind involves more than two variables, usually three variables and two equations. You may not find unique values for the variables, but that is not what such puzzles ask you to determine. As far as I can recall, there has been only one such previous puzzle in the current run of Problematics in HT. Here is my latest one, freshly created.
#Puzzle 137.1
A guesthouse is renovating its reception hall and 3 of its guestrooms. The staff estimate that the hall will require 5 cans of paint while each guestroom will require 2 cans. They also plan to replace the tube-lights (3 in the hall and 1 per guestroom) and the ceiling fans (again 3 in the hall and 1 per guestroom).
The manager, who wants to begin with the reception hall and 2 of the guestrooms, orders 9 cans of paint, 5 tube-lights and 5 ceiling fans. His assistant, however, wants to focus on the 3 guestrooms first. Unaware that the manager has already placed an order, the assistant orders 6 cans of paint, 3 tube-lights and 3 ceiling fans.
Anyone who has ever renovated part of a building knows that the budget will ultimately overshoot estimates, usually because of miscalculations and sometimes because of wastage. In this case, the painting exercise ends up consuming 14 cans instead of the estimated 11; the double order was not entirely pointless. Again, although the number of tube-lights and ceiling fans required was known for a fact (no estimates required here), the renovation still requires more than expected. This is because 1 tube-light and 1 ceiling fan are damaged during fitting by staff, who are aghast when the heartless manager tells them that the cost will be taken out of their salaries.
The stingy manager also wants to make sure that not a paisa is wasted on unused items. The original bills were for ₹15,250 (manager’s order) and ₹9,600 (assistant’s order). Now, the manager sends the unused and undamaged purchases back to the all-purpose store for a refund.
How much is the refund?
#Puzzle 137.2
Over the last two years and more, many of the Wordle puzzles I have created have turned out not quite the way I had intended them to be, with readers coming up with more than one possible answer when I had tried to ensure the solution was unique. This does not in any way make the puzzle less interesting, and in fact the analysis by readers can perhaps also help us form strategies in actual situations.
Although we have repeated the rules of Wordle several times, it is necessary to do so once again for the benefit of any reader who may not have played the game or solved earlier Problematics puzzles based on it. A five-letter word is “hidden” by the game, which you need to determine by entering your own five-letter words and deriving information from them. If you get a green cell, it means the letter appears in the same position in the five-letter hidden word. If you get yellow, it means the letter appears in a different position in the hidden word. Grey means that this letter is absent in the hidden word.
I think this week’s puzzle has a unique solution, although you may once again prove me wrong. Look at the position depicted in the illustration. Three words have been entered, with three more chances remaining.
Is there enough information to finish the game with certainty by entering the correct word on the fourth move itself? Or is there still more than one possible solution?
MAILBOX: LAST WEEK’S SOLVERS
As it turns out, both of last week’s puzzles have more than one solution.
#Puzzle 136.1
If the number of matchboxes and sachets is N each, the cost is 3N and selling price 3.3 N. The profit earned is , therefore, 0.3N
Suppose out of the 7 items left when entire cost is recovered, the number of matchboxes is A and the sachets (7 – A). Their sale price will be 1.1 A + 2.2x(7 – A) = 15.4 – 1.1A. We are told this is the profit, which gives us the equation:
15.4 – 1.1A = 0.3 N, or
11A = 154 – 3N
A has to be an integer. Therefore, N has to a multiple of 11. Further, A is less than N and less than 7. There are two solutions:
Number bought of each item is 33, A will be 5 and the profit ₹9.90; and
Number bought of each item is 44, A will be 2 and profit ₹13.20.
— Kanwarjit Singh, Chief Commissioner of Income-tax (retired)
#Puzzle 136.2
Dear Kabir
The number of students who fail in mathematics = 9, number who fail in English = 11, and the number who fail in both subjects = 3 x (number who pass in both). Thus, the number of students who fail in both subjects should be a multiple of 3 and cannot be more than 9.
Two solutions are possible, with either 3 or 6 students failing in both subjects.
If 3 students fail in both subjects, then 9 – 3 = 6 fail in mathematics only, 11 – 3 = 8 fail in English only, and 3/3 = 1 student passes in both subjects. Thus the total number of students = 3 + 6 + 8 + 1 = 18.
If 6 students fail in both subjects, then 9 – 6 = 3 fail in mathematics only, 11 – 6 = 5 fail in English only, and 6/3 = 2 student pass in both subjects. Thus the total number of students = 6 + 3 + 5 + 2 = 16.
— Y K Munjal, Delhi
Note: Sampath Kumar V and Shishir Gupta, separately, offer a third possibility with 14 students, of whom no one passes in English only, 2 pass in mathematics only, 3 pass in both, and 9 fail in both. Professor Anshul Kumar, on the other hand, proceeded on the assumption that all subgroups have nonzero values.
#Puzzle 135.1 (revisited)
Hi Kabir,
The trajectory of the stylus is an arc of a circle with its centre at the pivot point of the tonearm. It extends from a point (say P) on the outer circle (lead-in groove) to a point (say Q) on the inner circle. The length of arc PQ depends on the exact location of the pivot point. If the pivot point is at a location such that P and Q lie on a radial line of inner/outer circles, arc PQ has its minimum value. Let R be the midpoint of PQ, T be the pivot point of the tonearm and S be the centre of the inner/outer circles. From the given values, it can be shown that arc PQ = 90.49mm.
— Professor Anshul Kumar, Delhi
***
Dear Kabir,
90mm is the radial movement of the needle and 90.49mm is the movement along an arc made by the 250mm long pivoted tonearm.
— Yadvendra Somra, Sonipat
Apart from the two readers above, only Shishir Gupta and YK Munjal have approached the puzzle in this manner (their results were slightly different). The idea was to ignore the grooves and consider only the movement of the needle along the arc, which was hinted in the revised version although not explicitly stated.
Solved both puzzles: Kanwarjit Singh, Chief Commissioner of Income-tax, retired), Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat), Dr Sunita Gupta (Delhi), Shishir Gupta (Indore), Sampath Kumar V (Coimbatore), Shishir Gupta (Indore)
Solved Puzzle #136.2: YK Munjal (Delhi), Dr Vivek Jain (Delhi), Vinod Mahajan (Delhi), Ajay Ashok (Delhi)
Revisited #135.1 (with varying results): Professor Anshul Kumar, Yadvendra Somra, Shishir Gupta, YK Munjal, Ajay Ashok