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Problematics | Ribbon colours
A puzzle with four sisters, each wearing a ribbon of a bright colour. Whose ribbon colour can you work out?
One of last week’s mathematical puzzles proved tougher than I had thought, with a couple of veteran Problematics solvers stumbling on it. The crossword of Indian states, on the other hand, was probably taxing to fill up although it was a sitter to solve.
Both of this week’s puzzles come under the logic category, with the second involving numbers.
#Puzzle 185.1
Meet a family in which the parents have imaginatively named their four daughters Lily, Leela, Leena and Lulu. They are attending a birthday party, each wearing a bright ribbon of a solid colour (i.e. no dots, stripes or any kind of printed pattern).
(1) Lily has exactly two sisters wearing red ribbons.
(2) Leela has exactly two sisters who are not wearing a blue ribbon.
(3) Leena has exactly two sisters whose ribbons are either red or orange (this may be red-red, orange-orange, or red-orange).
(4) At least one of Lulu’s sisters is wearing an orange ribbon.
(5) At least one of Lulu’s sisters is wearing a blue ribbon.
It is not possible to determine the colour of all four ribbons with this information.
Whose ribbon colour(s) do you know for certain?
#Puzzle 185.2
In a game between two players, each one calls out a number in turn. The rule: the first player can call out any number from 1 to 10. The second player then calls a number which must be higher than the first number by a difference of at least 1 but not more than 10. For example, A calls 1, B calls 10, C calls 15, D calls 25 and so on. The person who calls 100 wins.
You have the first move. What number should you call first, and what next, depending on your opponent’s call?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 184.1
Hi Kabir,
Let S be the speed of the train boarded by the first puzzler. He meets train A after 120 km and train B after 140 km from X. Let the point where he meets train B be at a distance D km from Y. Let Sₐ and Sᵦ be the speeds of the trains running from Y to X. Therefore,
120/S = (D + 20)/Sₐ, and 140/S = D/Sᵦ
That is, Sₐ = (D + 20)S/120 and Sᵦ = D x S/140
=> Sₐ/Sᵦ = 7(D + 20)/(6D)
The second puzzler meets train B at the midpoint. Therefore, the speed of the train boarded by her is equal to Sᵦ. She meets train A at a point 126 km from Y. Therefore,
(D + 140 – 126)/Sᵦ = 126/Sₐ
=> Sₐ/Sᵦ = 126/(D + 14)
Equating the two expressions of Sₐ/Sᵦ we get,
126/(D + 14) = 7(D + 20)/(6D)
=> 108D = (D + 20)(D + 14)
=> 108D = D² + 34D + 280
=> D² – 74D + 280 = 0
=> (D – 70)(D – 4) = 0
There are two possible solutions: D = 70 or D = 4
The distance between X and Y = 120 + 20 + D, which is equal to either 210 km or 144 km
In the first solution, Sₐ = (70 + 20)S/120 = 3S/4, and Sᵦ = 70 x S/140 = S/2
In the second solution, Sₐ = (4 + 20)S/120 = S/5, and Sᵦ = 4 x S/140 = S/35
The first solution appears to be more reasonable.
— Professor Anshul Kumar, New Delhi
***
Dear Kabir,
Let the distance between P and Q be D km. Let the speed of the train on flat track be v km/min, so the uphill speed is 0.6v.
Let the train take T minutes to reach its destination as per its schedule.
Considering fully flat track we get
D= v*(T – 120)
Considering existing track we get
D= v*60 + 0.6v*(T – 60)
= 24v + 0.6vT
Considering that the uphill track starts 50km farther from the existing point:
D= 60v + 50 + 0.6v*(T – 80 – 60 – 50/v)
= 0.6vT – 23v + 20
Equating (2) and (3) we get
24v + 0.6vT = 0.6vT – 23v + 20
v = 5/12
Using this value in the first two equations, we get
D= (5/12)*(T-120) = 5T/12 – 50, and
D= 24*5/12 + 0.6*5T/12 = 10 + T/4
Equating (4) and (5) we get
5T/12 – 50 = 10 + T/4
=> T= 360 min. Therefore,
D = 5*360/12 – 50
= 150 – 50 = 100 km
— Yadvendra Somra, Sonipat
#Puzzle 184.2
In #Puzzle 184.1, the distance between X and Y is 210 km (there is also another solution, 144 km). The distance between P and Q is 100km
#Puzzle 184.2 requires ordinary skill to fill the crossword grid but exceptional skill of communication. So I pass.
— Kanwarjit Singh, Chief Commissioner of Income-tax, retired
The last point raised in the above mail is probably a valid one. One reader requested me to mail a larger image of the crossword grid, which I did. The filled crossword above is my own, and readers who solved it are credited as usual. In the list below, anyone who has solved any one of the two train puzzles is being credited.
Solved both puzzles: Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat), Sabornee Jana (Mumbai), Shishir Gupta (Indore), Vinod Mahajan (Delhi)
Solved Puzzle 184.1: Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Dr Sunita Gupta (Delhi)
Solved Puzzle 184.2: Ajay Ashok (Delhi)
