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# The hint’s right there

The letter ‘e’ could be added to all these 25 words, fulfilling the given conditions.

Published: 23rd December 2018 05:00 AM  |   Last Updated: 22nd December 2018 07:33 PM   |  A+A-

The next character in the sequence T N C I T S . . .is?

THROUGHPUT

(The problem was: “Check out the following list of words: AT, HAT, MAT, MAD, SCRAP, PET, BIT, RID, STRIP, SLIM, PIN, QUIT, SIT, ON, HOP, SLOP, COP, TAP, CAP, BREATH, CUB, TUB, US, HUG, RAG, Adding just one letter to each could make the word longer, different in sound, in pronunciation, meaning (or both) even though the letter itself is not pronounced ever, in any one of them.”)

The letter ‘e’ could be added to all these 25 words, fulfilling the given conditions. The resultant words are: ATE, HATE, MATE, MADE, SCRAPE, PETE, BITE, RIDE, STRIPE, SLIME, PINE, QUITE, SITE, ONE, HOPE, SLOPE, COPE, TAPE, CAPE, BREATHE, CUBE, TUBE, USE, HUGE, RAGE. -- J Vaseekhar Manuel, orcontactme@gmail.com

The magic alphabet is ‘e’. The answers are: AT/ATE, HAT/HATE, MAT/ ATE, MAD/MADE, SCRAP/SCRAPE, PET/PETE, BIT/BITE, RID/RIDE, STRIP/STRIPE,  SLIM/SLIME PIN/PINE, QUIT/QUITE,SIT/SITE, ON/ONE, HOP/ HOPE, SLOP/SLOPE, COP/COPE, TAP/TAPE, CAP/CAPE, BREATH/BREATHE, CUB/CUBE, TUB/TUBE, US/USE, HUG/HUGE, RAG/RAGE..PS::The silent ‘e’ in the above words is called the ‘marker’.

It is not pronounced in any of these words and changes the meaning and pronunciation. It also makes the ‘g’ soft in the above examples. I really enjoyed doing this challenge!! Thanks and Regards, Paripreet Kaur (10 years), d/o Charanjit Singh Pardesi, cspardesi@gmail.com (Yes, but what if I had given the target word as PAT? -- MS)(The other one was: “Three chess players agree to play a series of games for a prize which would go to the first player to win two consecutive games. The players draw lots to see which two shall play the first game and from then on the winner plays against the person who just sat out. Assuming the three players to have exactly equal skills, what are their respective chances?”)

If we consider the chances before lots are drawn, it is simply 1/3 for each player. The winning probability of the two friends who play first is equal and will be (1/2^2 + 1/2^5 + 1/2^8 + ....) + (1/2^4 + 1/2^7 +...) = 5/14. The winning probability of the third friend who sits idle in the first game is 2*(1/2^3 + 1/2^6 +....) = 2/7. -- Saifuddin S F Khomosi, DubaiSay A and B play the first game and C sits out waiting. Logically A and B have equal chances of winning the prize because skills and all conditions are the same for both.

C in his first game will be playing against a winner. So he has to win the first game to have any chance of winning the prize. Chance of losing the first game and still have the chance of winning the prize is available to A and B but not to C. Hence A and B have equal chances but better than C to win the prize.-- Raghavendra Rao Hebbani, rao.raghavendrah@gmail.com

Assume the correct time is T. T can be before 11:55, between 11:55 and 12:25, between 12:25 and 13:05, or after 13:05. Since the average timeouts is given as 30 min, a total of the timeouts = 90 min. If T is before 11:55, we have (11:55-T) + (12:25-T) + (13:05-T) = 90 min = 1 hr30min ; 3T = 11:55 + 12:25 + 13:05 -1:30 = 35:55; T = 11:58:20; This violates the assumption that T is before 11:55; So 11:58:20 can not be the correct time. if T is between 11:55 and 12:25, we have (T-11:55) + (12:25-T) + (13:05-T) = 1:30; T = -11:55+12:25+13:05-1:30 = 12:05; So 12:05 can be the correct time. if T is between 12:25 and 13:05 we have (T-11:55) + (T-12:25) + (13:05-T) = 1:30; T = 11:55+12:25-13:05+1:30 = 12:45;  Here timeouts are 50 min, 20 min and 20 min. This violates the condition that all three timeouts are different. So 12:45 cannot be the correct time. if T is after 13:05, we have (T-11:55) + (T-12:25) + (T-13:05) = 1:30min ; 3T = 11:55 + 12:25 + 13:05 +1:30 = 38:55; T = 12:58:20.  This violates the assumption that T is after13:05; So 12:58:20 cannot be the correct time. So the only correct solution is 12:05. -- Mohanachandran R, mohanachandran@gmail.com

Granted the log accelerating through water can never exceed 100 degrees C due to latent heat of vaporisation, what about the resultant envelop of steam surrounding the log? Can’t that temperature be high enough to leave scorch marks on the wood’s surface? (Submitted by Dhruv Narayan, dhruv510@gmail.com)

Sharma is a scriptwriter and former editor of Science Today magazine.              (mukul.mindsport@gmail.com)

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