# Clocking the watch... as time waits for no one!

At the precise instant when Mr H drove off from his residence in his car at a constant speed of 55 kmph, the minute hand of his 12 hour analogue wristwatch was exactly located on a minute mark.

Published: 12th August 2018 05:00 AM | Last Updated: 11th August 2018 06:48 PM | A+A A-

At the precise instant when Mr H drove off from his residence in his car at a constant speed of 55 kmph, the minute hand of his 12 hour analogue wristwatch was exactly located on a minute mark. After two kilometres of travel, Mr H stopped by a petrol pump to refuel when he observed that the minute hand and the hour hand of his wristwatch were exactly coincident with the minute hand not precisely situated on a minute mark. At what time did Mr H leave his residence?

THROUGHPUT

(The problem past its sell-by date was: “If New York and San Francisco seven days’ journey apart, and if trains start from both ends every day at noon, how many trains coming in an opposite direction will a train leaving NY meet before it arrives at its destination at SF?”)

The moment a train starts from New York there are six trains already on the way. Train #7 also starts at the same time. So it has to meet all these seven. But when the NY train is at point 1, an eighth train will be leaving San Francisco; when it is at point 2 a ninth train will be leaving; at point 3 a tenth; at point 4 an eleventh; at point 5 a twelfth; and at point 6 a thirteenth. Therefore it will meet 13 others on the journey. -- Dhruv Narayan, dhruv510@gmail.com

NY and SF are in different time zones. Thus at the instant we start, the intermediate trains won’t precisely occupy the numbered positions (as described above) but will be offset a bit. That changes the answer to 14! In effect, the “train that will be just starting” will be fairly out of the station when we arrive (if only just) and thus counts as an extra encounter. -- Vimala S, cryptoverse@gmail.com

(The second one was: “Four missiles are at the corners of a square of 20 kms sides. All are launched simultaneously with each homing in on the one on its left at 1 km per second. How long before they meet?”)

The ballistic missile on the left will always move perpendicular to the line between it and the missile homing in on it. It will not be moving away from the chasing missile. In fact, the chasing missile moves towards it at 1 km/s. Therefore, the distance between them decreases at 1 km/s. Thus, the time it takes for them to meet is 20/1 = 20 sec. -- Saifuddin S F Khomosi, Dubai

Time taken = 20 seconds. When they are fired the missile that is chased is leaning towards the missile chasing it. In effect both move such that the perpendicular distance chasing between them is reducing at the speed of travel and the chased missile is at standstill for the missile. Hence the time for all to meet is initial distance 20 km divided by speed 1 km per second. -- Raghavendra Rao Hebbani, rao.raghavendrah@gmail.com

(The third one was: “Know a 10-letter words in English that still remain a valid word even when each of their letters is successively removed?”)

This had me tearing my hair and hence I had SPLITTINGS. Removing one letter at a time from this 10-letter word we get: SLITTINGS, SITTINGS, SITTING, SITING, STING, SING, SIN, IN and finally left me brooding over myself (I). -- Ramakrishna Easwaran, drrke12@gmail.com

SPLITTINGS!. Remove S – SPLITTING; remove L – SPITTING; remove P – SITTING; remove T – SITING; remove I – STING; remove T – SING; remove G – SIN; remove S – IN; remove N – I. -- Rekha G, rekhapai@gmail.com(And now another Groucho Marx is back.)

Re: The pizza problem solution given by V K Bargah correctly states that the sum of the four slices must be 18/19. But, the first solution given by him: 1/76 + 1/171 + 1/228 + 1/342 adds up to 18/684 (or 1/36) only. The second solution is worse (one of the denominators of the fractions 522 is not a multiple of 19) and the fractions do not add up to 18/19. -- Dr P Gnanaseharan, gnanam.chithrabanu@gmail.com (Hey VKB, I think you have some fessing up to do. – MS)

BUT GOOGLE THIS NOW

1. Solve the alphabetical addition to replace the alphabets with their numerical values. C R O S S + R O A D S = D A N G E R. (Submitted by Balagopalan Nair. K, balagopalannair@gmail.com)

2. Aristotle used to teach that heavy things fall faster than light ones. In 1638, Galileo realised Ari was wrong even before conducting a single experiment himself. How?

Sharma is a scriptwriter and former editor of Science Today magazine.

(mukul.mindsport@gmail.com)