The One Where Technothlon Unpacks

Yeah! The Technothlon team has arrived the beautiful campus of IIT Guwahati. As the team members unpacked their bags (and ideas) to start up the work for the Mains, a short delay in this discussion might have depressed you. But now we are back with the next set of solutions, and some more questions. Tomorrow we will be discussing the section, "Ingenioso Matematica".

"Work In Progress"

 

Q-1:

            The basic idea to minimize the rope being used is to place the barricades such that they are equidistant from the ends to which they would get connected. Every time a new pillar is to be placed it's placed in the mid of the 2 existing pillars.

            So, first he places the barricade in the middle of the Left and Right Pillars, using 8m rope. Then he places the next barricade in the middle of this barricade and the Left Pillar, and then between this(middle) barricade and the Right pillar, using 4+4=8m rope (The order to place these two barricades can be interchanged).

Then rest of the barricades can be placed in any order, using a total of 2+2+2+2=8m of rope.

Hence the total rope used is:
8+2*(4+2+2) = 24m rope

8+2*(4+2+2) = 24m rope

Q-2:

            Since he has enough funds he doesn't have to strategies things out, and he will directly arrange the barricades one by one. The rope used for the first barricade is 8m, for the second is 7m, for the third is 6m, ..... for the seventh is 2m.
Hence the total rope used is:

8+7+6+5+4+3+2=35m rope.

Q-3:

(There was a typo in the question, and the correct question should have been:

            If he applies to another such stretch, 10m long, and with 9 barricades. Then            what would be his minimum wastage? (in case of 57 m long rope))

For this correct question:

            Check for the total rope required as in case of no bar on the budget and length of rope (similar to Ques-2). Following the procedure above, it would come out to be 54m.

Hence the minimum wastage would be of 57-54 = 3m.

(Add more to your knowledge: One might interpret that minimum wastage means the worst case as in Ques-2. Be cautious!

            You must check out the length given. In this question, if the total available length had been less than 54m (say 53m) then to solve the question it would have required to find an arrangement such that it uses length less than but nearest to  53 m)

Answers:

Question-1: option-(4) 24

Question-2: option-(3) 35

Question-3: option-(1) 3

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"Dcrypt D'crypto"

Ans-1: khulja sim sim
Reason: The clue for it lies in the words "wait for a second" which means we have to select the second letter from each of the words.

Since from here onwards all the questions are to be sequentially answered, i.e. the answer to a particular question will give the instruction for the next question, so the answer to all the questions must be some instruction except for the last one. (If you pay attention, you can cleverly decipher some questions directly too)

Ans-2: follow (the) arrow
Reason: Deciphered like Think outside the box, and as a hint the next question has arrow in the inscriptions.

Ans-3: next step swap even rows and read
Reason: Find the letters of "TECHNO" in the diamond. Now pick up only those letters written around these in the diamond where there is a line around the square boxes, and arrange them by following the arrow. For eg:, around T, pick up n,e,x,t in the ssense of the motion of the arrow around T, and hence 'NEXT'. Similarly for other characters.

Ans-4: divide half up half down and then alternate top & bottom
Reason: Follow the instruction fetched from the answer to the previous question. Swap the characters  in the even rows.

Ans-5: start right and alternate ends
Reason:  Distribute the 26 letters long code into 2 parts and write the second half just below the first. Now Pick first character of the upper code, second from the lower, third from the upper again, and so on.

Ans-6: reverse the abcd like a with z b with y

Reason: Pick one character from the right end and then from the left, then again from the right, and continue the same way (alternating ends).

Ans-7: ideate nurture kindle

Reason: Simply reverse a with z, b with y, and so on. (OR the hint: "Techniche speaks" can be used.)

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"Code Wars"

 

Let's tabulate all the information:

(First line of the Enciphered Text comes after taking into consideration: “DASY VJBYZ YL” and “GIFT DEATH TO”)

Q-1

Start checking for every word(city) one-by-one.

Kanpur, Nagpur, Mumbai could be enciphered to “CTCIBA”. Now reject Kanpur, and Nagpur as 'C' can be deciphered only in one way. Thus, MUMBAI is the only satisfactory choice for this.

Delhi, Patna could be enciphered to “OHYFB”. Now reject Delhi as 'Y' can be deciphered only to 'T'. Patna can be deciphered keeping in consideration all the rules.

Ahmedabad, Bengaluru, Hyderabad could be enciphered to “ZKVWNBIBE”. Now reject Ahmedabad, and Bengaluru (based on similar reasons as above). Thus, HYDERABAD is the only satisfactory choice for this.

All this helps us to fill the second line of the enciphered text row.

Hence the answer is: 3+5+8=16.

Q-2

Simply check for all those characters which can't be enciphered, and add the numbers corresponding to those words(city) which contain at-least one of these characters.

Hence the answer is: 1+2+4+6+7+8=28.

Q-3

Since 'E' can be ciphered in 2 ways: V,E; and for rest of the alphabets there is only one possible option.

Moreover since 'E' comes twice in 'EUROPE' therefore it can be enciphered in total 2*1*1*1*1*2=4 ways.

Answers:

Question-1: option-(4) 16

Question-2: option-(3) 28

Question-3: option-(1) 4

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