Q.62.

[Manish Salian]

Hi,

Since there are not many response to Saturday's question, I will extend its deadline by another day! So Monday, midnight is the deadline of that. . .

Surprisingly, there aren't any responses for Sunday's verbal question as well! I guess Sunday is not a good day to visit forum...or is it that everyone is afraid of negative marks? If latter is the case then it is fine as that is exactly what I propose you should be doing in CAT!

But, come on! Be a sport like Anil and Rishabh and start replying....

Here is the question for today:

Q.62.

I have a grocery shop! One day Anil comes to my shop and gives me a chain that has exactly 121 links. He says, the total weight of the link is 121 gm., with each link weighing 1 gm. He asks me to cut the links into smaller pieces so as to make weights out of them! He tells me that the weights should be in such denominations that I should be able to weigh any weight from 1 gm. to 121 gm. using my simple balance (with two pans).

“That’s 121 cuts!”, I tell myself! However, just when I am about to sever the first link, I realize that the number of links that I have to cut is far smaller than what I had imagined? Can you tell me what is the minimum number of links I must cut so as to satisfy the condition specified by Anil?

Be absolutely sure before you answer!

Deadline for Answer: Tuesday midnight!

Regards,

Manish sir

[AnilkumarBDaga]

Hello Sir,

The minimum number of cuts to satisfy the condition is 5. Hence we will have 6 sets containing 1, 3, 8, 21, 55 and 33 links. Now either by placing them in a single pan or different pans we can get all weights upto 121 grams.

117 = 33+55+21+8
101 = 55+33+21-8
50 = 33+21-3-1

etc..... etc..........

Thanks & Regards,
Anil Daga

[MinnatLalpuria]

hello sir

hav been a lil busy wid engg studies ..as u know.. n a lil inactive.. but i cldnt let go such an easy question

Answer : 5 links of weights 1,3,9,27,81 gms each

SInce thr r two pans hence i can use weights on both d pans inorder to obtain a certain weights... hence they have to b in powers of 3
The max weight i need is 121 gm
Hence the links hav to b of
1 , 3 , 9, 27 , 81 (all thsi totals upto 121 gm exactly)
all weights from 1 to 121 can b obtained
eg : for weighing 5 gms keep 9 gms on one pan and 1+3 gms on the other and so on....

hopefully no negative markings for this

although anil also answered as 5 but our links weigh differently.. now wat sir??


minnat

[MinnatLalpuria]

by the way anil..with those values how do i measure 101 gms yaar??

[PiyushTantia]

Well, people this is my 1st reply on the forum and after analysing both the answers of Minnat and Anil I go with Anil.

Minnat, don't take offense but i think he's already showed how can one measure 101 gms.

Piyush

[Manish Salian]

Hi Anil and Minnat,

Unfortunately, both of you are wrong this time! And Piyush, was only giving an 'outside support' to Anil, so doesn't lose any marks! But such tactics won't fetch you marks either (in case you are right that is!).

By the way, though it seemed that both of you were in agreement (Anil and Minnat), both of you gave different answers. Anil said, it required 6 links (hence 5 cuts), while Minnat you said 5 links (hence 4 cuts). So, Minnat, your answer was a shade better than Anil's answer!

The said denominations can be achieved by only 3 cuts! So both of you presented the right logic, but gave the wrong answer!

You need to cut the 10th link, 38th link and the 120th link...that is it!

Wht this will ensure is, you have 6 different individual links weighing 9 gms, 1 gm, 27 gms, 1 gm, 81 gms and 6 gm, respectively in that order.

You will be now able to weigh all weights from 1 to 121 gms. using both sides of the balance (the logic is correctly provided by the two of you).

So Anil, you lose 5 marks, Minnat you lose 2 marks.

Regards,

Manish sir

[Manish Salian]

Hi there,

Just a typrographic error in my solution! Don't worry the answer remains the same i.e. 3 cuts, but the last weight is 2 gms. and not 6 gms. as mentioned. Thus, the statement should read:

You need to cut the 10th link, 38th link and the 120th link...that is it!

Wht this will ensure is, you have 6 different individual links weighing 9 gms, 1 gm, 27 gms, 1 gm, 81 gms and 2 gm, respectively in that order.

Regards,

Manish sir