Latest TCS Aptitude Question SOLUTION: minimum no of weight required to wait from 1-40 kg..

• Placement Papers
• /
• TCS

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40025966) TCS question Keep an EYE

minimum no of weight required to wait from 1-40 kg..

Asked In TCS TargetTCS 2014 (10 years ago)

Unsolved Read Solution (6)

Is this Puzzle helpful?   (9)   (1) Submit Your Solution

Read Solution (6) :

A J V  (10 years ago)

The answer is 4 weights -- 1kg, 3kg, 9kg and 27 kg.

Explanation:
For 1 kg - Put 1 kg on one side of the plate
For 2 kg - Put 3 kg on one side of the plate, 1 kg on the other side of the plate
For 3 kg - Put 3kg on one side of the plate
For 4 kg - Put 3kg and 1 kg on one side of the plate
For 5 kg - Put 9kg on one side of the plate, and put 1kg and 3kg on the other side of the plate
...
Likewise,
For 40 kg - Put 1kg, 3kg, 9kg, 27kg on one side of the plate

Is this solution Helpfull? Yes (34) | No (4)  

manoj chauhan  (10 years ago)

case 1: answer is 4
if weight is allowed to keep both sides of weighing machine then required weight will be 4 like this 1kg,3kg,9kg,27kg.
case 2: answer is 6
if weight is allowed to keep only one side of weighing machine then required weight will be 6 like this 1kg,2kg,4kg,8kg,16kg,32kg

Is this solution Helpfull? Yes (7) | No (1)  

chandni  (10 years ago)

6
on taking the binary of 40, we will get 6 steps

Is this solution Helpfull? Yes (4) | No (3)  

Ashima Gupta  (10 years ago)

Answer is 4. The weights required would be 1,3,9,27.

For each weight, there are three things you can do: put it on the left
pan, the right pan, or not on the balance at all.

So, if you have n weights, there are 3^n things you can do with them.

One of those things is not putting any weights on the scale, which is
good if you want to weigh a 0-pound object, so really there are only
3^n - 1 arrangements.

Then, for each arrangement there's also its mirror image (where all
the weights are switched to the opposite pan of the scale), so there
are at MOST (3^n - 1)/2 arrangements of n weights.

That's enough to prove that 4 weights can weigh at most 40 different
things ... 40 is really the upper limit for 4 weights.

Is this solution Helpfull? Yes (4) | No   

Rahul Karki  (10 years ago)

weights of 1kg,2kg,5kg,10kg and 30 kg will be sufficient to weigh 1 to 40 kg. So minimum no. of weights will be 5.
In the solution given below with 4 weights you cannot weigh 16 kg.

Is this solution Helpfull? Yes (1) | No (1)  

anamika  (10 years ago)

@AJV,ashima gupta- i got it that it can be measured by 1,3,9,27.. but how did u got these values??????

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.