To measure the weight between 1 to 100 how many minimum weighs are required so that all weight can be measured
First understand its a pan balance here if we want to measure 40 kg then we have two ways
- Put 40 kg in single pan
- Put w1 and w2 weight on two pans where w1-w2=40
Now we need to measure all weights between 1 and 100 so we will start from 1 so we need to measure 1 kg so we need 1 kg weight
now we want to measure 2 kg so we can use a weight of 2 kg but we wan to use minimum weights so we can do that by using a weight of 3kg
as 3-1=2 so after 1 kg we need 3 kgs
now to measure 4 we have 1+3=4 so no extra weight needed now 5,but we will try to get it using differences
9-4=5 so we will instead use a 9 kg weight so now we have 3 weights
1,3,9 kgs now check pattern , we are getting powers of 3 ,the weights needed are 1,3,9,27,81,243,729… and so on
here we want to measure till 100 kg
and we have 1 + 3 + 9 + 27 + 81> = 100 so we need 5 weights only
3^k >= 100 so k =5 will be our answer
Now if spring balance then we have to use only sums because we don’t have two pans there in that case,
we need a 1kg weight then a 2kg weight so 3 could be measure with help of 1+2
so we needed a 4 kg weight ,for 5 we have 4+1 for 6=4+2 and for 7=4+2+1
so we now need an 8 kg weight
now check pattern – 1, 2, 4, 8, 16, 32, 64,……
so we will use power of 2 in case of spring balance
so 2^k > = 100 so k=7 so we need 7 weighing
“IF we are using Pan balance then we will work on power of 3 and if we are working on spring balance then we will work on power of 2. “
Q- What is the minimum number of weighing operations required to measure 63 kg of wheat , if only 1 weight of 1 kg is available
Direct Method- Only 1 weight of 1 kg is available means this is not pan ,this is spring balance
so 2^n > = 63 so n=6
1st time- 1kg
2nd time -2 kg (1 of weight and 1 of wheat)
3rd time-4 kg (1 of weight and 3 of wheat)
4th time-8 kg (1 of weight and 7 of wheat)
5th time 16 kg (1 of weight & 15 of wheat)
6th time-32 kg (1 of weight & 31 of wheat)
so 6 attempt
Given 5 coins out of which one coin is lighter.How many minimum number of weighing are required to figure out the odd coin.
3^k > = N
here minimum value of k will give minimum number of weighing
like here N=5 so 3^k>=5 so min value of k will be 2 so we need minimum 2 weighing
or if N=12 then 3^k>=12 so min k=3 so we need minimum 3
In a set of 400 balls, all balls except one, which is lighter than the rest, are of equal weight. What is the minimum number of weighing required to identify the lighter ball using a two pan balance?
3^n>=400 so n=6
Divide the balls into 3 groups and then weigh among themselves.
With maximum of 6 weighs you would know which ball is faulty.
For 3 Balls Just 1 weighing is required.(keep 1 in each pan and set the 3rd aside)
For 4-9 balls 2 weighing are required.
For 10-27 Balls, 3 weighing are required.
For 28 – 81 Balls, 4 weighing are required and so on.