We have an equal arms two pan balance and need to weigh objects with integral weights in the range 1

Option a) 4 → three std weights of 2 kg each and one std wt. of 5 kg

We just need to factorize the magnitude of the highest weight, i.e., 40.

40 = 2x2x2x5

to weigh 1kg → 2x2 kg in one pan and 5 on the other

to weigh 3 kgs → 2 kg in one pan and 5 kg on the other, and so on.
8 years agoHelpfull: Yes(36) No(13)

1,3,9,27..so 4

if both pan can be used for wights then 3^0,3^1,3^2,3^3,3^4...... weights are used if single pan can be used for weights then 2^0,2^1,2^2,2^3,2^4
9 years agoHelpfull: Yes(34) No(13)
We require minimum no. of weights to measure all the integral weights from (1-40)

i.e. min. no of weights whose addition or subtraction would result in a value from [1,40]

⇒ in order to measure 1 you need to have weight 1

Now you have one weight of 1Kg

⇒ for 2 you got choices i.e. either one more weight of 1 kg or a new weight of 2 kg or a new weight of 3 kg.

i.e. 1+1 =2 or 2=2 or 3-1 =2

So either of 1, 2, 3

For 3 you got choose among the choices before because you have to choose minimum weights

for 1 you can't have 3

for 2 you can have 3 i.e. 2+1 =3

for 3 you can have a single weight of 3 kg

so either 2 or 3

now for 4

1+2 will not result in 4

but 1+3 =4

so next weight would be of 3 kg

and hence you will be able to measure 1,2,3 & 4 with the help of 2 weights i.e 1 & 3

Now for 5 you will go for a big no. whose subtraction from 4 would result in 5

i.e. x - 4 =5 ⇒ x =9

with third weight will be 9

and hence you will be able to measure from 1 to 13 (i.e. 9+4) with the help of 3 weights i.e 1, 3 & 9

now for 14

y -13 =14

⇒ y =27

Hence 4th weight will be 27

and hence you will be able to measure from 1 to 40 with the help of 4 weights

i.e 1, 3, 9 & 27
6 years agoHelpfull: Yes(3) No(0)
@ARUSHIKHA_MISHRA can you plz explain how did you found out the series because it will be useful for us in the future
7 years agoHelpfull: Yes(0) No(2)

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