Menu Close

Suggest-minimum-number-of-weights-by-which-we-can-weigh-upto-40-kgs-in-whole-numbers-in-a-traditional-balance-If-possible-mention-the-process-also-

Tinku Tara 0 Comments
Pin




Question Number 3143 by Rasheed Soomro last updated on 06/Dec/15
Suggest minimum number of weights by which we can weigh upto 40 kgs(in whole numbers) in a traditional balance. If possible mention the process also.
$$\mathcal{S}{uggest}\:{minimum}\:{number}\:{of}\:{weights}\:{by}\:{which}\:{we}\:\:{can}\: \\ $$$${weigh}\:\:{upto}\:\mathrm{40}\:{kgs}\left({in}\:{whole}\:{numbers}\right)\:{in}\:{a}\:{traditional}\:{balance}. \\ $$$${If}\:{possible}\:{mention}\:{the}\:{process}\:{also}. \\ $$
Commented by Rasheed Soomro last updated on 06/Dec/15
In a traditional balance we can place weights on both sides.
$$\mathcal{I}{n}\:{a}\:{traditional}\:{balance}\:{we}\:{can}\:{place}\:{weights}\:{on}\:{both} \\ $$$${sides}. \\ $$
Answered by prakash jain last updated on 06/Dec/15
You start with 1kg For 2 kg select 1+2=3kg − you get 1 to 4 for 5 kg select 4+5=9kg you get 1to 13kgs for 14kg select 13+14=27kg you get 1 to 40. so you need following weight 1,3,9,27 to weight all upto 40kg (in whole numbers).
$$\mathrm{You}\:\mathrm{start}\:\mathrm{with}\:\mathrm{1kg} \\ $$$$\mathrm{For}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{select}\:\mathrm{1}+\mathrm{2}=\mathrm{3kg}\:−\:\mathrm{you}\:\mathrm{get}\:\mathrm{1}\:\mathrm{to}\:\mathrm{4} \\ $$$$\mathrm{for}\:\mathrm{5}\:\mathrm{kg}\:\mathrm{select}\:\mathrm{4}+\mathrm{5}=\mathrm{9kg}\:\mathrm{you}\:\mathrm{get}\:\mathrm{1to}\:\mathrm{13kgs} \\ $$$$\mathrm{for}\:\mathrm{14kg}\:\mathrm{select}\:\mathrm{13}+\mathrm{14}=\mathrm{27kg}\:\mathrm{you}\:\mathrm{get}\:\mathrm{1}\:\mathrm{to}\:\mathrm{40}. \\ $$$$\mathrm{so}\:\mathrm{you}\:\mathrm{need}\:\mathrm{following}\:\mathrm{weight} \\ $$$$\mathrm{1},\mathrm{3},\mathrm{9},\mathrm{27}\:\mathrm{to}\:\mathrm{weight}\:\mathrm{all}\:\mathrm{upto}\:\mathrm{40kg}\:\left(\mathrm{in}\:\mathrm{whole}\:\mathrm{numbers}\right). \\ $$
Commented by RasheedAhmad last updated on 06/Dec/15
N^V ice Sir!
$$\overset{\mathcal{V}} {\mathcal{N}}{ice}\:\mathcal{S}{ir}! \\ $$
Commented by RasheedAhmad last updated on 06/Dec/15
^(Rasheed Soomro) All are powers of 3. Next power is 81 Five weights(1,3,9,27,81) can weigh upto 1+3+9+27+81=121 kg. n weights 3^0 ,3^1 ,3^2 ,...3^(n−1) can weigh upto 3^0 +3^1 +3^2 +...+3^(n−1) =((3^n −1)/2)
$$\:^{{Rasheed}\:{Soomro}} \\ $$$$\mathcal{A}{ll}\:{are}\:{powers}\:{of}\:\mathrm{3}. \\ $$$$\mathcal{N}{ext}\:{power}\:{is}\:\mathrm{81} \\ $$$$\mathcal{F}{ive}\:{weights}\left(\mathrm{1},\mathrm{3},\mathrm{9},\mathrm{27},\mathrm{81}\right)\:{can} \\ $$$${weigh}\:{upto}\:\mathrm{1}+\mathrm{3}+\mathrm{9}+\mathrm{27}+\mathrm{81}=\mathrm{121}\:\mathrm{kg}. \\ $$$${n}\:{weights}\:\mathrm{3}^{\mathrm{0}} ,\mathrm{3}^{\mathrm{1}} ,\mathrm{3}^{\mathrm{2}} ,…\mathrm{3}^{{n}−\mathrm{1}} \:{can}\:{weigh} \\ $$$${upto}\:\mathrm{3}^{\mathrm{0}} +\mathrm{3}^{\mathrm{1}} +\mathrm{3}^{\mathrm{2}} +…+\mathrm{3}^{{n}−\mathrm{1}} \\ $$$$=\frac{\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{2}}\: \\ $$
Commented by prakash jain last updated on 06/Dec/15
Yes.
$$\mathrm{Yes}. \\ $$
Commented by Yozzi last updated on 06/Dec/15
Why is your answer correct? Why start with 1 kg,then 2kg, ...? I don′t understand your steps. (not doubting it′s wrong; I want to understand the idea behind the question.)
$${Why}\:{is}\:{your}\:{answer}\:{correct}?\:{Why} \\ $$$${start}\:{with}\:\mathrm{1}\:{kg},{then}\:\mathrm{2}{kg},\:…? \\ $$$${I}\:{don}'{t}\:{understand}\:{your}\:{steps}. \\ $$$$\left({not}\:{doubting}\:{it}'{s}\:{wrong};\:{I}\:{want}\:{to}\right. \\ $$$${understand}\:{the}\:{idea}\:{behind}\:{the} \\ $$$$\left.{question}.\right) \\ $$
Commented by prakash jain last updated on 06/Dec/15
With 1 kg you can weigh up to 1kg. To weigh 2 kg the maximum weight that we can use is 3kg. This gives the maximum range to 4 kg. 1, 3−1,3, 3+1. Similarly we use the largest possible weight every time to get the max range so that minimum number of weights are used. Since we are able to weigh 1 to 4 (with 1and3) adding 9kg will allow addition range (9−4) to (9+4). and so on. 1=1 2=3−1 3=3 4=3+1 5=9−4=9−(3+1) 6=9−3 7=9−2=9−(3−1)=9+1−3 8=9−1 9=9 10=9+1 11=9+3−1 12=9+3 13=9+3+1 When you add 27 you can get 27−(1 to 13) to 27+(1 to 13). Since 1 to 13 can be contructed without using 27kg weight. Since we can put weights on both sides of balancr subtraction is allowed.
$$\mathrm{With}\:\mathrm{1}\:\mathrm{kg}\:\mathrm{you}\:\mathrm{can}\:\mathrm{weigh}\:\mathrm{up}\:\mathrm{to}\:\mathrm{1kg}. \\ $$$$\mathrm{To}\:\mathrm{weigh}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{weight} \\ $$$$\mathrm{that}\:\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{is}\:\mathrm{3kg}.\:\mathrm{This}\:\mathrm{gives}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{range}\:\mathrm{to}\:\mathrm{4}\:\mathrm{kg}.\:\mathrm{1},\:\mathrm{3}−\mathrm{1},\mathrm{3},\:\mathrm{3}+\mathrm{1}. \\ $$$$\mathrm{Similarly}\:\mathrm{we}\:\mathrm{use}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{possible} \\ $$$$\mathrm{weight}\:\mathrm{every}\:\mathrm{time}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{max}\:\mathrm{range}\:\mathrm{so} \\ $$$$\mathrm{that}\:\mathrm{minimum}\:\mathrm{number}\:\mathrm{of}\:\mathrm{weights}\:\mathrm{are}\:\mathrm{used}. \\ $$$$\mathrm{Since}\:\mathrm{we}\:\mathrm{are}\:\mathrm{able}\:\mathrm{to}\:\mathrm{weigh}\:\mathrm{1}\:\mathrm{to}\:\mathrm{4}\:\left(\mathrm{with}\:\mathrm{1and3}\right) \\ $$$$\mathrm{adding}\:\mathrm{9kg}\:\mathrm{will}\:\mathrm{allow}\:\mathrm{addition}\:\mathrm{range} \\ $$$$\left(\mathrm{9}−\mathrm{4}\right)\:\mathrm{to}\:\left(\mathrm{9}+\mathrm{4}\right). \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{on}. \\ $$$$\mathrm{1}=\mathrm{1} \\ $$$$\mathrm{2}=\mathrm{3}−\mathrm{1} \\ $$$$\mathrm{3}=\mathrm{3} \\ $$$$\mathrm{4}=\mathrm{3}+\mathrm{1} \\ $$$$\mathrm{5}=\mathrm{9}−\mathrm{4}=\mathrm{9}−\left(\mathrm{3}+\mathrm{1}\right) \\ $$$$\mathrm{6}=\mathrm{9}−\mathrm{3} \\ $$$$\mathrm{7}=\mathrm{9}−\mathrm{2}=\mathrm{9}−\left(\mathrm{3}−\mathrm{1}\right)=\mathrm{9}+\mathrm{1}−\mathrm{3} \\ $$$$\mathrm{8}=\mathrm{9}−\mathrm{1} \\ $$$$\mathrm{9}=\mathrm{9} \\ $$$$\mathrm{10}=\mathrm{9}+\mathrm{1} \\ $$$$\mathrm{11}=\mathrm{9}+\mathrm{3}−\mathrm{1} \\ $$$$\mathrm{12}=\mathrm{9}+\mathrm{3} \\ $$$$\mathrm{13}=\mathrm{9}+\mathrm{3}+\mathrm{1} \\ $$$$\mathrm{When}\:\mathrm{you}\:\mathrm{add}\:\mathrm{27}\:\mathrm{you}\:\mathrm{can}\:\mathrm{get}\:\mathrm{27}−\left(\mathrm{1}\:\mathrm{to}\:\mathrm{13}\right)\:\mathrm{to} \\ $$$$\mathrm{27}+\left(\mathrm{1}\:\mathrm{to}\:\mathrm{13}\right).\:\mathrm{Since}\:\mathrm{1}\:\mathrm{to}\:\mathrm{13}\:\mathrm{can}\:\mathrm{be}\:\mathrm{contructed} \\ $$$$\mathrm{without}\:\mathrm{using}\:\mathrm{27kg}\:\mathrm{weight}. \\ $$$$\mathrm{Since}\:\mathrm{we}\:\mathrm{can}\:\mathrm{put}\:\mathrm{weights}\:\mathrm{on}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{of} \\ $$$$\mathrm{balancr}\:\mathrm{subtraction}\:\mathrm{is}\:\mathrm{allowed}. \\ $$
Commented by prakash jain last updated on 07/Dec/15
Suppose you have n weight. The possible weights that you can create (including −ve and duplicates) are. choose 1 and put a +/− sign before it=2^n C_1 choose 2 and put a +/− sign before each=2^2 ^n C_2 and so on So total number of sum that you can obtain with n +ve numbers are=Σ_(i=1) ^n 2^i ^n C_i (1+2)^n =^n C_0 +2∙^n C_1 +2^2 ^n C_2 +...+2^n ^n C_n 3^n =1+Σ_(i=1) ^n 2^i ^n C_i Σ_(i=1) ^n 2^i ^n C_i =3^n −1 with 3 +ve number you can create=3^3 −1=26 half of them will be −ve so 13 +ve number. with 4 +ve number you can create 3^4 −1=80 or 40 +ve numbers. This proves that you need a minimum of 4 weights to create 40 +ve results. Solving the question is about finding 4 numbers so the sums are 1 to 40.
$$\mathrm{Suppose}\:\mathrm{you}\:\mathrm{have}\:{n}\:\mathrm{weight}.\:\mathrm{The}\:\mathrm{possible} \\ $$$$\mathrm{weights}\:\mathrm{that}\:\mathrm{you}\:\mathrm{can}\:\mathrm{create}\:\left(\mathrm{including}\right. \\ $$$$\left.−\mathrm{ve}\:\mathrm{and}\:\mathrm{duplicates}\right)\:\mathrm{are}. \\ $$$$\mathrm{choose}\:\mathrm{1}\:\mathrm{and}\:\mathrm{put}\:\mathrm{a}\:+/−\:\mathrm{sign}\:\mathrm{before}\:\mathrm{it}=\mathrm{2}\:^{{n}} {C}_{\mathrm{1}} \\ $$$$\mathrm{choose}\:\mathrm{2}\:\mathrm{and}\:\mathrm{put}\:\mathrm{a}\:+/−\:\mathrm{sign}\:\mathrm{before}\:\mathrm{each}=\mathrm{2}^{\mathrm{2}} \:^{{n}} {C}_{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{on}\: \\ $$$$\mathrm{So}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{sum}\:\mathrm{that}\:\mathrm{you}\:\mathrm{can} \\ $$$$\mathrm{obtain}\:\mathrm{with}\:{n}\:+{ve}\:\mathrm{numbers}\:\mathrm{are}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{i}} \:^{{n}} {C}_{{i}} \\ $$$$\left(\mathrm{1}+\mathrm{2}\right)^{{n}} =\:^{{n}} {C}_{\mathrm{0}} +\mathrm{2}\centerdot\:^{{n}} {C}_{\mathrm{1}} +\mathrm{2}^{\mathrm{2}} \:^{{n}} {C}_{\mathrm{2}} +…+\mathrm{2}^{{n}} \:^{{n}} {C}_{{n}} \\ $$$$\mathrm{3}^{{n}} =\mathrm{1}+\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{2}^{{i}} \:^{{n}} {C}_{{i}} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{2}^{{i}} \:^{{n}} {C}_{{i}} =\mathrm{3}^{{n}} −\mathrm{1} \\ $$$${with}\:\mathrm{3}\:+\mathrm{ve}\:{number}\:{you}\:{can}\:{create}=\mathrm{3}^{\mathrm{3}} −\mathrm{1}=\mathrm{26}\: \\ $$$$\mathrm{half}\:\mathrm{of}\:\mathrm{them}\:\mathrm{will}\:\mathrm{be}\:−\mathrm{ve}\:\mathrm{so}\:\mathrm{13}\:+\mathrm{ve}\:\mathrm{number}. \\ $$$$\mathrm{with}\:\mathrm{4}\:+\mathrm{ve}\:\mathrm{number}\:\mathrm{you}\:\mathrm{can}\:\mathrm{create}\:\mathrm{3}^{\mathrm{4}} −\mathrm{1}=\mathrm{80} \\ $$$$\mathrm{or}\:\mathrm{40}\:+\mathrm{ve}\:\mathrm{numbers}. \\ $$$$\mathrm{This}\:\mathrm{proves}\:\mathrm{that}\:\mathrm{you}\:\mathrm{need}\:\mathrm{a}\:\mathrm{minimum}\:\mathrm{of} \\ $$$$\mathrm{4}\:\mathrm{weights}\:\mathrm{to}\:\mathrm{create}\:\mathrm{40}\:+\mathrm{ve}\:\mathrm{results}. \\ $$$$\mathrm{Solving}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{about}\:\mathrm{finding}\:\mathrm{4}\:\mathrm{numbers} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{sums}\:\mathrm{are}\:\mathrm{1}\:\mathrm{to}\:\mathrm{40}. \\ $$
Commented by Yozzi last updated on 06/Dec/15
Ah, I now understand. It′s the thought behind it that I wanted to grasp.
$${Ah},\:{I}\:{now}\:{understand}.\:{It}'{s}\:{the}\:{thought} \\ $$$${behind}\:{it}\:{that}\:{I}\:{wanted}\:{to}\:{grasp}.\: \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *