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Mathematics 9 Online
OpenStudy (anonymous):

In a factory there are 7 machines that produces tablets(medicines).Out of the7 ,6 of them are produce tablets of exact weight, the other one producestablets that weigh 1gm less. All the machines are numbered from 1 to 7.Inhow many minimum chances (chances refers to the number of times you usethe balance) can you find out the faulty machine if you are given a balancethat displays the weight digitally.

OpenStudy (anonymous):

I came across many questions like this any method

OpenStudy (anonymous):

1

OpenStudy (anonymous):

im not good at these problems :( ask more like the ones last night <.<

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

2

OpenStudy (anonymous):

I don't know whether I am right or Wrong .... But I think if defect in weight is an event independent of Machine used then It should be 1(Answer) as there are seven Machines .

OpenStudy (anonymous):

but i really need this

OpenStudy (anonymous):

these type

OpenStudy (anonymous):

ill post that type ques later

OpenStudy (anonymous):

k

OpenStudy (anonymous):

Tatiana I would Like to know .. How'd u got 2

OpenStudy (anonymous):

i think that we dont know weight of one initially

OpenStudy (anonymous):

is this a 2 sided scale? (like normal?)

OpenStudy (anonymous):

so we have to cross check if thats the one were looking for

OpenStudy (anonymous):

@joe i dont know check attachment

OpenStudy (anonymous):

can you solve any of them

OpenStudy (anonymous):

its digital...so it might only have one place to put stuff on it at a time. im thinking this: pick one of them to hold in your hand. Place 3 on the scale and weight them. Then place the other 3 on the scale. If they are both equal in weight, then the one you have in your hand is the lighter one. if one group weighs less than the other, then the lighter one is in that group (thats 2 uses of the scale so far).

OpenStudy (anonymous):

For the Second Question .. In ur attachment i think the answer is at least 6

OpenStudy (anonymous):

So now you have 3 pills left that might be the lighter one. again, pick one to hold in your hand. and weight the other 2 seperatly. if those 2 are equal, the one in your hand is the lighter one. if not, the lighter one will be....the lighter one lol. (thats another 2 uses) so i think the answer is 4. you would need to use the scale at least 4 times.

OpenStudy (anonymous):

can u explain a little ishaan

OpenStudy (anonymous):

if we had been using a normal 2 sided scale, the answer would be 2, but since this is a digital scale with only one place to measure something at a time, the answer is 4.

OpenStudy (anonymous):

5 tracks 25 horses. at every trial 4 horses would be disqualified ....24 horses should leave ..and 4x6=24

OpenStudy (anonymous):

cant it be like this that u keep 1 pill on the machine note its weight then put another one and note then other one and note and another one and note (thats 3 times) the odd one is lighter????

OpenStudy (anonymous):

sry did 4 times shud be 3

OpenStudy (anonymous):

ishaan we have to find fastest 3

OpenStudy (anonymous):

i'm with joemath on the first question ...tht would be the best way

OpenStudy (anonymous):

sry .. i'l read the question again

OpenStudy (anonymous):

for the first problem - about 8 coins 4-4 (one of them will be lighter) take lighter 4 2-2 (one of them will be lighter) take lighter 2 1-1 take lighter Answer - 3

OpenStudy (anonymous):

i believe thats 4 times problem ( im counting where you put the first pill on the scale).

OpenStudy (anonymous):

and i agree with tatiana on the coin problem, i think thats right.

OpenStudy (anonymous):

no i wrote it 4 times by mistake i sud have written to keep pills 4 times by mistake

OpenStudy (anonymous):

@titina Thank u

OpenStudy (anonymous):

for 2nd 11 maybe

OpenStudy (anonymous):

if u dont have a stopwatch then 11 should be fastest way to take fastest 3 because every time u will need 2 exits ..... so tht sould make 22 horses to exit and 11 trials

OpenStudy (anonymous):

btw, what class are these problems from? there are by far the most interesting problems ive seen in a while <.<

OpenStudy (anonymous):

About pills 1) 3-3 (if equal, so light pill found (in hand)- and realy then answet 1), (if one is lighter, then take lighter 3, go to the 2)) 2) 1-1 (and 1 in hand) (if equal - in hand is lighter) (if one is lighter then other, then we find lighter as well) - and answer 2 In this problem the most problem is question - how many MINIMUM chanses...

OpenStudy (anonymous):

On first question it should be 3 with tatiana on this

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

@tatiana its a digital scale being used in the pill problem, i dont think digital scales are double sided. Usually you can only weight one group at a time. Thats why I say it takes 4. Otherwise you are 100% correct.

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

1st qustion is solved

OpenStudy (anonymous):

may I delete about pills then?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

let it be

OpenStudy (anonymous):

In 2nd one problemsolver there are 25 horses race horses exit horses remain 1 2 3 2 2 3 . . . 11 2 3 the fastest so it should cont. until 22 horses are disqualified so tht should make it 11 if u dont have a stopwatch

OpenStudy (anonymous):

ok thanks very much

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