Ram

• Oct 4th, 2005

 

if n=3 then the answer is 4 chances

vicky_boyinoasis

 Profile Answers by vicky_boyinoasis 

• Mar 31st, 2006

 

for n=3first put five boxes in each side of balance....the side which goes lower is sure to contain the box having the lighter ball......weigh 1divide those five into 2-2-1 group......weigh putting any 2-2 on each side of balance.....if equal weight results,the last one is containing the lighter ball,elsethe side weighing less in the just performed weigh contains the lighter ball...weigh 2having found the box containing the 90 gms. ball.....divide the balls in 1-1-1 groups....repeat the previos box method with ball in place instead....weigh 3 so total 3 weighs

Guest

• Jul 17th, 2006

 

n = 3  => chances = 6

n = 10 => chances = 13

n = 9  => chances = 12

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dheepti

• Sep 15th, 2006

 

to me

the chances are

when n=3

(i) nC1= 3C1 =3 for 10 boxes .. 10*3=30

(ii) 10C1=10 for 10 boxes ....10*10=100

(iii)9C1=9 for 10 boxes .....10*9=90

mushez

• Apr 22nd, 2007

 

According to me its n=3

---> 6 chances.
first keep 5 boxes in each in a balance.
then the 1 with low weight.
take it divide 5 boxes as 2-2-1.
keep 2-2 in each of balance.
if both weigh same..
then take the one left. in which we have 3 balls.1-1-1
keep any 2 balls in balance.
if both weigh the same. the one left is of low weight..!
else if both don't weigh the same.... then the one with low weight is the one
req.
as such is for n=10.

Krithikaa.K.R

• May 7th, 2007

 

The minimum no. of chances is three.
The boxes are made into three groups:

g1: 1 2 3
g2: 4 5 6
g3: 7 8 9 10

ch 1:
g1, g2

outcome 1: g1
outcome 2: g1>g2
outcome 3: g1 =g2 (g3 has 90 gms ball)

ch2:

if outcome 1 in ch1 then
1,2
outcome 1: 1<2
outcome 2: 1>2
outcome 3: 1=2 (3 has 90 gm ball)

if outcome 2 in ch1 then
4,5
outcome 1: 4<5
outcome 2: 4>5
outcome 3: 4=5 (6 has 90 gm ball)

if outcome 3 in ch1 then 
7 8,9 10
outcome 1: 7 8< 9 10
outcome 2: 7 8>9 10

ch :3

if outcome 1 and 2 in ch1 and any of the outcomes wrt to them in ch2
then take two balls from the box and weigh them .the lesser weighing ball is the
90 gms one. if both are equal then the remaining one is the 90 gms ball.

if outcome 3 in ch1 and outcome 1 wrt to it in ch2  7,8
outcome 1: 7>8
outcome 2 :

 7<8
else
9,10

outcome 1 :9>10
outcome 2: 9<10

ch:4(optional ie outcome 3 in ch1)

take two balls from the box and weigh them .the lesser weighing ball is the
90 gms one. if both are equal then the remaining one is the 90 gms ball.

muddu krishna

• May 12th, 2007

 

If n=3 then answer is:1
 
 Take two balls and weigh. If both are equal then remaning ball is 90 gms
else if second is less than first one then second ball is 90gms
else if first is less than second on then first ball is 90 gms

If n=10 answer is: 3

Divide the n=10 into two 5-sub groups

compare the both the groups

If 1group<2group then 90 gms ball is present in the first group
else it is present in the second group.

Take the less weight group and divide it into 2-2-1 subgroup.
Now compare the 2-subgroups first. If first 2-subgroup is less than the second 2-subgroup then 90gms is present in the first 2-subgroup otherwise it is present in the second 2-subgroup. If both are equal then the remaing 1-subgroup ball is 90gms ball.

Now take less weigth subgroup of 2-subgroups. Now compare the two balls that are existing in the 2-subgroup group. Then we can find the 90 gms ball.

So we require 3 chances for detection of 90 gms ball when n=10

If n=9 then answer is :3

chandrasekar

• Jun 13th, 2007

 

chances 7

vaidees

• Jun 14th, 2007

 

For all the three options
The answer is 3

Bhavya Shah

• Jun 26th, 2007

 

For N = 3
First > Balance 5 : 5 boxes
Second > Balance lover weight side and reduce 1 box mean
Lower weight boxes 2 : 2 and remain 1 box as it is ..if that 2 : 2 are equal then remain 1 box has 90gm ball.
So,
A. Compare that 1 : 1 ball and remain 1 as it is. You have to find result of 90gm ball.
Third > Find out lower weight side and compare 1 : 1
Fourth > In lower weight box compare ball to 1 : 1 and u will find lower weight 90 gm ball. if with comparison it not found then remaining 1 is lower weight ball.

chaitanyabijur

• Jul 30th, 2007

 

total of 30 balls.
now 3^3 < 30 < 3^4
so  4 attempts
the base is 3 since from 3 objects the ddefective can be determined in one attempt.
so alwasys divide the ball into  nearly 3 parts and compare the two of them which r equal

KIRANMAYEE

• Sep 2nd, 2007

 

My answer is
1) 30 chances, 2) 100 chances, 3) 90 chances.
These answer is solved by probability formula nc1
E.g, n=3--nc1=3c1=3, and total boxes are 10. So 3*10=30

sarat_619

 Profile Answers by sarat_619 

• May 18th, 2008

 

30,100,90 times

shekhar mehta

 Profile Answers by shekhar mehta 

• Sep 2nd, 2010

 

Answer is 3
1st divide boxes into 5 5
then make choice which one is
heavy, neglect that bunch of boxes!

2nd now divide lighter group of boxes into 2 2 1
by this we will get which box is lightest.. then get the 3 balls of the lightest box;

3rd measure 1 ball with 1 ball
You will get which one among is lightest!