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There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum number of weighings required to identify the heavier bag is
- a)2
- b)3
- c)4
- d)8
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
There are eight bags of rice looking alike, seven of which have equal ...
We will categorize the 8 bags in three groups as :
(i) A1A2A3 (ii) B1B2B3 (iii) C1C<2
Weighting will be done as bellow :
1st weighting → A1A2A3 will be on one side of balance and B1B2B3 on the other. It
may have three results as described in the following cases.
Case 1 : A1A2A3 = B1B2B3
(i) A1A2A3 (ii) B1B2B3 (iii) C1C<2
Weighting will be done as bellow :
1st weighting → A1A2A3 will be on one side of balance and B1B2B3 on the other. It
may have three results as described in the following cases.
Case 1 : A1A2A3 = B1B2B3
This results out that either C1 or C2 will heavier for which we will have to perform
weighting again.
2nd weighting " C1 is kept on the one side and C2 on the other.
if C1 > C2 then C1 is heavier.
C1 < C2 then C2 is heavier.
weighting again.
2nd weighting " C1 is kept on the one side and C2 on the other.
if C1 > C2 then C1 is heavier.
C1 < C2 then C2 is heavier.
Case 2 : A1A2A3> B1B2B3it means one of the A1A2A3 will be heavier So we will perform next weighting as:
2nd weighting → A1 is kept on one side of the balance and A2 on the other.
if A1 = A2 it means A3 will be heavier
A1 > A2 then A1 will be heavier
A1 < A2 then A2 will be heavier
2nd weighting → A1 is kept on one side of the balance and A2 on the other.
if A1 = A2 it means A3 will be heavier
A1 > A2 then A1 will be heavier
A1 < A2 then A2 will be heavier
Case 3 : A1A2A3 < B1B2B3
This time one of the B1B2B3 will be heavier, So again as the above case weighting
will be done.
2nd weighting " B1 is kept one side and B2 on the other
if B, = B2 B3 will be heavier
B1 > B2 B1 will be heavier
B1 < B2 B2 will be heavier
So, as described above, in all the three cases weighting is done only two times to
give out the result so minimum no. of weighting required = 2.
This time one of the B1B2B3 will be heavier, So again as the above case weighting
will be done.
2nd weighting " B1 is kept one side and B2 on the other
if B, = B2 B3 will be heavier
B1 > B2 B1 will be heavier
B1 < B2 B2 will be heavier
So, as described above, in all the three cases weighting is done only two times to
give out the result so minimum no. of weighting required = 2.
Most Upvoted Answer
There are eight bags of rice looking alike, seven of which have equal ...
To identify the heavier bag in the minimum number of weighings, we can use a binary search approach.
1. **Divide the bags into two groups**: Start by dividing the eight bags into two groups of four each. Weigh these two groups against each other on the balance.
a. If the two groups weigh the same, the heavier bag must be in the remaining four bags.
b. If one group is heavier than the other, the heavier bag must be in that group.
2. **Divide the remaining bags**: Take the group of four bags that is heavier and divide them into two groups of two each. Weigh these two groups against each other on the balance.
a. If the two groups weigh the same, the heavier bag must be one of the remaining two bags.
b. If one group is heavier than the other, the heavier bag must be in that group.
3. **Identify the heavier bag**: Take the group of two bags that is heavier and weigh them against each other on the balance.
a. If one bag is heavier than the other, that bag is the heavier bag.
b. If they weigh the same, the remaining bag is the heavier bag.
Using this binary search approach, we can identify the heavier bag in a minimum of two weighings. In the worst case scenario, the heavier bag is always on the side that is heavier in each weighing, and we eliminate half of the bags at each step. Thus, the minimum number of weighings required is two.
Therefore, the correct answer is option 'A' - 2 weighings.
1. **Divide the bags into two groups**: Start by dividing the eight bags into two groups of four each. Weigh these two groups against each other on the balance.
a. If the two groups weigh the same, the heavier bag must be in the remaining four bags.
b. If one group is heavier than the other, the heavier bag must be in that group.
2. **Divide the remaining bags**: Take the group of four bags that is heavier and divide them into two groups of two each. Weigh these two groups against each other on the balance.
a. If the two groups weigh the same, the heavier bag must be one of the remaining two bags.
b. If one group is heavier than the other, the heavier bag must be in that group.
3. **Identify the heavier bag**: Take the group of two bags that is heavier and weigh them against each other on the balance.
a. If one bag is heavier than the other, that bag is the heavier bag.
b. If they weigh the same, the remaining bag is the heavier bag.
Using this binary search approach, we can identify the heavier bag in a minimum of two weighings. In the worst case scenario, the heavier bag is always on the side that is heavier in each weighing, and we eliminate half of the bags at each step. Thus, the minimum number of weighings required is two.
Therefore, the correct answer is option 'A' - 2 weighings.
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There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum number of weighings required to identify the heavier bag isa)2b)3c)4d)8Correct answer is option 'A'. Can you explain this answer?
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