Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same.
The total number of coins in each column is also the same. Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied.
i) The minimum among the numbers of coins in the three sacks in the box is 1.
ii) The median of the numbers of coins in the three sacks is 1.
iii) The maximum among the numbers of coins in the three sacks in the box is 9.
For how many boxes are the average and median of the numbers of coins contained in the three sacks in that box the same?
Approach Solution - 1
The correct answer is 4.
Approach Solution -2
(note:Median is termed as middle here,bag and pack are interchangeably used, Normal=Average))We are given that each container contains three sacks. Each sack has a specific number of coins, somewhere in the range of 1 and 9, both comprehensive. The typical number of coins per sack in the crates are unmistakable numbers. The complete number of coins in each line is something similar. The complete number of coins in every segment is additionally something similar.
=> The complete number of coins in a case range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9) Since, it is given that the normal number of coins per sack in the crates are particular numbers
=> The all out number of coins in a container would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => midpoints of 1, 2, 3, 4,....,9 => Total = 45.
=> Amount of midpoints coins in a container in succession or segment = 45/3 = 15 [The all out number of coins in each column is something similar. The complete number of coins in every segment is likewise the same.]
==> First, let us use the following boxes to represent the final arrangement of the sacks:
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | |||
| R-2 | |||
| R-3 |
We are given that each container contains three sacks. Each sack has a specific number of coins, somewhere in the range of 1 and 9, both comprehensive. The typical number of coins per sack in the crates are unmistakable numbers. The complete number of coins in each line is something similar. The complete number of coins in every segment is additionally something similar. => The complete number of coins in a case range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9) Since, it is given that the normal number of coins per sack in the crates are particular numbers => The all out number of coins in a container would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => midpoints of 1, 2, 3, 4,....,9 => Total = 45. => Amount of midpoints coins in a container in succession or segment = 45/3 = 15 [The all out number of coins in each column is something similar. The complete number of coins in every segment is likewise the same.] ==> First, let us use the following boxes to represent the final arrangement of the sacks:
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | 3,9,9(7) | ||
| R-2 | 1,2,9(4) | ||
| R-3 | 7,8,9(8) |
From (1), The average in bag (1,1) is 15 - 4 - 8 = 3.
From (1), The average in bag (1,3) is 15-3-7 = 5.
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | 3,9,9(7) | Avg=5 | |
| R-2 | 1,2,9(4) | ||
| R-3 | 7,8,9(8) |
Think about pack (1,1)
Avg = 3, 1 sack has more than 5 and ** => 2 circumstances are being fulfilled. => (can't be condition-3 => 9 coins as the absolute amount of coins is it self 3*3 = 9)
=> pack (1,1) has 1, 1, 7 coins with normal = 3.
Think about sack (1,3)
Avg. 5> Sum = 15.
Middle = 6 and 2 sacks have more than 5 and * => (1 condition is fulfilled) Not condition ii as the middle is 6 and Not condition iii as the amount of 2 sacks itself will become 6 + 9 = 15
=> 1, 6, c are the coins => For aggregate = 15 => c = 15-1-6=8
=> pack (1,3) has 1, 6, 8 coins with average = 5.
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | 1,1,7(3) | 3,9,9(7) | 1,6,8(5) |
| R-2 | 1,2,9(4) | ||
| R-3 | 7,8,9(8) |
Take into consideration bag (3,3):
0 sacks contain more than 5 coins, and ** indicates that conditions i and ii are met,
1,1,c are the coins. Presently c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a different of 3. Be that as it may, c = 1 as no other sack has the likelihood to get avg. = 1 as sack (2,2) ought to have 1, b, c coins and b and c ought to be more than 1 as just 1* => pack (3,3) has 1, 1, 1 coins with normal = 1. Presently, we can fill the midpoints in every one of the sacks.
| Tables | |||
| C-1 | C-2 | C-3 | |
| R-1 | 1,1,7(3) | 3,9,9(7) | 1,6,8(5) |
| R-2 | 1,2,9(4) | avg=2 | avg=9 |
| R-3 | 7,8,9(8) | avg=6 | 1,1,1(1) |
In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins.
In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens-t should be 1.
=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2 + 3 = 5 is the only possibility.
=> 1, 2, 3 are the coins with average = 2.
Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | 1,1,7(3) | 3,9,9(7) | 1,6,8(5) |
| R-2 | 1,2,9(4) | 1,2,3(2) | 9,9,9(9) |
| R-3 | 7,8,9(8) | Avg=6 | 1,1,1(1) |
Considering bag (3,2)
Avg. = 6=> Sum = 18. 2 sacks more than 5 coins and **
=> 2 sacks have 1 and 9 coins.
=> bag (3,2) has 1, c, 9 coins and c = 18-1-9=8
=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.
==> Final required table, bracket number => average coins per sack in the bag.
Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | 1,1,7(3) | 3,9,9(7) | 1,6,8(5) |
| R-2 | 1,2,9(4) | 1,2,3(2) | 9,9,9(9) |
| R-3 | 7,8,9(8) | 1,8,9(6) | 1,1,1(1) |
Average = Median in boxes (3,1), (2,2), (2,3) and (3,3): => 4 boxes.The correct answer is 4.
How many sacks have exactly one coin?
Approach Solution - 1
The correct answer is 9.
Approach Solution -2
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same. The averages are 1, 2, 3, 4,.....9. The totals are these into 3. Total number of coins = 3(1 + 2 + 3....9) = 135. Each row and each column adds up to 45. Also, the total number coins in each box should be a multiple of 3.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9. Chase the extremes. Which box will have average 1? Which one will have average 9
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9. Chase the extremes. Which box will have average 1? Which one will have average 9
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9. Which box could have average 2?
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9. Now let us finish this off
Table 1 gives the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied. i) The minimum among the numbers of coins in the three sacks in the box is 1. ii) The median of the numbers of coins in the three sacks is 1. iii) The maximum among the numbers of coins in the three sacks in the box is 9. Now let us finish this off
Hence the Answer is 9
In how many boxes do all three sacks contain different numbers of coins?
Approach Solution - 1
The correct answer is 5.
Approach Solution -2
We are given that each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
=>The total number of coins in a box range from 3 (1 + 1 + 1) to 27 (9+9+9)
Since, it is given that the average number of coins per sack in the boxes are all distinct integers => The total number of coins in a box would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => averages of 1, 2, 3, 4,...,9 => Sum = 45.
=> Sum of averages coins in a box in a row or column = 45/3 = 15 [The total number of coins in each row is the same. The total number of coins in each column is also the same.] ==> (1) Let us represent the final configuration of the sacks in boxes as follows:
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | |||
| R-2 | |||
| R-3 |
Also a bag (x,y) => bag in xth row and yth column.
We are given 2 clues => Table-1 & Table-2
Consider bag (3,1)
=> From Table-1 => Median = 8 & From Table-2 all 3 sacks have more than 5 coins. Also * => There is a 9 in one of the sacks.
=> c, 8, 9 are the coins in bag (3,1), now c>5 & c + 8+ 9 should be a multiple of 3 => c = 7 is the only possiblility.
=> bag (3,1) has 7, 8, 9 coins with average = 8.
Consider bag (2,1)
Median = 2 and 1 sack has more than 5 coins. Also **=> conditions i & iii should be satisfied.
=> 1, 2, 9 are the coins in bag (2,1) with average = 4
Consider bag (1,2)
Median = 9 and 2 elements are more than 5. Also * => (9 is present & 1 is not present)
=> c, 9, 9 are the coins in bag (1,2) and c is not equal to 1 and less than 5
=> c = 3 for c + 18 to be a multiple of 3.
=> 3, 9, 9 are the coins in bag (1,2) with average = 7.
Capturing this info. in the table:
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | 3,9,9(7) | ||
| R-2 | 1,2,9(4) | ||
| R-3 | 7,8,9(8) |
From (1), The average in bag (1,1) is 15 - 4-8 = 3.
From (1), The average in bag (1,3) is 15-3-7= 5.
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | Avg=3 | 3,9,9(7) | Avg=5 |
| R-2 | 1,2,9(4) | ||
| R-3 | 7,8,9(8) |
Consider bag (1,1)
Avg = 3, 1 sack has more than 5 and ** => 2 conditions are being satisfied. => (can't be condition-3 => 9 coins as the total sum of coins is it self 3*3 = 9)
=> bag (1,1) has 1, 1, 7 coins with average = 3.
Consider bag (1,3)
Avg. 5 => Sum = 15.
Median = 6 and 2 sacks have more than 5 and * => (1 condition is satisfied)
Not condition ii as the median is 6 & Not condition iii as the sum of 2 sacks itself will become 6+ 9 = 15
=> 1, 6, c are the coins => For sum = 15 => c = 15-1-6=8
=> bag (1,3) has 1, 6, 8 coins with average = 5.
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | 1,1,7(3) | 3,9,9(7) | 1,6,8(5) |
| R-2 | 1,2,9(4) | ||
| R-3 | 7,8,9(8) |
Consider bag (3,3)
O sacks have more than 5 coins and ** => conditions i & ii are being satisfied.
=> >1,1,c are the coins. Now c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a multiple of 3.
But c = 1 as no other bag has the possibility to get avg. = 1 as bag (2,2) should have 1, b, c coins and b and c should be more than 1 as only 1*
=> bag (3,3) has 1, 1, 1 coins with average = 1.
Now, we can fill the averages in all the bags.
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | 1,1,7(3) | 3,9,9(7) | 1,6,8(5) |
| R-2 | 1,2,9(4) | Avg=2 | Avg=9 |
| R-3 | 7,8,9(8) | Avg=6 | 1,1,1(1) |
In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins.
In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elements should be 1.
=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2+3 = 5 is the only possibility.
=> 1, 2, 3 are the coins with average = 2.
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | 1,1,7(3) | 3,9,9(7) | 1,6,8(5) |
| R-2 | 1,2,9(4) | 1,2,3(2) | 9,9,9(9) |
| R-3 | 7,8,9(8) | Avg=6 | 1,1,1(1) |
Considering bag (3,2)
Avg. = 6=> Sum = 18.
2 sacks more than 5 coins and ** => 2 sacks have 1 and 9 coins.
=> bag (3,2) has 1, c, 9 coins and c = 18-1-9=8
=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.
==> Final required table, bracket number => average coins per sack in the bag.
| Table | |||
| C-1 | C-2 | C-3 | |
| R-1 | 1,1,7(3) | 3,9,9(7) | 1,6,8(5) |
| R-2 | 1,2,9(4) | 1,2,3(2) | 9,9,9(9) |
| R-3 | 7,8,9(8) | 1,8,9(6) | 1,1,1(1) |
Bags with different number of coins in all 3 sacks are (2,1), (3,2), (2,2), (3,2), (1,3) => 5 bags.
How many boxes have at least one sack containing 9 coins?
- 3
- 4
- 5
- 8
The Correct Option is C
Solution and Explanation
Here's the extracted information from the image:
1. Analyzing Conditions for Each Box:
- From Table 2, any box marked with "\" satisfies multiple conditions, one of which could include the maximum sack containing 9 coins.
2. Identifying Boxes with Sacks Containing 9 Coins:
- By checking each box and applying the given conditions, we find that 5 boxes have at least one sack containing 9 coins.
Thus, there are 5 boxes that have at least one sack containing 9 coins.
What is the total number of coins in all the boxes in the 3rd row?
- 45
- 15
- 36
- 30
The Correct Option is A
Solution and Explanation
1. Understanding the Constraints:
- The total number of coins in each row and column is the same.
- The average coins per sack in each box are distinct integers.
2. Analyzing the Third Row:
- Based on the distinct integer values for the average coins and conditions in Tables 1 and 2, the sum of coins in the third row is calculated to satisfy the row and column balance requirements.
3. Conclusion:
- By verifying possible values that satisfy all conditions, we find that the total number of coins in all the boxes in the 3rd row is 45.
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