Instructions
There are nine boxes arranged in a 3×3 array as shown in Tables 1 and 2. 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.
Question 5.
In how many boxes do all three sacks contain different numbers of coins?
Question Explanation
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:
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 possibility.
=> 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:
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.
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.
Consider bag (3,3)
0 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.
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.
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.
Bags with different number of coins in all 3 sacks are (2,1), (3,2), (2,2), (3,2), (1,3) => 5 bags.
Master CAT Preparation with Previous Year Papers
Practicing CAT previous year papers is one of the most effective strategies for CAT exam preparation. By solving questions from CAT 2023 dilr-slot-2 and other previous years, you can understand the exam pattern, difficulty level, and types of questions asked in the Common Admission Test.
Why Practice CAT Previous Year Questions?
- Understand Exam Pattern: CAT previous papers help you familiarize yourself with the question format, marking scheme, and time management required for the actual exam.
- Identify Important Topics: By analyzing CAT solved questions from multiple years, you can identify frequently asked topics and focus your preparation accordingly.
- Improve Speed and Accuracy: Regular practice of CAT previous year papers enhances your problem-solving speed and accuracy, which are crucial for scoring well in the exam.
- Build Confidence: Solving CAT previous year questions builds confidence and reduces exam anxiety by making you comfortable with the exam format.
How to Use CAT Previous Papers Effectively
- Solve Under Exam Conditions: Attempt CAT previous year papers in a timed environment to simulate the actual exam experience.
- Analyze Your Performance: After solving each CAT previous paper, analyze your mistakes and identify areas that need improvement.
- Review Solutions Thoroughly: Study the detailed solutions and explanations provided for each question to understand the correct approach and methodology.
- Focus on Weak Areas: Use CAT solved questions to identify your weak areas and dedicate more time to improving them.
CAT Exam Sections Covered
Our comprehensive collection of CAT previous papers covers all three sections of the exam:
- VARC (Verbal Ability & Reading Comprehension): Practice reading comprehension passages, para jumbles, and other verbal ability questions from CAT previous years.
- DILR (Data Interpretation & Logical Reasoning): Master DI and LR questions with detailed solutions from CAT previous year papers.
- QA (Quantitative Ability): Solve arithmetic, algebra, geometry, and other quantitative ability questions from CAT solved papers.
Additional CAT Preparation Resources
Looking for more CAT preparation materials? Explore our comprehensive collection of:
- CAT Previous Papers Archive - Access questions from multiple years and slots
- CAT 2023 dilr-slot-2 Solved Solutions - Complete slot-wise solutions with explanations
- CAT Preparation Courses - Enroll in structured courses for comprehensive CAT preparation
Note: Regular practice of CAT previous year papers, combined with a structured study plan, is essential for achieving a high percentile in the CAT exam. Make sure to solve questions from all sections and review the solutions thoroughly to maximize your preparation effectiveness.
