Distribution Puzzle: Practice Reasoning Questions
Distribution puzzles are generally the difficult part fo the reasoning syllabus for comeptitive exams specially CAT. You need both you mathemtical and reasonign skills in the puzzle to be able to solve it in strict exam time limits. Let’s practice some puzzles:
Puzzle #1
Five employees each logged hours on one of three projects; the hours are whole numbers from 1 to 5.
Constraint 1: Each employee worked on exactly one project and recorded an integer number of hours between 1 and 5.
Constraint 2: Every project has at least one employee contributing hours.
Constraint 3: The average hours contributed on Project 2 is exactly 1 hour.
Constraint 4: The smallest hours recorded by any employee on Project 3 is 4 hours.
Constraint 5: Exactly one employee contributed hours to Project 1.
Constraint 6: Employee 1 recorded the highest number of hours among all employees, and Employee 5 recorded the lowest number of hours among all employees.
Constraint 7: The average hours for each project is an integer number of hours.
Based on given clues, please complete the table
| Employee | Project | Hours |
|---|---|---|
| 1 | ||
| 2 | ||
| 3 | ||
| 4 | ||
| 5 |
Solution
Project 2’s average is 1 and hours are positive integers, so every employee on Project 2 would have to have 1 hour.
Constraint 6 says Employee 5 has the lowest number of hours (uniquely), so h5 = 1 and no one else can have 1; therefore Project 2 has exactly one contributor, Employee 5 with 1 hour.
Constraint 5 says Project 1 has exactly one contributor, so the remaining distribution must be (Project1:1, Project2:1, Project3:3). Project 3’s smallest recorded hours is 4, so each Project 3 contributor has 4 or 5 hours.
The three Project 3 hours must average to an integer; the only possible sums for three numbers from {4,5} that give an integer average are 12 (all 4s) or 15 (all 5s). The minimum on Project 3 must be 4, so the three Project 3 contributors are all 4 hours.
That leaves one remaining employee (not Employee 5 and not on Project 3) who must be the unique maximum per Constraint 6, so they must have 5 hours and be the sole Project 1 contributor.
Constraint 6 requires that unique maximum be Employee 1, so Employee 1 is the Project 1 contributor with 5 hours. This assignment satisfies all constraints and is forced, so the solution is unique.
Mapping:
| Employee | Project | Hours |
|---|---|---|
| 1 | 1 | 5 |
| 2 | 3 | 4 |
| 3 | 3 | 4 |
| 4 | 3 | 4 |
| 5 | 2 | 1 |
Puzzle #2
Three bowlers: A, B and C, each bowled three overs; record how many wickets each took in each over using the numbers 1–9.
Constraint 1: Each of the nine wicket counts is a different integer from 1 through 9, used exactly once.
Constraint 2: In the first over the average wickets per bowler is 2 (the three values for that over sum to 6); in the second over the average per bowler is 8 (those three values sum to 24).
Constraint 3: Each bowler finished the three overs with the same total number of wickets.
Constraint 4: Bowler C took 6 wickets in the third over.
Constraint 5: In the first over, Bowler B’s wicket count is the second-lowest of all nine values.
Based on given clues, please complete the table
| Bowler | Over 1 | Over 2 | Over 3 |
|---|---|---|---|
| A | |||
| B | |||
| C |
Solution
B1 must be 2 (the second-lowest of 1–9). Over 1 sums to 6, so the three values there are 1,2,3; with B1=2 the other two are A1 and C1 as 1 and 3.
The total of all numbers 1–9 is 45, so over 3 sums to 15; with C3=6 we have A3 + B3 = 9, so the only available pair (from unused numbers) is 4 and 5.
Thus over 3 is {4,5,6} and over 2 must be the remaining {7,8,9} (sum 24). Requiring each bowler’s column to total 15 forces A1=3 and C1=1, and A3=5, B3=4; then A2=7, B2=9, C2=8.
All nine numbers 1–9 are used exactly once and all constraints are satisfied; this solution is unique.
| Bowler | Over 1 | Over 2 | Over 3 |
|---|---|---|---|
| A | 3 | 7 | 5 |
| B | 2 | 9 | 4 |
| C | 1 | 8 | 6 |