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CAT 2025 Lesson : Numerical Reasoning - Case 6
Example Case 6
Read the following and answer the questions that follow
You are given an square matrix to be filled with numerals so that no two adjacent cells have the same numeral. Two cells are called adjacent if they touch each other horizontally, vertically or diagonally. So a cell in one of the four corners has three cells adjacent to it, and a cell in the first or last row or column which is not in the corner has five cells adjacent to it. Any other cell has eight cells adjacent to it.
[CAT 2018 S1]
18) What is the minimum number of different numerals needed to fill a 3×3 square matrix?
Answer:
19) What is the minimum number of different numerals needed to fill a 5×5 square matrix?
Answer:
20) Suppose you are allowed to make one mistake, that is, one pair of adjacent cells can have the same numeral. What is the minimum number of different numerals required to fill a 5×5 matrix?
(1) 9
(2) 25
(3) 4
(4) 16
21) Suppose that all the cells adjacent to any particular cell must have different numerals. What is the minimum number of different numerals needed to fill a 5×5 square matrix?
(1) 9
(2) 25
(3) 16
(4) 4
The table below is filled by first filling different numbers (1, 2, 3 and 4) for the 4 adjacent cells. Starting from top left corner, the 2nd row acts as a barrier for 3rd row and 1st row, similarly the 2nd column acts as a barrier for 1st and 3rd column. So, we can reuse the numbers which are on either side on the barrier.
Therefore, we can fill a square matrix with a minimum of four numerals.
Answer: 4
19)
The table below is filled by first filling different numbers (1, 2, 3 and 4) for the 4 adjacent cells starting from top left corner. The 2nd row acts as a barrier for 3rd row and 1st row, similarly the 2nd column acts as a barrier for 1st and 3rd column. So, we can reuse the numbers which are on either side on the barrier.
Therefore, we can fill a square matrix with minimum of four numerals.
Answer: 4
20)
In the above question, if we make one mistake it will not reduce the numerals. If you look at the answer options, we have 4 which is the minimum number of numerals that can be used. This is the same as the last question.
Answer: (3) 4
21) The approach is the same as above. But in this question, a cell should not have the same numeral in any of the adjacent cells. So, here we use a double barrier (eg. 2nd row and 3rd row acts as a barrier for 4th row and 1st row). The table below is filled by first filling different numbers (1, 2, 3, 4, 5, 6, 7, 8 and 9) for the 9 adjacent cells starting from the middle cell. The 2nd row and 3rd row act as a double barrier for 4th row and 1st row, similarly the 2nd and 3rd column act as a double barrier for 1st and 4th column. So, we can reuse the numbers which are on either side on the double barrier.
Therefore, we can fill a square matrix with a minimum of nine numerals.
Answer: (1) 9
Answer:
18) 4
19) 4
20) (3) 4
21) (1) 9
You are given an square matrix to be filled with numerals so that no two adjacent cells have the same numeral. Two cells are called adjacent if they touch each other horizontally, vertically or diagonally. So a cell in one of the four corners has three cells adjacent to it, and a cell in the first or last row or column which is not in the corner has five cells adjacent to it. Any other cell has eight cells adjacent to it.
[CAT 2018 S1]
18) What is the minimum number of different numerals needed to fill a 3×3 square matrix?
Answer:
19) What is the minimum number of different numerals needed to fill a 5×5 square matrix?
Answer:
20) Suppose you are allowed to make one mistake, that is, one pair of adjacent cells can have the same numeral. What is the minimum number of different numerals required to fill a 5×5 matrix?
(1) 9
(2) 25
(3) 4
(4) 16
21) Suppose that all the cells adjacent to any particular cell must have different numerals. What is the minimum number of different numerals needed to fill a 5×5 square matrix?
(1) 9
(2) 25
(3) 16
(4) 4
Solution
18)The table below is filled by first filling different numbers (1, 2, 3 and 4) for the 4 adjacent cells. Starting from top left corner, the 2nd row acts as a barrier for 3rd row and 1st row, similarly the 2nd column acts as a barrier for 1st and 3rd column. So, we can reuse the numbers which are on either side on the barrier.
Therefore, we can fill a square matrix with a minimum of four numerals.
Answer: 4
19)
The table below is filled by first filling different numbers (1, 2, 3 and 4) for the 4 adjacent cells starting from top left corner. The 2nd row acts as a barrier for 3rd row and 1st row, similarly the 2nd column acts as a barrier for 1st and 3rd column. So, we can reuse the numbers which are on either side on the barrier.
Therefore, we can fill a square matrix with minimum of four numerals.
Answer: 4
20)
In the above question, if we make one mistake it will not reduce the numerals. If you look at the answer options, we have 4 which is the minimum number of numerals that can be used. This is the same as the last question.
Answer: (3) 4
21) The approach is the same as above. But in this question, a cell should not have the same numeral in any of the adjacent cells. So, here we use a double barrier (eg. 2nd row and 3rd row acts as a barrier for 4th row and 1st row). The table below is filled by first filling different numbers (1, 2, 3, 4, 5, 6, 7, 8 and 9) for the 9 adjacent cells starting from the middle cell. The 2nd row and 3rd row act as a double barrier for 4th row and 1st row, similarly the 2nd and 3rd column act as a double barrier for 1st and 4th column. So, we can reuse the numbers which are on either side on the double barrier.
Therefore, we can fill a square matrix with a minimum of nine numerals.
Answer: (1) 9
Answer:
18) 4
19) 4
20) (3) 4
21) (1) 9
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