CAT — Matrix / Grid Logic
17 questions, free to view. Click any question to see the answer and explanation.
The above is a schematic diagram of walkways (indicated by all the straight-lines) and lakes(3 of them, each in the shape of rectangles – shaded in the diagram) of a gated area. Different points on the walkway are indicated by letters (A through P) with distances being OP = 150 m, ON = MN = 300 m, ML = 400 m, EL = 200 m, DE = 400 m.
The following additional information about the facilities in the area is known.
1. The only entry/exit point is at C.
2. There are many residences within the gated area; all of them are located on the path AH and ML with four of them being at A, H, M, and L.
3. The post office is located at P and the bank is located at B.
One resident whose house is located at L, needs to visit the post office as well as the bank. What is the minimum distance (in m) he has to walk starting from his residence and returning to his residence after visiting both the post office and the bank?
[Note: In actual paper CAT 2024, two options (3) and (4) i.e., 3000 were same.]
3200
2700
3000
3400
3200
One person enters the gated area and decides to walk as much as possible before leaving the area without walking along any path more than once and always walking next to one of the lakes. Note that he may cross a point multiple times. How much distance (in m) will he walk within the gated area?
3200
3000
2800
3800
3800
One resident takes a walk within the gated area starting from A and returning to A without going through any point (other than A) more than once. What is the maximum distance (in m) she can walk in this way?
Visitors coming for morning walks are allowed to enter as long as they do not pass by any of the residences and do not cross any point (except C) more than once.What is the maximum distance (in m) that such a visitor can walk within the gated area?
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.
What is the total number of coins in all the boxes in the 3rd row?
45
36
30
15
45
Step 1:
The average number of coins per sack in each box is given to be a distinct integer. Hence, the nine boxes must have average values 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order.
Since the sum of these averages is
(9 × 10) / 2 = 45,
and every row and column has the same total number of coins, each row and each column must also contain a total of 45 coins.
Further, because the average number of coins in a box is an integer, the sum of the three sacks in every box must be divisible by 3.
An average of 1 is possible only with the combination (1, 1, 1).
This combination can be placed only in the 3rd row, 3rd column, as it satisfies both conditions:
- the minimum value is 1, and
- the median is also 1.
Similarly, an average of 9 is possible only with (9, 9, 9).
Since the box in 3rd row, 1st column has a median of 8, it cannot have an average of 9.
The box in 3rd row, 1st column satisfies condition (iii). Hence, the only valid combination is (7, 8, 9).
The box in 3rd row, 2nd column must satisfy conditions (i) and (ii). Therefore, its minimum and maximum values are 1 and 9, respectively.
The possible combinations are (1, 5, 9) and (1, 8, 9). However, only (1, 8, 9) allows the total of the third row to become 45.
Thus, the third row is:
- 3rd row, 1st column: (7, 8, 9) → Average = 8
- 3rd row, 2nd column: (1, 8, 9) → Average = 6
- 3rd row, 3rd column: (1, 1, 1) → Average = 1
Step 2:
If the average of a box is 2, then the total number of coins in that box must be 6.
Since each sack must contain fewer than 5 coins, this is possible only for the box in the 2nd row, 2nd column.
Among the possible combinations, only (1, 2, 3) satisfies the required conditions.
(The combination (1, 1, 4) satisfies only two conditions and therefore cannot be used.)
Hence:
- 2nd row, 1st column: (1, 2, 9) → Average = 4
- 2nd row, 2nd column: (1, 2, 3) → Average = 2
- 2nd row, 3rd column: (9, 9, 9) → Average = 9
Similarly, the first row is determined as:
- 1st row, 1st column: (1, 1, 7) → Average = 3
- 1st row, 2nd column: (3, 9, 9) → Average = 7
- 1st row, 3rd column: (1, 6, 8) → Average = 5
The total number of coins in all the boxes in the 3rd row is 45.
How many boxes have at least one sack containing 9 coins?
4
8
3
5
5
Step 1:
The average number of coins per sack in each box is given to be a distinct integer. Hence, the nine boxes must have average values 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order.
Since the sum of these averages is
(9 × 10) / 2 = 45,
and every row and column has the same total number of coins, each row and each column must also contain a total of 45 coins.
Further, because the average number of coins in a box is an integer, the sum of the three sacks in every box must be divisible by 3.
An average of 1 is possible only with the combination (1, 1, 1).
This combination can be placed only in the 3rd row, 3rd column, as it satisfies both conditions:
- the minimum value is 1, and
- the median is also 1.
Similarly, an average of 9 is possible only with (9, 9, 9).
Since the box in 3rd row, 1st column has a median of 8, it cannot have an average of 9.
The box in 3rd row, 1st column satisfies condition (iii). Hence, the only valid combination is (7, 8, 9).
The box in 3rd row, 2nd column must satisfy conditions (i) and (ii). Therefore, its minimum and maximum values are 1 and 9, respectively.
The possible combinations are (1, 5, 9) and (1, 8, 9). However, only (1, 8, 9) allows the total of the third row to become 45.
Thus, the third row is:
- 3rd row, 1st column: (7, 8, 9) → Average = 8
- 3rd row, 2nd column: (1, 8, 9) → Average = 6
- 3rd row, 3rd column: (1, 1, 1) → Average = 1
Step 2:
If the average of a box is 2, then the total number of coins in that box must be 6.
Since each sack must contain fewer than 5 coins, this is possible only for the box in the 2nd row, 2nd column.
Among the possible combinations, only (1, 2, 3) satisfies the required conditions.
(The combination (1, 1, 4) satisfies only two conditions and therefore cannot be used.)
Hence:
- 2nd row, 1st column: (1, 2, 9) → Average = 4
- 2nd row, 2nd column: (1, 2, 3) → Average = 2
- 2nd row, 3rd column: (9, 9, 9) → Average = 9
Similarly, the first row is determined as:
- 1st row, 1st column: (1, 1, 7) → Average = 3
- 1st row, 2nd column: (3, 9, 9) → Average = 7
- 1st row, 3rd column: (1, 6, 8) → Average = 5
5 boxes have atleast one sack containing 9 coins.
For how many boxes are the average and median of the numbers of coins contained in the three sacks in that box the same?
Step 1:
The average number of coins per sack in each box is given to be a distinct integer. Hence, the nine boxes must have average values 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order.
Since the sum of these averages is
(9 × 10) / 2 = 45,
and every row and column has the same total number of coins, each row and each column must also contain a total of 45 coins.
Further, because the average number of coins in a box is an integer, the sum of the three sacks in every box must be divisible by 3.
An average of 1 is possible only with the combination (1, 1, 1).
This combination can be placed only in the 3rd row, 3rd column, as it satisfies both conditions:
- the minimum value is 1, and
- the median is also 1.
Similarly, an average of 9 is possible only with (9, 9, 9).
Since the box in 3rd row, 1st column has a median of 8, it cannot have an average of 9.
The box in 3rd row, 1st column satisfies condition (iii). Hence, the only valid combination is (7, 8, 9).
The box in 3rd row, 2nd column must satisfy conditions (i) and (ii). Therefore, its minimum and maximum values are 1 and 9, respectively.
The possible combinations are (1, 5, 9) and (1, 8, 9). However, only (1, 8, 9) allows the total of the third row to become 45.
Thus, the third row is:
- 3rd row, 1st column: (7, 8, 9) → Average = 8
- 3rd row, 2nd column: (1, 8, 9) → Average = 6
- 3rd row, 3rd column: (1, 1, 1) → Average = 1
Step 2:
If the average of a box is 2, then the total number of coins in that box must be 6.
Since each sack must contain fewer than 5 coins, this is possible only for the box in the 2nd row, 2nd column.
Among the possible combinations, only (1, 2, 3) satisfies the required conditions.
(The combination (1, 1, 4) satisfies only two conditions and therefore cannot be used.)
Hence:
- 2nd row, 1st column: (1, 2, 9) → Average = 4
- 2nd row, 2nd column: (1, 2, 3) → Average = 2
- 2nd row, 3rd column: (9, 9, 9) → Average = 9
Similarly, the first row is determined as:
- 1st row, 1st column: (1, 1, 7) → Average = 3
- 1st row, 2nd column: (3, 9, 9) → Average = 7
- 1st row, 3rd column: (1, 6, 8) → Average = 5
How many sacks have exactly one coin?
Step 1:
The average number of coins per sack in each box is given to be a distinct integer. Hence, the nine boxes must have average values 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order.
Since the sum of these averages is
(9 × 10) / 2 = 45,
and every row and column has the same total number of coins, each row and each column must also contain a total of 45 coins.
Further, because the average number of coins in a box is an integer, the sum of the three sacks in every box must be divisible by 3.
An average of 1 is possible only with the combination (1, 1, 1).
This combination can be placed only in the 3rd row, 3rd column, as it satisfies both conditions:
- the minimum value is 1, and
- the median is also 1.
Similarly, an average of 9 is possible only with (9, 9, 9).
Since the box in 3rd row, 1st column has a median of 8, it cannot have an average of 9.
The box in 3rd row, 1st column satisfies condition (iii). Hence, the only valid combination is (7, 8, 9).
The box in 3rd row, 2nd column must satisfy conditions (i) and (ii). Therefore, its minimum and maximum values are 1 and 9, respectively.
The possible combinations are (1, 5, 9) and (1, 8, 9). However, only (1, 8, 9) allows the total of the third row to become 45.
Thus, the third row is:
- 3rd row, 1st column: (7, 8, 9) → Average = 8
- 3rd row, 2nd column: (1, 8, 9) → Average = 6
- 3rd row, 3rd column: (1, 1, 1) → Average = 1
Step 2:
If the average of a box is 2, then the total number of coins in that box must be 6.
Since each sack must contain fewer than 5 coins, this is possible only for the box in the 2nd row, 2nd column.
Among the possible combinations, only (1, 2, 3) satisfies the required conditions.
(The combination (1, 1, 4) satisfies only two conditions and therefore cannot be used.)
Hence:
- 2nd row, 1st column: (1, 2, 9) → Average = 4
- 2nd row, 2nd column: (1, 2, 3) → Average = 2
- 2nd row, 3rd column: (9, 9, 9) → Average = 9
Similarly, the first row is determined as:
- 1st row, 1st column: (1, 1, 7) → Average = 3
- 1st row, 2nd column: (3, 9, 9) → Average = 7
- 1st row, 3rd column: (1, 6, 8) → Average = 5
In how many boxes do all three sacks contain different numbers of coins?
Step 1:
The average number of coins per sack in each box is given to be a distinct integer. Hence, the nine boxes must have average values 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order.
Since the sum of these averages is
(9 × 10) / 2 = 45,
and every row and column has the same total number of coins, each row and each column must also contain a total of 45 coins.
Further, because the average number of coins in a box is an integer, the sum of the three sacks in every box must be divisible by 3.
An average of 1 is possible only with the combination (1, 1, 1).
This combination can be placed only in the 3rd row, 3rd column, as it satisfies both conditions:
- the minimum value is 1, and
- the median is also 1.
Similarly, an average of 9 is possible only with (9, 9, 9).
Since the box in 3rd row, 1st column has a median of 8, it cannot have an average of 9.
The box in 3rd row, 1st column satisfies condition (iii). Hence, the only valid combination is (7, 8, 9).
The box in 3rd row, 2nd column must satisfy conditions (i) and (ii). Therefore, its minimum and maximum values are 1 and 9, respectively.
The possible combinations are (1, 5, 9) and (1, 8, 9). However, only (1, 8, 9) allows the total of the third row to become 45.
Thus, the third row is:
- 3rd row, 1st column: (7, 8, 9) → Average = 8
- 3rd row, 2nd column: (1, 8, 9) → Average = 6
- 3rd row, 3rd column: (1, 1, 1) → Average = 1
Step 2:
If the average of a box is 2, then the total number of coins in that box must be 6.
Since each sack must contain fewer than 5 coins, this is possible only for the box in the 2nd row, 2nd column.
Among the possible combinations, only (1, 2, 3) satisfies the required conditions.
(The combination (1, 1, 4) satisfies only two conditions and therefore cannot be used.)
Hence:
- 2nd row, 1st column: (1, 2, 9) → Average = 4
- 2nd row, 2nd column: (1, 2, 3) → Average = 2
- 2nd row, 3rd column: (9, 9, 9) → Average = 9
Similarly, the first row is determined as:
- 1st row, 1st column: (1, 1, 7) → Average = 3
- 1st row, 2nd column: (3, 9, 9) → Average = 7
- 1st row, 3rd column: (1, 6, 8) → Average = 5
Alia, Badal, Clive, Dilshan, and Ehsaan played a game in which each asks a unique question to all the others and they respond by tapping their feet, either once or twice or thrice. One tap means “Yes”, two taps mean “No”, and three taps mean “Maybe”.
A total of 40 taps were heard across the five questions. Each question received at least one “Yes”, one “No”, and one “Maybe.”
The following information is known.
1. Alia tapped a total of 6 times and received 9 taps to her question. She responded “Yes” to the questions asked by both Clive and Dilshan.
2. Dilshan and Ehsaan tapped a total of 11 and 9 times respectively. Dilshan responded “No” to Badal.
3. Badal, Dilshan, and Ehsaan received equal number of taps to their respective questions.
4. No one responded “Yes” more than twice.
5. No one’s answer to Alia’s question matched the answer that Alia gave to that person’s question. This was also true for Ehsaan.
6. Clive tapped more times in total than Badal.
How many taps did Clive receive for his question?
Step 1:
Here, one tap represents "Yes", two taps represent "No", and three taps represent "Maybe".
Since every question received at least one "Yes", one "No", and one "Maybe", each person must have received more than 6 taps in total.
Using Conditions (1) and (4), Alia gave 2 "Yes" responses and 2 "No" responses.
From Condition (6), the combined number of taps made by Badal and Clive is:
40 − (6 + 11 + 9) = 14
Also, Clive made more taps than Badal. Along with Condition (4), this implies that Badal made 6 taps, while Clive made 8 taps.
Now, applying Condition (3), let the number of taps received by Badal, Dilshan, and Ehsaan be x each.
Therefore, the taps received by Clive are:
40 − 9 − 3x = 31 − 3x
Since every person received more than 6 taps,
31 − 3x > 6
which gives x < 8.33. Hence, x can only be 7 or 8.
- If x = 7, then Clive would receive 31 − 21 = 10 taps, which is not possible because Badal answered either "Yes" or "No" to Clive's question.
- Therefore, x = 8, giving Clive = 31 − 24 = 7 taps, which satisfies all the conditions.
The deductions obtained so far are summarized in the table below.
Step 2:
Using Condition (5), no person's response to Alia's question matched the response that Alia gave to that person's question. The same condition also applies to Ehsaan.
Applying this constraint completes the entire arrangement, which is shown in the final table below.
Hence, Clive received 7 taps for his question.
Which two people tapped an equal number of times in total?
Alia and Badal
Clive and Ehsaan
Dilshan and Clive
Badal and Dilshan
Alia and Badal
Step 1:
Here, one tap represents "Yes", two taps represent "No", and three taps represent "Maybe".
Since every question received at least one "Yes", one "No", and one "Maybe", each person must have received more than 6 taps in total.
Using Conditions (1) and (4), Alia gave 2 "Yes" responses and 2 "No" responses.
From Condition (6), the combined number of taps made by Badal and Clive is:
40 − (6 + 11 + 9) = 14
Also, Clive made more taps than Badal. Along with Condition (4), this implies that Badal made 6 taps, while Clive made 8 taps.
Now, applying Condition (3), let the number of taps received by Badal, Dilshan, and Ehsaan be x each.
Therefore, the taps received by Clive are:
40 − 9 − 3x = 31 − 3x
Since every person received more than 6 taps,
31 − 3x > 6
which gives x < 8.33. Hence, x can only be 7 or 8.
- If x = 7, then Clive would receive 31 − 21 = 10 taps, which is not possible because Badal answered either "Yes" or "No" to Clive's question.
- Therefore, x = 8, giving Clive = 31 − 24 = 7 taps, which satisfies all the conditions.
The deductions obtained so far are summarized in the table below.
Step 2:
Using Condition (5), no person's response to Alia's question matched the response that Alia gave to that person's question. The same condition also applies to Ehsaan.
Applying this constraint completes the entire arrangement, which is shown in the final table below.
Alia and Badal tapped an equal number of times in total.
What was Clive’s response to Ehsaan’s question?
No
Maybe
Cannot be determined
Yes
No
Step 1:
Here, one tap represents "Yes", two taps represent "No", and three taps represent "Maybe".
Since every question received at least one "Yes", one "No", and one "Maybe", each person must have received more than 6 taps in total.
Using Conditions (1) and (4), Alia gave 2 "Yes" responses and 2 "No" responses.
From Condition (6), the combined number of taps made by Badal and Clive is:
40 − (6 + 11 + 9) = 14
Also, Clive made more taps than Badal. Along with Condition (4), this implies that Badal made 6 taps, while Clive made 8 taps.
Now, applying Condition (3), let the number of taps received by Badal, Dilshan, and Ehsaan be x each.
Therefore, the taps received by Clive are:
40 − 9 − 3x = 31 − 3x
Since every person received more than 6 taps,
31 − 3x > 6
which gives x < 8.33. Hence, x can only be 7 or 8.
- If x = 7, then Clive would receive 31 − 21 = 10 taps, which is not possible because Badal answered either "Yes" or "No" to Clive's question.
- Therefore, x = 8, giving Clive = 31 − 24 = 7 taps, which satisfies all the conditions.
The deductions obtained so far are summarized in the table below.
Step 2:
Using Condition (5), no person's response to Alia's question matched the response that Alia gave to that person's question. The same condition also applies to Ehsaan.
Applying this constraint completes the entire arrangement, which is shown in the final table below.
Clive’s response to Ehsaan’s question was ‘No’.
How many “Yes” responses were received across all the questions?
Step 1:
Here, one tap represents "Yes", two taps represent "No", and three taps represent "Maybe".
Since every question received at least one "Yes", one "No", and one "Maybe", each person must have received more than 6 taps in total.
Using Conditions (1) and (4), Alia gave 2 "Yes" responses and 2 "No" responses.
From Condition (6), the combined number of taps made by Badal and Clive is:
40 − (6 + 11 + 9) = 14
Also, Clive made more taps than Badal. Along with Condition (4), this implies that Badal made 6 taps, while Clive made 8 taps.
Now, applying Condition (3), let the number of taps received by Badal, Dilshan, and Ehsaan be x each.
Therefore, the taps received by Clive are:
40 − 9 − 3x = 31 − 3x
Since every person received more than 6 taps,
31 − 3x > 6
which gives x < 8.33. Hence, x can only be 7 or 8.
- If x = 7, then Clive would receive 31 − 21 = 10 taps, which is not possible because Badal answered either "Yes" or "No" to Clive's question.
- Therefore, x = 8, giving Clive = 31 − 24 = 7 taps, which satisfies all the conditions.
The deductions obtained so far are summarized in the table below.
Step 2:
Using Condition (5), no person's response to Alia's question matched the response that Alia gave to that person's question. The same condition also applies to Ehsaan.
Applying this constraint completes the entire arrangement, which is shown in the final table below.
Six ‘’Yes” responses were received across all the questions.
Anu, Bijay, Chetan, Deepak, Eshan, and Faruq are six friends. Each of them uses a mobile number from exactly one of the two mobile operators - Xitel and Yocel. During the last month, the six friends made several calls to each other. Each call was made by one of these six friends to another. The table below summarizes the number of minutes of calls that each of the six made to (outgoing minutes) and received from (incoming minutes) these friends, grouped by the operators. Some of the entries are missing.
It is known that the duration of calls from Faruq to Eshan was 200 minutes.
Also, there were no calls from:
i. Bijay to Eshan,
ii. Chetan to Anu and Chetan to Deepak,
iii. Deepak to Bijay and Deepak to Faruq,
iv. Eshan to Chetan and Eshan to Deepak.
What was the duration of calls (in minutes) from Bijay to Anu?
What was the total duration of calls (in minutes) made by Anu to friends having mobile numbers from Operator Yocel?
What was the total duration of calls (in minutes) made by Faruq to friends having mobile numbers from Operator Yocel?
What was the duration of calls (in minutes) from Deepak to Chetan?
50
125
0
100
100
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