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CAT 2023 Slot 3 DILR · Set 4 of 4Data InterpretationEasy DifficultyQ16–Q22 · 3 MCQ + 2 TITA

CAT 2023 Slot 3 DILR Set 4: Tables (5 Questions) with Solutions

The Tables set from the official CAT 2023 Slot 3 DILR paper: the caselet, all 5 questions, the official answer key and a step-by-step solution for each. 100% free, no sign-up.

The Set: Read the Caselet First

CAT 2023 Slot 3 DILR · Questions 16–22
In a coaching class, some students register online, and some others register offline. No student registers both online and offline; hence the total registration number is the sum of online and offline registrations. The following facts and table pertain to these registration numbers for the five months, January to May of 2023. The table shows the minimum, maximum, median registration numbers of these five months, separately for online, offline and total number of registrations. The following additional facts are known.
1
In every month, both online and offline registration numbers were multiples of 10.
2
In January, the number of offline registrations was twice that of online registrations.
3
In April, the number of online registrations was twice that of offline registrations.
4
The number of online registrations in March was the same as the number of offline registrations in February.
5
The number of online registrations was the largest in May.
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In the exam this set competed with 3 others for 40 minutes. Budget about 10 minutes if you attempt it.

Q16. Tables

TITA · +3 / 0
What was the total number of registrations in April?
Correct Answer
120
Step-by-step solution
📌 Core Concept
Facts 2 and 3 fix each of January and April as a 1:2 or 2:1 split, so the total is three times one registration number. The table's total range of 110 to 130 then pins that number down.
🔢 Step-by-Step Solution
1
In April, let offline =y= y. Then online =2y= 2y (fact 3), so the April total is 3y3y.
2
Every registration number is a multiple of 10, so yy is a multiple of 10.
3
The table shows every monthly total lies between 110 (minimum) and 130 (maximum). So 110≤3y≤130110 \le 3y \le 130, which gives 36.7≤y≤43.336.7 \le y \le 43.3.
4
The only multiple of 10 in that range is y=40y = 40. April has 80 online and 40 offline.
5
The April total is 80+40=12080 + 40 = 120.
Quick Check
Total =3×40=120= 3 \times 40 = 120, which lies within 110 to 130 and equals the median total.
Correct Answer: 120

Q17. Tables

TITA · +3 / 0
What was the number of online registrations in January?
Correct Answer
40
Step-by-step solution
📌 Core Concept
January's offline figure is twice its online figure (fact 2), so the January total is three times the online number. The table's total range of 110 to 130 then fixes it.
🔢 Step-by-Step Solution
1
Let January online =x= x. Then offline =2x= 2x (fact 2) and the total is 3x3x.
2
Every monthly total lies between the table's minimum of 110 and maximum of 130. So 110≤3x≤130110 \le 3x \le 130, which gives 36.7≤x≤43.336.7 \le x \le 43.3.
3
Fact 1 makes xx a multiple of 10, so x=40x = 40.
4
January has 40 online and 80 offline, and the total is 120.
Quick Check
January online is 40, which equals the table's online minimum. Offline is 80, which equals the offline maximum.
Correct Answer: 40

Q18. Tables

MCQ · +3 / −1
Which of the following statements can be true?
I
The number of offline registrations was the smallest In May.
II
The total number of registrations was the smallest in February.
  1. A) Only I
  2. B) Neither I nor II
  3. C) Only II
  4. D) Both I and II
Correct Answer
A) Only I
Step-by-step solution
📌 Core Concept
The table's minimum, maximum and median values, together with the five facts, fix every month's online and offline figures. Each statement can then be checked against them.
🔢 Step-by-Step Solution
1
Every total is a multiple of 10 and lies between 110 and 130, so each total is 110, 120 or 130.
2
January: offline = 2 × online, so the total is 3 × online, a multiple of 30. Only 120 fits, so January is online 40, offline 80.
3
April: online = 2 × offline, so the total is again a multiple of 30, namely 120. April is online 80, offline 40.
4
May: online is the largest, which is 100. The total is at most 130, so offline is at most 30. The offline minimum is 30, so May offline = 30 and the total = 130.
5
Offline values so far are 80 (Jan), 40 (Apr) and 30 (May). The median is 50, so one of February or March offline is exactly 50 and the other is at least 50. Let February offline = aa and March offline = bb. Fact 4 gives March online = aa.
6
Try b=50b = 50 with a≥50a \ge 50. Online values are 40, 80, 100, Feb online xx and aa. The median is 80, so x≥80x \ge 80, because a≤80a \le 80. February's total x+a≥130x + a \ge 130 must be at most 130, so x=80x = 80 and a=50a = 50. March's total would then be 50+50=10050 + 50 = 100, which is below the minimum 110. This case fails.
7
So a=50a = 50. March online = 50, and the online median of 80 forces x≥80x \ge 80. February's total x+50≤130x + 50 \le 130 gives x=80x = 80 and a total of 130. A total of 110 is needed for the minimum, so March's total is 110, giving b=60b = 60.
8.
Statement I
May offline is 30, the smallest offline figure (others are 80, 50, 60, 40). This is true.
9.
Statement II
February's total is 130, the largest, not the smallest. This is false.
Quick Check
Offline values 30, 40, 50, 60, 80 have median 50. Online values 40, 50, 80, 80, 100 have median 80. Totals 110, 120, 120, 130, 130 have median 120.
Correct Answer: A

Q21. Tables

MCQ · +3 / −1
What was the number of offline registrations in February?
  1. A) 5050 or 8080
  2. B) 3030 or 50 or 8080
  3. C) 5050
  4. D) 8080
Correct Answer
C) 5050
Step-by-step solution
📌 Core Concept
Fix January, April and May from the table and the facts. Then test each possible February offline value against the medians and the total limits.
🔢 Step-by-Step Solution
1
January is 40 online and 80 offline. April is 80 online and 40 offline. Both totals are 120, since 3x3x must lie between 110 and 130.
2
May online is the maximum, 100 (fact 5). May's total is at most 130, so May offline is at most 30. The offline minimum is 30, so May offline is 30 and the May total is 130.
3
Let February offline be ff. March online is also ff (fact 4). The offline values 80, 40 and 30 are already placed. For the median to be 50, 50 must appear in February or March offline. Neither f<50f<50 nor f=40f=40 is possible.
4
If f=80f = 80: February online is 30 or more, so the February total is at least 110. A total of 110 needs February online 30, below the online minimum of 40. February online 40 gives 120. March offline must be 50, so March totals 130. No month totals 110, and the table minimum would be missed.
5
If f=60f = 60 or 70: March online is ff, which is below 80. The online median of 80 then needs February online of 80 or more. The February total is then at least 140, above the maximum of 130.
6
If f=50f = 50: February online must be 80, so the February total is 130. March offline is 60, so the March total is 110, which supplies the minimum. All table values match.
7
February offline is 50.
Correct Answer: C

Q22. Tables

MCQ · +3 / −1
Which pair of months definitely had the same total number of registrations? I. January and April II. February and May
  1. A) Neither I nor II
  2. B) Only II
  3. C) Both I and II
  4. D) Only I
Correct Answer
C) Both I and II
Step-by-step solution
📌 Core Concept
The table bounds every monthly total between 110 and 130. The clues then fix January, April and May directly, and the remaining unknowns are tested case by case.
🔢 Step-by-Step Solution
1
Totals lie between 110 and 130, so a monthly total can only be 110, 120 or 130 (all values are multiples of 10).
2
January (fact 2): offline = 2 × online, so the total is 3 × online. The only multiple of 30 in the 110–130 range is 120, which gives online 40 and offline 80.
3
April (fact 3): online = 2 × offline, so the total is 3 × offline. The total is again 120, giving offline 40 and online 80.
4
So January and April both total 120. Statement I is definitely true.
5
May (fact 5): online is the largest in May, so it equals the online maximum of 100. The total is at most 130, so offline is at most 30. The offline minimum is 30, so May offline = 30 and the May total is 130.
6
Let February online = aa, February offline = bb, March offline = cc. By fact 4, March online = bb. Offline values are 80, 40, 30, bb, cc with median 50. This forces min⁡(b,c)=50\min(b,c)=50 and the other value to be between 50 and 80. So b≥50b \ge 50.
7
Online values are 40, 80, 100, aa, bb with median 80. This needs at least one of a,ba, b to be ≤80\le 80, and not both below 80. In particular, if b<80b<80 then a≥80a \ge 80.
8
Case b<80b<80 (so b=50b = 50, 60 or 70): a≥80a \ge 80, so the February total is a+b≥80+ba+b \ge 80+b. Since this must be at most 130, b=50b=50 and a=80a=80. For b=60b = 60 or 70 the total would be at least 140, which is impossible. The February total is then 130.
9
Case b=80b=80: this makes c=50c=50. March has online 80 and offline 50, a total of 130. The February total is a+80a+80, which is at most 130, so a≤50a \le 50 and February is 120 or 130. Then no month totals 110, but the table's minimum total is 110. This case is impossible.
10
Only the case in step 8 is left: b=50b=50, a=80a=80, February total 130. March must then supply the 110 minimum (c=60c=60), which is consistent.
11
February and May both total 130. Statement II is definitely true.
Why Not the Others
A: Statement I holds from steps 2–4.
B: Statement II holds, but I holds as well.
D: Statement I holds, but II holds as well.
Quick Check
The totals are 120, 130, 120, 110, 130 for January to May. The minimum is 110, the maximum is 130 and the median is 120, matching the table.
Correct Answer: C
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CAT 2023 Slot 3 DILR at a Glance

Sets in this section4
Questions in this section20
Section time40 minutes
Convenor IIMIIM Lucknow

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