CAT 2023 Slot 3 · DILR · Set 3: Online and Offline Registrations

Solving the set

Step 1 – Every month's total is 110, 120 or 130 (min 110, max 130, median 120 means the five totals are 110, x, 120, y, 130 with x, y between them, all multiples of 10).

Step 2 – January: offline = 2 × online, so the total is a multiple of 30 ⇒ 120: online 40, offline 80. April: online = 2 × offline ⇒ total 120: online 80, offline 40.

Step 3 – Online values are 40 (min), 100 (max), median 80. 100 is in May (largest). January has 40 and April 80, so the online numbers are 40, ?, 80, ?, 100 with February and March holding one value ≤ 80 and one = 80 (median).

Step 4 – Offline values: min 30, max 80 (January), median 50. April = 40. May's online is 100, so May's offline is 10, 20 or 30 to keep the total ≤ 130 ⇒ 30 (the minimum), total 130.

Step 5 – February and March. The offline numbers so far are 80 (Jan), 40 (Apr), 30 (May); the median 50 must be February's or March's. March's online = February's offline (fact 4).

• If February's offline = 50, March's online = 50, so February's online = 80 (median) ⇒ February total 130. March's total must then be 110 (we need one 110): March offline = 60. Offline list: 30, 40, 50, 60, 80 – median 50 ✓.

• If March's offline = 50 instead, February's offline f ≥ 50 and March's online = f. One of February/March must total 110. March total 110 ⇒ f = 60, and then February's online must be 80 or more (online median), giving February a total above 130 – impossible. February total 110 ⇒ either February's online ≥ 80 (so f ≤ 30, contradiction) or March's online f ≥ 80 (so February's online ≤ 30, below the online minimum 40) – impossible. So this case fails.

This question

From the final table, April: online 80 + offline 40 = 120.

Hence, the answer is 120.

Solving the set

Step 1 – Every month's total is 110, 120 or 130 (min 110, max 130, median 120 means the five totals are 110, x, 120, y, 130 with x, y between them, all multiples of 10).

Step 2 – January: offline = 2 × online, so the total is a multiple of 30 ⇒ 120: online 40, offline 80. April: online = 2 × offline ⇒ total 120: online 80, offline 40.

Step 3 – Online values are 40 (min), 100 (max), median 80. 100 is in May (largest). January has 40 and April 80, so the online numbers are 40, ?, 80, ?, 100 with February and March holding one value ≤ 80 and one = 80 (median).

Step 4 – Offline values: min 30, max 80 (January), median 50. April = 40. May's online is 100, so May's offline is 10, 20 or 30 to keep the total ≤ 130 ⇒ 30 (the minimum), total 130.

Step 5 – February and March. The offline numbers so far are 80 (Jan), 40 (Apr), 30 (May); the median 50 must be February's or March's. March's online = February's offline (fact 4).

• If February's offline = 50, March's online = 50, so February's online = 80 (median) ⇒ February total 130. March's total must then be 110 (we need one 110): March offline = 60. Offline list: 30, 40, 50, 60, 80 – median 50 ✓.

• If March's offline = 50 instead, February's offline f ≥ 50 and March's online = f. One of February/March must total 110. March total 110 ⇒ f = 60, and then February's online must be 80 or more (online median), giving February a total above 130 – impossible. February total 110 ⇒ either February's online ≥ 80 (so f ≤ 30, contradiction) or March's online f ≥ 80 (so February's online ≤ 30, below the online minimum 40) – impossible. So this case fails.

This question

From Step 2, January's online registrations = 40.

Hence, the answer is 40.

Solving the set

Step 1 – Every month's total is 110, 120 or 130 (min 110, max 130, median 120 means the five totals are 110, x, 120, y, 130 with x, y between them, all multiples of 10).

Step 2 – January: offline = 2 × online, so the total is a multiple of 30 ⇒ 120: online 40, offline 80. April: online = 2 × offline ⇒ total 120: online 80, offline 40.

Step 3 – Online values are 40 (min), 100 (max), median 80. 100 is in May (largest). January has 40 and April 80, so the online numbers are 40, ?, 80, ?, 100 with February and March holding one value ≤ 80 and one = 80 (median).

Step 4 – Offline values: min 30, max 80 (January), median 50. April = 40. May's online is 100, so May's offline is 10, 20 or 30 to keep the total ≤ 130 ⇒ 30 (the minimum), total 130.

Step 5 – February and March. The offline numbers so far are 80 (Jan), 40 (Apr), 30 (May); the median 50 must be February's or March's. March's online = February's offline (fact 4).

• If February's offline = 50, March's online = 50, so February's online = 80 (median) ⇒ February total 130. March's total must then be 110 (we need one 110): March offline = 60. Offline list: 30, 40, 50, 60, 80 – median 50 ✓.

• If March's offline = 50 instead, February's offline f ≥ 50 and March's online = f. One of February/March must total 110. March total 110 ⇒ f = 60, and then February's online must be 80 or more (online median), giving February a total above 130 – impossible. February total 110 ⇒ either February's online ≥ 80 (so f ≤ 30, contradiction) or March's online f ≥ 80 (so February's online ≤ 30, below the online minimum 40) – impossible. So this case fails.

This question

Statement I: May's offline (30) is the smallest of all months – true, so it can be true.

Statement II: February's total is 130; the smallest total is March's 110 – cannot be true.

Hence, the answer is option 4.

Solving the set

Step 1 – Every month's total is 110, 120 or 130 (min 110, max 130, median 120 means the five totals are 110, x, 120, y, 130 with x, y between them, all multiples of 10).

Step 2 – January: offline = 2 × online, so the total is a multiple of 30 ⇒ 120: online 40, offline 80. April: online = 2 × offline ⇒ total 120: online 80, offline 40.

Step 3 – Online values are 40 (min), 100 (max), median 80. 100 is in May (largest). January has 40 and April 80, so the online numbers are 40, ?, 80, ?, 100 with February and March holding one value ≤ 80 and one = 80 (median).

Step 4 – Offline values: min 30, max 80 (January), median 50. April = 40. May's online is 100, so May's offline is 10, 20 or 30 to keep the total ≤ 130 ⇒ 30 (the minimum), total 130.

Step 5 – February and March. The offline numbers so far are 80 (Jan), 40 (Apr), 30 (May); the median 50 must be February's or March's. March's online = February's offline (fact 4).

• If February's offline = 50, March's online = 50, so February's online = 80 (median) ⇒ February total 130. March's total must then be 110 (we need one 110): March offline = 60. Offline list: 30, 40, 50, 60, 80 – median 50 ✓.

• If March's offline = 50 instead, February's offline f ≥ 50 and March's online = f. One of February/March must total 110. March total 110 ⇒ f = 60, and then February's online must be 80 or more (online median), giving February a total above 130 – impossible. February total 110 ⇒ either February's online ≥ 80 (so f ≤ 30, contradiction) or March's online f ≥ 80 (so February's online ≤ 30, below the online minimum 40) – impossible. So this case fails.

This question

From Step 5, February's offline registrations are exactly 50.

Hence, the answer is option 4.

Solving the set

Step 1 – Every month's total is 110, 120 or 130 (min 110, max 130, median 120 means the five totals are 110, x, 120, y, 130 with x, y between them, all multiples of 10).

Step 2 – January: offline = 2 × online, so the total is a multiple of 30 ⇒ 120: online 40, offline 80. April: online = 2 × offline ⇒ total 120: online 80, offline 40.

Step 3 – Online values are 40 (min), 100 (max), median 80. 100 is in May (largest). January has 40 and April 80, so the online numbers are 40, ?, 80, ?, 100 with February and March holding one value ≤ 80 and one = 80 (median).

Step 4 – Offline values: min 30, max 80 (January), median 50. April = 40. May's online is 100, so May's offline is 10, 20 or 30 to keep the total ≤ 130 ⇒ 30 (the minimum), total 130.

Step 5 – February and March. The offline numbers so far are 80 (Jan), 40 (Apr), 30 (May); the median 50 must be February's or March's. March's online = February's offline (fact 4).

• If February's offline = 50, March's online = 50, so February's online = 80 (median) ⇒ February total 130. March's total must then be 110 (we need one 110): March offline = 60. Offline list: 30, 40, 50, 60, 80 – median 50 ✓.

• If March's offline = 50 instead, February's offline f ≥ 50 and March's online = f. One of February/March must total 110. March total 110 ⇒ f = 60, and then February's online must be 80 or more (online median), giving February a total above 130 – impossible. February total 110 ⇒ either February's online ≥ 80 (so f ≤ 30, contradiction) or March's online f ≥ 80 (so February's online ≤ 30, below the online minimum 40) – impossible. So this case fails.

This question

From the table: January 120 = April 120 and February 130 = May 130. Both pairs definitely match.

Hence, the answer is option 1.