Question

How many viewers in the city C watch less than one movie a week?

• A. 2040
• B. 13600
• C. 16000
• D. 3600
• A. City E
• B. City D
• C. City B
• D. City C
• A. City E
• B. City D
• C. City B
• D. City C
• A. 50000
• B. 36000
• C. 18500
• D. 45250

Answer 1

The Correct Option is B

Step 1 — Data for City C:
From the table:
• % of viewers in C watching < 1 movie/week = 85%.
• Number of viewers in C watching ≥ 1 movie/week = 2400.

Step 2 — Express the situation mathematically:
Let the total number of surveyed viewers in city C = T.
Then,
Viewers watching < 1 movie/week = 85% of T = 0.85T.
Viewers watching ≥ 1 movie/week = 15% of T = 0.15T.
But we know 0.15T = 2400.

Step 3 — Solve for T:
0.15T = 2400 ⇒ T = 2400 ÷ 0.15.
T = 2400 × (100/15) = 2400 × 6.666… = 16000.

Step 4 — Find the number watching < 1 movie/week:
= 0.85 × T = 0.85 × 16000.
= 13600.

Final Answer:
The number of viewers in city C who watch less than one movie a week is 13600. This corresponds to Option (B).

Answer 2

Step 1 — Recall formula:
For each city,
Viewers watching < 1 movie/week = (Percentage given in Column I ÷ (100 − Percentage)) × (Column II).
Because Column II = viewers watching ≥ 1 movie/week.

Step 2 — Compute city by city:
• City A:
% < 1 = 60%, so % ≥ 1 = 40%.
0.40T = 2400 ⇒ T = 2400 ÷ 0.40 = 6000.
Viewers < 1 = 60% of 6000 = 3600.

• City B:
% < 1 = 20%, so % ≥ 1 = 80%.
0.80T = 3000 ⇒ T = 3000 ÷ 0.80 = 3750.
Viewers < 1 = 20% of 3750 = 750.

• City C:
% < 1 = 85%, so % ≥ 1 = 15%.
0.15T = 2400 ⇒ T = 2400 ÷ 0.15 = 16000.
Viewers < 1 = 85% of 16000 = 13600.

• City D:
% < 1 = 55%, so % ≥ 1 = 45%.
0.45T = 2700 ⇒ T = 2700 ÷ 0.45 = 6000.
Viewers < 1 = 55% of 6000 = 3300.

• City E:
% < 1 = 75%, so % ≥ 1 = 25%.
0.25T = 8000 ⇒ T = 8000 ÷ 0.25 = 32000.
Viewers < 1 = 75% of 32000 = 24000.

Step 3 — Compare results:
• A → 3600
• B → 750
• C → 13600
• D → 3300
• E → 24000

Step 4 — Identify maximum:
The highest number of viewers who watch < 1 movie/week is from City E (24000).

Final Answer:
The city with the highest number of viewers who watch less than one movie a week is City E. This corresponds to Option (A).

Answer 3

Step 1 — Understand the requirement:
We need the city with the second lowest number of movie watchers (i.e., viewers who watch ≥ 1 movie/week).
Column II of the table already gives the number of viewers watching ≥ 1 movie/week for each city.

Step 2 — Extract data from Column II:
• City A → 2400
• City B → 3000
• City C → 2400
• City D → 2700
• City E → 8000

Step 3 — Arrange in ascending order:
2400 (A)
2400 (C)
2700 (D)
3000 (B)
8000 (E)

Step 4 — Identify the second lowest:
• The lowest = 2400 (Cities A and C).
• The next after 2400 = 2700 (City D).

Final Answer:
The city with the second lowest number of movie watchers is City D. This corresponds to Option (B).

Answer 4

Objective: Compute the total number of viewers (across A, B, C, D, E) who watch < 1 movie per week.

Given (from the table):
• Column I = % of viewers watching < 1 movie/week in that city.
• Column II = absolute number of viewers watching ≥ 1 movie/week in that city.
Let each city’s total viewers be T_city. Then:
  • % ≥ 1 movie/week = (100 − Column I).
  • So (100 − Column I)% of T_city = Column II ⇒ T_city = Column II ÷ ((100 − Column I)/100).
  • Viewers < 1 movie/week = Column I% of T_city.

Quick direct formula (for each city):
Viewers < 1 per week = (Column I / (100 − Column I)) × Column II.

City-wise calculations:
A: I = 60%, II = 2400 ⇒ T_A = 2400 ÷ 0.40 = 6000 ⇒ < 1 = 0.60 × 6000 = 3600.
B: I = 20%, II = 3000 ⇒ T_B = 3000 ÷ 0.80 = 3750 ⇒ < 1 = 0.20 × 3750 = 750.
C: I = 85%, II = 2400 ⇒ T_C = 2400 ÷ 0.15 = 16000 ⇒ < 1 = 0.85 × 16000 = 13600.
D: I = 55%, II = 2700 ⇒ T_D = 2700 ÷ 0.45 = 6000 ⇒ < 1 = 0.55 × 6000 = 3300.
E: I = 75%, II = 8000 ⇒ T_E = 8000 ÷ 0.25 = 32000 ⇒ < 1 = 0.75 × 32000 = 24000.

Summation across all five cities:
Total (< 1 per week) = A + B + C + D + E
= 3600 + 750 + 13600 + 3300 + 24000
= (3600 + 750) = 4350
= 4350 + 13600 = 17950
= 17950 + 3300 = 21250
= 21250 + 24000 = 45250.

Final Answer: 45250 viewers watch < 1 movie per week across the five cities. (Matches Option D)