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At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, \(\frac{4}{7}\) were donors, \(\frac{1}{5}\) were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?
A. \(\frac{4}{7}\)
B. \(\frac{20}{39}\)
C. \(\frac{4}{9}\)
D. \(\frac{2}{5}\)
E. \(\frac{10}{39}\)
A. \(\frac{4}{7}\)
B. \(\frac{20}{39}\)
C. \(\frac{4}{9}\)
D. \(\frac{2}{5}\)
E. \(\frac{10}{39}\)
31 Jul 2025, 09:01
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Official Solution:
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, \(\frac{4}{7}\) were donors, \(\frac{1}{5}\) were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?
A. \(\frac{4}{7}\)
B. \(\frac{20}{39}\)
C. \(\frac{4}{9}\)
D. \(\frac{2}{5}\)
E. \(\frac{10}{39}\)
Let total invited = 35 (LCM of 7 and 5, since fractions are \(\frac{4}{7}\) and \(\frac{1}{5}\)).
• Donors = \(\frac{4}{7} * 35 = 20\)
• Press = \(\frac{1}{5} * 35 = 7\)
• Remaining = 35 - 20 - 7 = 8, split equally: 2 each to volunteers, entertainers, organizers, sponsors
Actual attendees:
• Donors = 20 (unchanged)
• Press = 2 * 7 = 14
• Volunteers = 1 (half of 2)
• Entertainers = 0 (boycott)
• Organizers = 2
• Sponsors = 2
Total actual attendees = 20 + 14 + 1 + 0 + 2 + 2 = 39
Donors = 20, so the fraction of actual attendees who were donors is \(\frac{20}{39}\).
Answer: B
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, \(\frac{4}{7}\) were donors, \(\frac{1}{5}\) were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?
A. \(\frac{4}{7}\)
B. \(\frac{20}{39}\)
C. \(\frac{4}{9}\)
D. \(\frac{2}{5}\)
E. \(\frac{10}{39}\)
Let total invited = 35 (LCM of 7 and 5, since fractions are \(\frac{4}{7}\) and \(\frac{1}{5}\)).
• Donors = \(\frac{4}{7} * 35 = 20\)
• Press = \(\frac{1}{5} * 35 = 7\)
• Remaining = 35 - 20 - 7 = 8, split equally: 2 each to volunteers, entertainers, organizers, sponsors
Actual attendees:
• Donors = 20 (unchanged)
• Press = 2 * 7 = 14
• Volunteers = 1 (half of 2)
• Entertainers = 0 (boycott)
• Organizers = 2
• Sponsors = 2
Total actual attendees = 20 + 14 + 1 + 0 + 2 + 2 = 39
Donors = 20, so the fraction of actual attendees who were donors is \(\frac{20}{39}\).
Answer: B
